Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 2 Work, Energy and Power

Exercise 1(A)

Solution 1.

Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity.

Solution 2.

(i) When force is in direction of displacement, then work done , W = F x S
(ii) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ

Solution 3.

(a) When force is at an angle θ to the direction of displacement, then work done, W= F S cos θ
(b) (i) For zero work done, the angle between force and displacement should be 90o as cos 90= 0
W = FS cos90o= FS x 0 = 0
(ii) For maximum work done, the angle between force and displacement should be 0o as cos0= 1
Hence, W=FScos 0o=FS

Solution 4.

Two conditions when the work done is zero are:

  1. When there is no displacement (S = 0) and,
  2. When the displacement is normal to the direction of the force (θ = 90o).

Solution 5.

(i) If the displacement of the body is in the direction of force, then work done is positive.
Hence, W= F x S
For example: A coolie does work on the load when he raises it up against the force of gravity. The force exerted by coolie (=mg) and displacement, both are in upward direction.

(ii) If the displacement of the body is in the direction opposite to the force, then work done is negative.
Hence, W =- F x S
For example: When a body moves on a surface, the frictional forces between the body and the surface is in direction opposite to the motion of the body and so the work done by the force of friction is negative.

Solution 6.

Work is done against the force.

Solution 7.

When a body moves in a circular path, no work is done since the force on the body is directed towards the centre of circular path (the body is acted upon by the centripetal force), while the displacement at all instants is along the tangent to the circular path, i.e., normal to the direction of force.

Solution 8.

Work done by the force of gravity (which provides the centripetal force) is zero as the force of gravity acting on the satellite is normal to the displacement of the satellite.

Solution 9.

Work is done only in case of a boy climbing up a stair case.

Solution 10.

When a coolie carrying some load on his head moves, no work is done by him against the force of gravity because the displacement of load being horizontal, is normal to the direction of force of gravity.

Solution 11.

Force applied by the fielder on the ball is in opposite direction of displacement of ball. So, work done by the fielder on the ball is negative.

Solution 12.

When a coolie carries a load while moving on a ground, the displacement is in the horizontal direction while the force of gravity acts vertically downward. So the work done by the force of gravity is zero.

Solution 13.

S.I unit of work is Joule.
C.G.S unit of work is erg.
Relation between joule and erg :
1joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm = 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 14.

S. I unit of work is Joule.
1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.
Previous

Solution 15.

Relation between joule and erg :
1 joule = 1N x 1m
But 1N = 105dyne
And 1m = 100 cm= 10cm
Hence, 1 joule = 105dyne x 102cm
= 107dyne x cm = 107erg
Thus, 1 Joule = 107 erg

Solution 16.

Let a body of mass m fall down through a vertical height h either directly or through an inclined plane e.g. a hill, slope or staircase. The force of gravity on the body is F = mg acting vertically downwards and the displacement in the direction of force (i.e., vertical) is S=h. Therefore the work done by the force of gravity is
W = FS = mgh

Solution 17.

Let a boy of mass m climb up through a vertical height h either through staircase of using a lift. The force of gravity on the boy is F=mg acting vertically downwards and the displacement in the direction opposite to force (i.e., vertical) is S=-h. Therefore the work done by the force of gravity on the boy is
W= FS =-mgh
or,the work W=mgh is done by the boy against the force of gravity.

Solution 18.

The energy of a body is its capacity to do work. Its S.I unit is Joule (J).

Solution 19.

eV measures the energy of atomic particles.
1eV= 1.6 x 10-19J

Solution 20.

1 J = 0.24 calorie

Solution 21.

Calorie measures heat energy.
1calorie = 4.18 J

Solution 22.

1kWh is the energy spent (or work done) by a source of power 1kW in 1 h.
1kWh = 3.6 x 106J

Solution 23.

The rate of doing work is called power. The S.I. unit of power is watt (W).

Solution 24.

Power spent by a source depends on two factors:
(i) The amount of work done by the source, and
(ii) The time taken by the source to do the said work.
Example: If a coolie A takes 1 minute to lift a load to the roof of a bus, while another coolie B takes 2 minutes to lift the same load to the roof of the same bus, the work done by both the coolies is the same, but the power spent by the coolie A is twice the power spent by the coolie B because the coolie A does work at a faster rate.

Solution 25.

Work Power
1. Work done by a force is equal to the product of force and the displacement in the direction of force. 1. Power of a source is the rate of doing work by it.
2. Work done does not depend on time. 2. Power spent depends on the time in which work is done.
3. S.I unit of work is joule (J). 3. S.I unit of power is watt (W).

Solution 26.

Energy Power
1. Energy of a body is its capacity to do work. 1. Power of a source is the energy spent by it in 1s.
2. Energy spent does not depend on time. 2. Power spent depends on the time in which energy is spent.
3. S.I unit of energy is joule (J). 3. S.I unit of power is watt (W).

Solution 27.

S.I unit of power is watt (W).
If 1 joule of work is done in 1 second, the power spent is said to be 1 watt.

Solution 28.

Horse power is another unit of power, largely used in mechanical engineering. It is related to the S.I unit watt as :
1 H.P =746 W

Solution 29.

Watt (W) is the unit of power, while watt hour (Wh) is the unit of work, since power x time = work.

Solution 30.

a. Energy is measured in kWh
b. Power is measure in kW
c. Energy is measured in Wh
d. Energy is meaused in eV
Concept insight: Energy has bigger units like kWh (kilowatt hour) and Wh (watt hour). Similarly bigger unit of power is kW (kilo watt).
The energy of atomic particles is very small, and hence, it is measured in eV (electron volt).

Solution 1 (MCQ).

746 W

Solution 2 (MCQ).

The unit kWh is the unit of energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 1

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 2

Solution 3.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 3

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 4

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 5

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 6

Solution 7.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 7

Solution 8.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 8

Solution 9.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 9

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 10

Solution 12.
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Solution 13.

(i) The work done by persons A and B is independent of time. Hence both A and B will do the same amount of work. Hence,
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 12
(ii) Power developed by the person A and B is calculated as follows:
A takes 20 s to climb the stairs while B takes 15 s, to do the same. Hence B does work at a much faster rate than A; more power is spent by B.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 13

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 14

Exercise 2(B)

Solution 1.

Two forms of mechanical energy are:

  1. Kinetic energy
  2. Potential energy

Solution 2.

Elastic potential energy is possessed by wound up watch spring.

Solution 3.

(a) Kinetic energy (K)
(b) Potential energy (U)
(c) Kinetic energy (K)
(d) Potential energy (U)
(e) Kinetic energy (K)
(f) Potential energy (U)

Solution 4.

Potential energy: The energy possessed by a body by virtue of its specific position (or changed configuration) is called the potential energy.
Different forms of P.E. are as listed below:

  1. Gravitational potential energy: The potential energy possessed by a body due to its position relative to the centre of Earth is called its gravitational potential energy.
    Example: A stone at a height has gravitational potential energy due to its raised height.
  2. Elastic potential energy: The potential energy possessed by a body in the deformed state due to change in its configuration is called its elastic potential energy.
    Example: A compressed spring has elastic potential energy due to its compressed state.

Solution 5.

Potential energy is possessed by the body even when it is not in motion. For example: a stone at a height has the gravitational potential energy due to its raised position.

Solution 6.

Gravitational potential energy is the potential energy possessed by a body due to its position relative to the centre of earth.
For a body placed at a height above the ground, the gravitational potential energy is measured by the amount of work done in lifting it up to that height against the force of gravity.
Let a body of mass m be lifted from the ground to a vertical height h. The least upward force F required to lift the body (without acceleration) must be equal to the force of gravity (=mg) on the body acting vertically downwards. The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
= mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 7.

The work done W on the body in lifting it to a height h is
W= force of gravity (mg) x displacement (h)
=mgh
This work is stored in the body when it is at a height h in the form of its gravitational potential energy.
Gravitational potential energy U= mgh

Solution 8.

A body in motion is said to possess the kinetic energy. The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

Solution 9.

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 15

Solution 10.

According to the work-energy theorem, the work done by a force on a moving body is equal to the increase in its kinetic energy.

Solution 11.

Body of mass m is moving with a uniform velocity u. A force is applied on the body due to which its velocity changes from u to v and produces an acceleration a in moving a distance S.Then,
Work done by the force= force x displacement
W = F x S———(i)
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 16

Solution 12.

Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 17
Both the masses have same momentum p. The kinetic energy, K is inversely proportional to mass of the body.
Hence light mass body has more kinetic energy because smaller the mass, larger is the kinetic energy.

Solution 13.

Kinetic energy is related to momentum and mass as
p = √2mK
As the kinetic energy of both bodies are same, momentum is directly proportional to square root of mass.
Now, mass of body B is greater than that of body A.
Hence, body B will have more momentum than body A.

Solution 15.
The three forms of kinetic energy are:

  1. Translational kinetic energy- example: a freely falling body
  2. Rotational kinetic energy-example: A spinning top.
  3. Vibrational kinetic energy-example: atoms in a solid vibrating about their mean position.

Solution 16.

Potential energy (U)

Kinetic energy (K)
    1. The energy possessed by a body by virtue of its specific position or changed configuration is called potential energy.

    1.The energy possessed by a body by virtue of its state of motion is called the kinetic energy.

     2.Two forms of potential energy are gravitational potential energy and elastic potential energy.

2. Forms of kinetic energy are translational, rotational and vibrational kinetic energy.
     3.Example: A wound up watch spring has potential energy.

3. For example: a moving car has kinetic energy.

Solution 17.

(a) Motion.
(b) Position.

Solution 18.

When the string of a bow is pulled, some work is done which is stored in the deformed state of the bow in the form of its elastic potential energy. On releasing the string to shoot an arrow, the potential energy of the bow changes into the kinetic energy of the arrow which makes it move.

Solution 19.

The compressed spring has elastic potential energy due to its compressed state. When it is released, the potential energy of the spring changes into kinetic energy which does work on the ball if placed on it and changes into kinetic energy of the ball due to which it flies away.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 18

Solution 20.

When water falls from a height, the potential energy stored in water at a height changes into the kinetic energy of water during the fall. On striking the ground, a part of the kinetic energy of water changes into the heat energy due to which the temperature of water rises.

Solution 21.

Yes, when force is normal to displacement, no transfer of energy takes place.

Solution 22.

Kinetic energy.

Solution 23.

The six different forms of energy are:

  1. Solar energy
  2. Heat energy
  3. Light energy
  4. Chemical or fuel energy
  5. Hydro energy
  6. Nuclear energy

Solution 24.

(a) Potential energy of wound up spring converts into kinetic energy.
(b) Chemical energy of petrol or diesel converts into mechanical energy (kinetic energy)
(c) Kinetic energy to potential energy
(d) Light energy changes into chemical energy
(e) Electrical energy changes into chemical energy
(f) Chemical energy changes into heat energy
(g) Chemical energy changes into heat and light energy
(h) Chemical energy changes into heat, light and sound energy

Solution 25.

(a) Electrical energy into sound energy
(b) Heat energy into mechanical energy
(c) Sound energy into electrical energy
(d) Electrical energy to mechanical energy
(e) Electrical energy into light energy
(f) Chemical energy to heat energy
(g) Light energy into electrical energy
(h) Chemical energy into heat energy
(i) Chemical energy into electrical energy
(j) Chemical energy to mechanical energy
(k) Electrical energy into heat energy
(l) Light energy into electrical energy
(m) Electrical energy into magnetic energy.

Solution 27.

No. This is because, whenever there is conversion of energy from one form to another apart of the energy is dissipated in the form of heat which is lost to surroundings.

Solution 1 (MCQ).

Potential energy
Hint: P.E. is the energy possessed by a body by virtue of its position.

Solution 2 (MCQ).

Chemical to electrical
Hint: When current is drawn from an electric cell, the chemical energy stored in it changes into electrical energy.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 19

Solution 2.

Mass , m=1kg
Height, h=5m
Gravitational potential energy = mgh
=1 x 10 x 5=50J

Solution 3.

Gravitational potential energy=14700 J
Force of gravity = mg= 150 x 9.8N/kg= 1470N
Gravitational potential energy= mgh
14700 =1470 x h
h=10m

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 20

Solution 5.

Mass =0.5 kg
Energy= 1 J
Gravitational potential energy= mgh
1=0.5 x10 x h
1=5h
Height, h= 0.2 m

Solution 6.

Force of gravity on boy=mg= 25 x 10 =250N
Increase in gravitational potential energy= Mg (h2-h1)
= 250 x (9-3)
=250 x6=1500 J

Solution 7.

Mass of water, m= 50kg
Height, h=15m
Gravitational potential energy= mgh
=50 x10 x 15
=7500 J

Solution 8.

Mass of man=50kg
Height of ladder, h2=10m

(i) Work done by man =mgh2
=50 x 9.8 x10= 4900J

(ii) increase in his potential energy:
Height, h2= 10m
Reference point is ground, h1=0m
Gravitational potential energy= Mg (h2-h1)
= 50 x9.8x (10-0)
= 50 x 9.8 x10= 4900J

Solution 9.

F=150N

(a) Work done by the force in moving the block 5m along the slope =Force x displacement in the direction of force
=150 x 5=750 J.

(b) The potential energy gained by the block U =mgh where h =3m
=200 x 3=600 J
he potential energy gained by the block

(c) The difference i.e., 150 J energy is used in doing work against friction between the block and the slope, which will appear as heat energy.

Solution 10.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 21

Solution 11.

If the speed is halved (keeping the mass same), the kinetic energy decreases, it becomes one-fourth (since kinetic energy is proportional to the square of velocity).

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 22

Solution 14.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 23

Solution 15.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 24

Solution 16.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 25

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 26

Solution 18.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 27

Solution 19.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 28

Solution 20.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 29

Solution 21.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 30

Solution 22.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 31

Solution 23.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 32

Exercise 2(C)

Solution 1.

According to the law of conservation of energy, energy can neither be created nor can it be destroyed. It only changes from one form to another.

Solution 2.

According to the law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy (i.e., the sum of kinetic energy K and potential energy U) remains constant i.e., K + U = constant when there are no frictional forces.

Mechanical energy is conserved only when there are no frictional forces for a given system (i.e. between body and air). Thus, conservation of mechanical energy is strictly valid only in vacuum, where friction due to air is absent.

Solution 3.

Motion of a simple pendulum and motion of a freely falling body.

Solution 4.

Kinetic energy of the body changes to potential energy when it is thrown vertically upwards and its velocity becomes zero.

Solution 5.

(a) Potential energy
(b) Potential energy and kinetic energy
(c) Kinetic energy

Solution 6.

Let a body of mass m be falling freely under gravity from a height h above the ground (i.e., from position A). Let us now calculate the sum of kinetic energy K and potential energy U at various positions, say at A (at height h above the ground), at B (when it has fallen through a distance x) and at C (on the ground).
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 33

(i) At the position A (at height h above the ground):
Initial velocity of body= 0 (since body is at rest at A)
Hence, kinetic energy K =0
Potential energy U = mgh
Hence total energy = K + U= 0 + mgh =mgh—–(i)

(ii) At the position B (when it has fallen a distance x):
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 34
Thus from equation (i), (ii) and (iii), we note that the total mechanical energy i.e., the sum of kinetic energy and potential energy always remain constant at each point of motion and is equal to initial potential energy at height h.

Solution 7.

When the bob swings from A to B, the kinetic energy decreases and the potential energy becomes maximum at B where it is momentarily at rest.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 35
From B to A, the potential energy again changes into the kinetic energy and the process gets repeated again and again.
Thus while swinging, the bob has only the potential energy at the extreme position B or C and only the kinetic energy at the resting position A. At an intermediate position (between A and B or between A and C), the bob has both the kinetic energy and potential energy, and the sum of both the energies (i.e., the total mechanical energy) remains constant throughout the swing.

Solution 8.

(a) At position A, pendulum has maximum kinetic energy and its potential energy is zero at its resting position. Hence, K=mgh and U= 0.
(b) At B, kinetic energy decreases and potential energy increases. Hence, K= 0 and U=mgh
(c) At C also, kinetic energy K= 0 and potential energy U=mgh.

Solution 9.

a) Extreme position: Potential energy
b) Mean position: Kinetic energy
c) Between mean and extreme: Both kinetic energy and potential energy

Solution 10.

The gradual decrease of useful energy due to friction etc. is called the degradation of energy.
Examples:

  1. When we cook food over a fire, the major part of heat energy from the fuel is radiated out in the atmosphere. This radiated energy is of no use to us.
  2. When electrical appliances are run by electricity, the major part of electrical energy is wasted in the form of heat energy.

Solution 1 (MCQ).

Potential energy of the ball at the highest point is mgh.
Hint: At the highest point, the ball momentarily comes to rest and thus its kinetic energy becomes zero.

Solution 2 (MCQ).

The sum of its kinetic and potential energy remains constant throughout the motion.
Hint: In accordance with law of conservation of mechanical energy, whenever there is an interchange between the potential energy and kinetic energy, the total mechanical energy remains constant.

Numericals

Solution 1.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 36

Solution 2.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 37

Solution 3.

(a)Potential energy of the ball =mgh
=2 x 10 x 5=100J
(b)Kinetic energy of the ball just before hitting the ground = Initial potential energy= mgh=2x10x5=100J
(c)Mechanical energy converts into heat and sound energy.

Solution 4.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 38

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 39

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Work, Energy and Power 40

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Selina Concise Chemistry Class 10 ICSE Solutions Practical Work

Selina Concise Chemistry Class 10 ICSE Solutions Practical Work

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Chemistry Chapter 13 Practical Work. You can download the Selina Concise Chemistry ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Chemistry for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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ICSE SolutionsSelina ICSE Solutions

Selina ICSE Solutions for Class 10 Chemistry Chapter 13 Practical Work

Exercise 1

Solution 1.

(a) (i) Chemical test for ammonia:
If a rod dipped in concentrated hydrochloric acid is brought near ammonia gas, dense white fumes of ammonium chloride (NH4Cl) are formed.

(ii) Chemical test for Sulphur dioxide:
It decolorizes pink coloured potassium permanganate solution.

(iii) Chemical test for HCl:
When HCl gas is passed through AgNO3 solution, white precipitates of AgCl are formed which gets dissolved in excess of NH4OH.

(iv) Chemical test for Chlorine:
It turns moist starch iodide paper (KI + starch solution) blue black.

(v) Chemical test for Carbon dioxide:
When this gas is passed through lime water, it turns milky due to the formation of white precipitates of CaCO3 and on passing excess of carbon dioxide gas, this milkiness disappears.

(vi) Chemical test for oxygen:
This gas is absorbed in colourless alkaline solution of pyrogallol and turns it dark brown.

(vii) Chemical test for hydrogen:
It burns with a pop sound when a burning taper is brought near it.

(b) Ammonia is a basic gas and its basic nature is suspected through litmus paper test because it changes the colour of red litmus paper to blue.
(c) Chlorine, carbon dioxide, hydrogen chloride, hydrogen sulphide and sulphur dioxide are acidic gases since they convert blue litmus to red.
(d) A is chlorine and B is Sulphur dioxide.
(e) Water vapour.

Solution 2.

(a) O2
(b) NH3
(c) Water vapour
(d) SO2

Solution 3.

(a) Na2CO3 and K2CO3
(b) SO2
(c) CO2
(d) Cl2
(e) H2S

Solution 4.

Silver nitrate and ammonium nitrate.

Solution 5.

(a) Cl
(b) SO42-
(c) CO32-
(d) SO32-

Solution 6.

(a) Since the salt solution turned blue litmus red hence the salt may be an acid.
(b) Since addition of barium chloride into the solution of salt gave white precipitate so the salt may contain SO42-, SO32-, CO32 anion.
(c) The flame test of the salt gives persistent golden yellow colourisation which suggests presence of Na+ ion.

Solution 7.

(a) Ca2+
(b) Cu+
(c) The three ways are:

  1. Ammonia gas turns moist red litmus blue.
  2. If a rod dipped in concentrated HCl is brought near the gas, dense white fumes of NH4Cl are formed.
  3. The gas turns colourless Nessler’s reagent i.e. K2HgI4 brown.

Solution 8.

Hydrogen sulphide Ammonia Sulphur dioxide Hydrogen chloride
Shake the gas with red litmus solution No change in the colour of litmus solution Red litmus solution becomes blue in colour. No change in the colour of litmus solution No change in the colour of litmus solution
Shake the gas with blue litmus solution Blue litmus solution becomes red in colour. No change in the colour of blue litmus solution. Blue litmus solution becomes red in colour. Blue litmus solution becomes red in colour
Apply a burning splint to a gas No reaction. No reaction. No reaction. No reaction.

Solution 9.

(P) Ammonium chloride
(Q) Calcium
(R) Calcium hydroxide
(S) Lead (II) Nitrate
(T) Calcium Oxide
(U) Lead (II) Oxide
(V) Chlorine
(W) Hydrogen chloride

Solution 10.

Carbonate Colour of residue on cooling
Zinc Carbonate white
Lead Carbonate yellow
Copper Carbonate black

Solution 11.

(i) Sodium carbonate and sodium sulphite can be distinguished by using acidified K2Cr2O7:
Take a small quantity of salt in a test tube; add dil. H2SO4.and warm if necessary. Now if on bringing a filter paper moistened with acidified K2Cr2O7 near the gas evolved, the orange colour of the paper turns green then it is sodium sulphite.

(ii) Sodium thiosulphate and sodium sulphite:
The salts can be distinguished by using silver acetate. To the salt silver acetate and dil. HNO3 are added. If there is formation of a white precipitate which slowly turns black then it is thiosulphate anion since silver acetate forms Ag2S2O3 which being unstable in acid solution gets converted to black Ag2S.

(iii) Sodium hydroxide solution and ammonium hydroxide solution:
These salts can be distinguished by using a metal cation like calcium. When we add calcium salt to sodium hydroxide and ammonium hydroxide, then a white curdy ppt. is formed only in case of sodium hydroxide.

(iv) Ammonium sulphate and sodium sulphate:
These salts can be distinguished by using KOH. When KOH is added to ammonium sulphate, ammonia gas is evolved. Whereas there is no evolution of ammonia gas in case of sodium sulphate.

(v) Add barium chloride solution to sulphuric acid, nitric acid and hydrochloric acid. A white precipitate is formed in dilute sulphuric acid, and no such precipitate is formed in nitric acid and hydrochloric acid.
BaCl2(aq)  + H2SO4(aq) → BaSO4(s) + 2HCl(aq)

Solution 12.

(a) Lead chloride as precipitate and sodium nitrite are formed.
(b)

  Zinc chloride Zinc nitrate Zinc sulphate
Barium chloride No reaction No reaction White ppt. is obtained
Lead nitrate No reaction No reaction No reaction

(c) Dilute sulphuric acid liberates carbon dioxide from metallic carbonates and bicarbonates. Carbon dioxide when bubbled into a test tube containing calcium hydroxide solution turns it milky.

Solution 1 (2004).

Aqueous salt solution Colour of the precipitate when NaOH is added in small quantity Nature of the (soluble or insoluble) when NaOH is added in excess
copper (II) sulphate

zinc nitrate

lead nitrate

calcium chloride

iron (III) sulphate

(i) Pale blue

(ii) White gelatinous

(iii) White chalky

(iv)White curdy

(v) Reddish brown

(vi) Insoluble

(viii) Soluble

(viii) Soluble

(ix) Insoluble

(x) Insoluble

Solution 1 (2005).

(I) Iron (II) Sulphate and Magnesium sulphate
(II) Iron (III) chloride and Zinc Chloride
(III) Lead nitrate
(IV) Copper nitrate.
(V) Lead nitrate.

Solution 1 (2006).

(a) When alkaline phenolphthalein solution is added to acids then the colourless solution remains colourless.
(b) Orange colour of methyl orange indicator turns pink when the indicator is added to acids.
(c) Neutral litmus solution turns red on addition to acids.

Solution 1 (2007).

Salt Anion
A Cl
B S2-
C NO3
D SO32-
E CO32-

Solution 1 (2008).

(a) Iron (II) sulphate
(b) (I) Ammonia (NH3)
(II) Dilute nitric acid (HNO3)
(III) H2S
(IV) Chlorine (Cl2)
(V) Ethanol

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Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes)

Selina Concise Biology Class 10 ICSE Solutions The Excretory System (Elimination of Body Wastes)

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes). You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 8 The Excretory System (Elimination of Body Wastes)

Exercise 1

Solution A.1.
(c) Removal of nitrogenous wastes

Solution A.2.
(a) Proximal convoluted tubule

Solution A.3.
(c) Sweating

Solution A.4.
Liver

Solution B.1.
(a) Liver
(b) Cortex
(c) Glomerulus
(d) Collecting duct
(e) Renal artery (Renal vein has urea but renal artery has higher concentration of urea as compared to renal vein).

Solution B.2.
(a) Afferent arteriole, glomerulus, efferent arteriole, capillary network, renal vein
(b) Renal artery, kidney, ureter, urinary bladder, urethra

Solution B.3.
(a) Ultrafiltration
(b) Excretion
(c) Osmoregulation
(d) Excretion

Solution C.1.
(a) Glomerulus is involved in the process of ultrafiltration.The liquid part of the blood which is plasma including urea, salts, glucose filters out from the glomerulus into the renal tubule.
(b) Henle’s loop is involved in reabsorption of water and sodium ions.
(c) Ureter carries urine to the urinary bladder by ureteral peristalsis.
(d) Renal artery supplied blood to the kidney.
(e) Urethra is involved in the process of micturition i.e. expelling urine out of the body.

Solution C.2.
Excretion helps in removing toxic wastes from our body and it also plays an important role osmoregulation i.e. the maintenance of the homeostasis of the body.
Carbon dioxide, water, nitrogenous compounds such as urea, uric acid and excess salts are some common excretory products.

Solution C.3.
A uriniferous tubule also known as the kidney tubule is the structural and functional unit of the kidney.
It takes in impure blood from the renal artery and removes wastes in the form of urine. It also provides a larger surface area for reabsorption of salts and water.

Solution C.4.
Maintaining a normal osmotic concentration in the body means regulating the percentage of water and salts. If this regulation mechanism fails we either end up losing vital salts and water or may accumulate unwanted salts and excess water in our body.

Solution C.5.
If one kidney is donated to a needy patient, the other kidney alone is sufficient for removing wastes or excretion. Thus, the donor can live a normal life.

Solution C.6.
During summer, a considerable part of water is lost through perspiration so the kidneys have to reabsorb more water from the urine. This makes the urine thicker in summer than in winters.

Solution C.7.

(a) Bowman’s capsule is a thin walled cup containing the glomerulus. This Bowman’s capsule along with the glomerulus is known as malpighian capsule.

(b) The renal cortex is the outer darker region of the kidney whereas the renal medulla is the inner lighter region of the kidney.

(c) Renal pelvis is the expanded front end of the ureter in the kidney whereas the renal papilla is the apex of the renal pyramid which projects into the pelvis.

(d) Urea is the chief excretory product which is excreted in the form of urine whereas urine is the filtrate left after reabsorption and tubular secretion which contains 95% water and 5% solid wastes.

(e) Excretion is the process of removal of chemical wastes especially nitrogenous wastes from the body.
Catabolism on the other hand is the set of metabolic pathways which break down molecules into smaller units and release energy.

Solution C.8.
Urea, creatinine, uric acid

Solution C.9.

Column I Column II 
(a)   Bowman’s Capsule Glomerulus
(b)   Contains more CO2 and less urea Renal Vein
(c)    Anti-diuretic hormone Regulates amount of water excreted
(d)   Contains more urea Renal artery

Solution C.10.
In a nephron, the blood flows through the glomerulus under great pressure. The reason for this great pressure is that the efferent (outgoing) arteriole is narrower than the afferent arteriole (incoming). This high pressure causes the liquid part of the blood to filter out from the glomerulusinto the renal capsule.

Solution D.1.

(a) Ultrafiltration – The process of the filtration of blood in the glomerulus under great pressure during which the liquid part of the blood i.e. plasma along with urea, glucose, amino acids and other substances enter the renal tubule.

(b) Micturition – The process of expelling urine out of the body through urethra by opening the sphincter muscles passing of urine involving relaxation of sphincter muscles between the urinary bladder and urethra.

(c) Renal pelvis – Renal pelvis is the expanded front end of the ureters into the kidney.

(d) Urea – A nitrogenous waste produced primarily in the liver due to the break down dead protein remains and extra amino acids.

(e) Osmoregulation – It is a process of maintaining the blood composition of the body i.e. the normal osmotic concentration of water and salts in the body.

Solution D.2.

Ultrafiltration – Ultrafiltration involves filtration of the blood which takes place in the glomerulus. The blood containing urea from the afferent arteriole enters the glomerulus under high pressure. The high pressure is created because the efferent arteriole is narrower than the afferent arteriole. The high pressure causes the liquid part of the blood to filter out from the glomerulus into the renal tubule. This filtrate is known as ‘glomerular filtrate’.
Glomerular filtrate consists of water, urea, salts, glucose and other plasma solutes. Blood corpuscles, proteins and other large molecules remain behind in the glomerulus. Therefore the blood which is carried away by the efferent arteriole is relatively thick.

Selective absorption – The Glomerular filtrate entering the renal tubule contains a lot of usable materials such as glucose and sodium. As this filtrate passes down the renal tubule, a lot of water along with these usable materials is reabsorbed. Such reabsorption is called ‘selective absorption’. The reabsorption occurs only to the extent that the normal concentration of the blood is undisturbed.

Solution D.3.
Dialysis involves the use of artificial kidney or a dialysis machine. The patient’s blood is from the radial artery is led through the machine where excess salts and urea is removed. The purified blood is then returned to a vein in the same arm.
Dialysis is carried out in case of failure of both the kidneys. In case there is a permanent damage, then the dialysis is to be repeated for about 12 hours twice a week.

Solution E.1.
(a) The image shown can be left or right kidney. As the right kidney is slightly lower than the left one, so we need to have the images of both the kidneys for comparison.
(b) It is a longitudinal section of the kidney.
(c) 1-renal artery, 2-renal vein, 3-ureter, 4-cortex, 5-pelvis
(d) (i) 4/cortex
(ii) medulla
(iii) 5/pelvis

Solution E.2.
(a) Excretory system and Circulatory system.
(b) 1-kidney, 2-renal artery, 3-ureter, 4-urinary bladder, 5-urethera
(c) Nephron
(d) Urea and ammonia
(e) Ultrafiltration and selective reabsorption

Solution E.3.
(a) 4/Glomerulus
(b) 2/Efferent arteriole
(c) 1/ Afferent arteriole from renal artery
(d) 7/Collecting tubule
(e) 5/ Proximal convoluted tubule with blood capillaries

Solution E.4.
(a) The process of removal of chemical wastes especially nitrogenous waste from the body is known as excretion.
(b) Nephrons
(c) As the cortex region contains numerous nephrons or kidney tubules, therefore, it shows a dotted appearance.
(d) Kidneys help in removing wastes or excretion and osmoregulation.
(e) The blood vessel ‘B’ is renal artery and the blood vessel ‘A’ is renal vein.

So the blood vessel ‘B’ contains oxygenated blood with high concentration of urea and glucose whereas the blood vessel ‘A’ contains deoxygenated blood with low concentration of urea and glucose as compared to renal artery.

Solution E.5.
a. The structure is a Bowman’s capsule, which is part of the nephron. The Bowman’s capsule is found in the cortex of the kidney.

b.

  1. Afferent arteriole
  2. Glomerulus
  3. Bowman’s capsule
  4. Efferent arteriole

c. Urine formation occurs in two steps – ultrafiltration and reabsorption.

d. The process occurring in 2 and 3 is known as ultrafiltration.
In the glomerulus, the blood flows under high pressure because of the narrow lumen of the capillary network of the glomerulus. This forces most of the components (both waste and useable materials) of the blood out of the capillaries. This process of the filtration of blood under high pressure in the Bowman’s capsule is known as ultrafiltration.

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Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles

Circles Exercise 17A – Selina Concise Mathematics Class 10 ICSE Solutions

Circles Class 10 Question 1.
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 2

Question 2.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 4

Question 3.
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠ OCA,
(ii) ∠OAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 6

Question 4.
In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 8

Question 5.
In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 10
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 11

Question 6.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centres of two circles.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 13

Question 7.
In the figure given beow, find :
(i) ∠ BCD,
(ii) ∠ ADC,
(iii) ∠ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 14
Show steps of your workng.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 15

Question 8.
In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :
(i) ∠ ACB,
(ii) ∠ OBC,
(iii) ∠ OAB,
(iv) ∠CBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 16
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 17

Question 9.
Calculate :
(i) ∠ CDB,
(ii) ∠ ABC,
(iii) ∠ ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 18
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 19

Question 10.
In the figure given below, ABCD is a eyclic quadrilateral in which ∠ BAD = 75°; ∠ ABD = 58° and ∠ADC = 77°. Find:
(i) ∠ BDC,
(ii) ∠ BCD,
(iii) ∠ BCA.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 20
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 21

Question 11.
In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find :
(i) ∠ ADB
(ii) ∠ AEB
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 22
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 23

Question 12.
Given—∠ CAB = 75° and ∠ CBA = 50°. Find the value of ∠ DAB + ∠ ABD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 24
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 25

Question 13.
ABCD is a cyclic quadrilateral in a circle with centre O.
If ∠ ADC = 130°; find ∠ BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 27

Question 14.
In the figure given below, AOB is a diameter of the circle and ∠ AOC = 110°. Find ∠ BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 29

Question 15.
In the following figure, O is centre of the circle,
∠ AOB = 60° and ∠ BDC = 100°.
Find ∠ OBC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 30

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 31

Question 16.
ABCD is a cyclic quadrilateral in which ∠ DAC = 27°; ∠ DBA = 50° and ∠ ADB = 33°.
Calculate :
(i) ∠ DBC,
(ii) ∠ DCB,
(iii) ∠ CAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 32
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 33

Question 17.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°. Find the number of degrees in:
(i) ∠DCE;
(ii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 34
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 35

Question 17 (old).
In the figure given below, AB is diameter of the circle whose centre is O. Given that:
∠ ECD = ∠ EDC = 32°.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 36
Show that ∠ COF = ∠ CEF.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 37

Question 18.
In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 38
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 39

Question 19.
In the following figure,
(i) if ∠BAD = 96°, find BCD and
(ii) Prove that AD is parallel to FE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 41

Question 20.
Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 42
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 43

Question 21.
In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 44
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 45

Question 22.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 46

Question 23.
The figure given below, shows a circle with centre O. Given: ∠ AOC = a and ∠ ABC = b.
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 47
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 48

Question 24.
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 49

Question 24 (old).
ABCD is a quadrilateral inscribed in a circle having ∠A = 60°; O is the centre of the circle. Show that: ∠OBD + ∠ODB = ∠CBD + ∠CDB
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 50

Question 25.
In the figure given RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°
Calculate:
(i) ∠RNM;
(ii) ∠NRM.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 51
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 52

Question 26.
In the figure given alongside, AB || CD and O is the centre of the circle. If ∠ ADC = 25°; find the angle AEB. Give reasons in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 54

Question 27.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at Cand D. Prove that AC is parallel to BD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 56

Question 28.
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 57

Question 29.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 58
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 59

Question 30.
In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 60
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 61

Question 31.
In the figure, O is the centre of the circle, ∠AOE = 150°, DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 62
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 63

Question 32.
In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 65

Question 33.
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that ∠APB = a°. Calculate, in terms of a°, the value of:
(i) obtuse ∠AOB
(ii) ∠ACB
(iii) ∠ADB.
Give reasons for your answers clearly.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 67

Question 34.
In the given figure, O is the centre of the circle and ∠ ABC = 55°. Calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 68
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 69

Question 35.
In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that ∠BCD = 2∠ABE
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 70
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 71

Question 36.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate:
(i) ∠ DAB,
(ii) ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 72
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 73

Question 37.
∠ In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠ EAB = 63°; calculate:
(i) ∠EBA,
(ii) BCD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 74
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 75

Question 38.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°; calculate:
(i) ∠ DAB,
(ii) ∠ DBA,
(iii) ∠ DBC,
(iv) ∠ ADC.
Also, show that the ∆AOD is an equilateral triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 76
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 77

Question 39.
In the given figure, I is the incentre of the ∆ ABC. Bl when produced meets the circumcirle of ∆ ABC at D. Given ∠BAC = 55° and ∠ ACB = 65°, calculate:
(i) ∠DCA,
(ii) ∠ DAC,
(iii) ∠DCI,
(iv) ∠AIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 78
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 79

Question 40.
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ABC = 2 ∠APQ
(ii) ∠ACB = 2 ∠APR
(iii) ∠QPR = 90° – \(\frac{1}{2}\)BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 80
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 81

Question 40 (old).
The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB are produced to meet at F. If ∠BEC = 42° and ∠BAD = 98°; calculate:
(i) ∠AFB,
(ii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 82
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 83

Question 41.
Calculate the angles x, y and z if: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 84
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 85

Question 42.
In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate:
(i) Angle ABC
(ii) Angle BEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 86
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 87

Question 43.
In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, and r in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 88
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 89

Question 44.
In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80° and angle ACE = 10°. Calculate:
(i) Angle BEC;
(ii) Angle BCD;
(iii) Angle CED.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 90
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 91

Question 45.
In the given figure, AE is the diameter of circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 92
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 93

Question 46.
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 94
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 95

Question 47.
Use the given figure to find
(i) ∠BAD
(ii) ∠DQB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 96
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 97

Question 48.
In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:
(i) ∠COB
(ii) ∠DOC
(iii) ∠DAC
(iv) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 98
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 99

Question 49.
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 100
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 101

Question 50.
In the given figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°; calculate ∠RTS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 102
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 103

Question 51.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°; calculate
(i) ∠RPQ
(ii) ∠STP.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 104
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 105

Question 52.
AOD = 60°; calculate the numerical values of:
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°; calculate the numerical values of:
(i) ∠ABD,
(ii) ∠DBC,
(iii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 106
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 107

Question 53.
In the given figure, the centre of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40″; find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 108
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 109

Question 54.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°; find:
(i) ∠BCD,
(ii) ∠ACB.
Hence, show that AC is a diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 110
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 111

Question 55.
In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 112

Question 56.
The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 195
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 196

Question 57.
In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠ MAD =x and ∠BAC = y.
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that : x = y
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 197
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 198

Question 61 (old).
In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 199

Circles Exercise 17B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 200

Question 2.
In the following figure, AD is the diameter of the circle with centre 0. chords AB, BC and CD are equal. If ∠DEF = 110°, calculate:
(i) ∠ AFE,
(ii) ∠FAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 201

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 202

Question 3.
If two sides of a cycli-quadrilateral are parallel; prove thet:
(i) its other two side are equal.
(ii) its diagonals are equal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 203

Question 4.
The given figure show a circle with centre O. also, PQ = QR = RS and ∠PTS = 75°. Calculate:
(i) ∠POS,
(ii) ∠ QOR,
(iii) ∠PQR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 204
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 205

Question 5.
In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠ AOB,
(ii) ∠ ACB,
(iii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 206
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 207

Question 6.
In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 208

Question 7.
In the given figure. AB = BC = CD and ∠ABC = 132°, calculate:
(i) ∠AEB,
(ii) ∠ AED,
(iii) ∠COD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 209
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 210

Question 8.
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find:
(i) ∠ CAB,
(ii) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 211
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 212

Question 9.
The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 213
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 214

Question 10.
In the given figire, BD is a side of a regularhexagon, DC is a side of a regular pentagon and AD is adiameter. Calculate:
(i) ∠ ADC
(ii) ∠BAD,
(iii) ∠ABC
(iv) ∠ AEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 215
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 216
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 217

Circles Exercise 17C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the given circle with diametre AB, find the value of x.

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 218
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 219
∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a A)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180° ⇒ x + 120° = 180°
∴ x= 180° – 120° = 60° Ans.

Question 1.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 220
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 221

Question 2.
In the given figure, ABC is a triangle in which ∠ BAC = 30° Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 222
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 223

Question 3.
Prove that the circle drawn on any one a the equalside of an isoscele triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 224

Question 3 (old).
The given figure show two circles with centres A and B; and radii 5 cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 225
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 226

Question 4.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠ CBE = 65°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 227
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 228

Question 5.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 229

Question 6.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 230
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 231

Question 7.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Provet that the points B, C, E and D are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 232

Question 7 (old).
Chords AB and CD of a circle intersect each other at point P such that AP = CP.
Show that: AB = CD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 233
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 234

Question 8.
In the given rigure, ABCD is a cyclic eqadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ ADC = 92°, ∠ FAE = 20°; determine ∠ BCD. Given reason in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 235
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 236

Question 9.
If I is the incentre of triangle ABC and Al when produced meets the cicrumcircle of triangle ABC in points D. if ∠ BAC = 66° and ∠ = 80o.calculate:
(i) ∠ DBC
(ii) ∠ IBC
(iii) ∠ BIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 237
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 238
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 239

Question 10.
In the given figure, AB = AD = DC = PB and ∠ DBC = x°. Determine, in terms of x:
(i) ∠ ABD,
(ii) ∠ APB.
Hence or otherwise, prove thet AP is parallel to DB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 240
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 241

Question 11.
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠ CQE are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 242
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 243

Question 12.
In the given, AB is the diameter of the circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 244
If ∠ ADC = 32°, find angle BOC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 245

Question 13.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ prouduced meet at point A: whereas sides PQ and SR produced meet at point B.
If ∠A: ∠B = 2 : 1;find angles A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 246
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 247

Question 17 (old).
If the following figure, AB is the diameter of a circle with centre O and CD is the chord with lengh equal radius OA.
If AC produced and BD produed meet at point p; show that ∠APB = 60°
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 248
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 249
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 250

Question 14.
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 251
If the bisector of angle A meet BC at point E and the given circle at point F, prove that:
(i) EF = FC
(ii) BF =DF
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 252

Question 15.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point e; whereas sides BC and AD produced meet at point F. I f ∠ DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 253
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 254

Question 16.
The following figure shows a cicrcle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimetre of the cyclic quadrilateral PORS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 255
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 256

Question 17.
In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD.prove that:
(i) arc BC = arc DB
(ii) AB is bisector of ∠ CAD.
Further if the lenghof arc AC is twice the lengthof arc BC find :
(a) ∠ BAC
(b) ∠ ABC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 257
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 258
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 259

Question 18.
In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;
(i) ∠ BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ ADC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 260
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 261

Question 19.
In the given figure, ∠ACE = 43° and ∠ = 62°; find the values of a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 262
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 263

Question 20.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠ BAC = 25°
Find
(i) ∠ CAD
(ii) ∠ CBD
(iii) ∠ ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 264
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 265

Question 21.
ABCD is a cyclic quadrilateral of a circle with centre o such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle..if AD and BC produced meet at P, show that APB =60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 266

Question 22.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 267
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 268

Question 23.
In the figure, given below, AD = BC, ∠ BAC = 30° and ∠ = 70° find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 269
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 270

Question 24.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 271

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 272

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 273

Question 25.
In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of
i. ∠BCD
ii. ∠BOD
iii. ∠OBD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 274
Solution:
∠DAE and ∠DAB are linear pair
So,
∠DAE + ∠DAB = 180°
∴∠DAB = 110°

Also,
∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC
∴∠BCD = 70°
∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD…angles subtended by an arc on the centre and on the circle
∴∠BOD = 140°

In ΔBOD,
OB = OD……radii of same circle
So,
∠OBD =∠ODB……isosceles triangle theorem
∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle
2∠OBD = 40°
∠OBD = 20°

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere (Surface Area and Volume)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume)

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The height of a circular cylinder is 20 cm and the radius of its base is 7 cm. Find :
(i) the volume
(ii) the total surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere ex 20a q1

Question 2.
The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 1
Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 3

Question 4.
How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter? Also, find the cost of plastering its inner surface at Rs 4.50 per sq meter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 4

Question 5.
What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 5

Question 6.
A cylinder has a diameter of 20 cm. The area of curved surface is 100 sq cm. Find:
(i) the height of the cylinder correct to one decimal place.
(ii) the volume of the cylinder correct to one decimal place.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 6

Question 7.
A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm3 of the metal weights 7.7 g.
Solution:
Inner radius of the pipe = r =\(\frac{5}{2}\) = 2.5 cm
External radius of the pipe = R = Inner radius of the pipe + Thickness of the pipe
= 2.5 cm + 0.5 cm
= 3 cm
Length of the pipe = h = 2 m= 200 cm
Volume of the pipe = External Volume – Internal Volume
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 7
Since 1cm3 of the metal weights 7.7 9,
∴ Weight of the pipe = (1728.6 × 7.7)g = \(\left(\frac{1728.6 \times 7.7}{1000}\right)\) kg = 13.31 kg

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 8

Question 9.
A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wetted surface of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 9

Question 10.
Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 10

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 153
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 11

Question 12.
The radius of a solid right circular cylinder increases by 20% and its height decreases by 20%. Find the percentage change in its volume.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 12

Question 13.
The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its :
(i) volume
(ii) curved surface area
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 13
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 14

Question 14.
Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.
Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 15

Question 15.
3080 cm3 of water is required to fill a cylindrical vessel completely and 2310 cm3 of water is required to fill it upto 5 cm below the top. Find :
(i) radius of the vessel.
(ii) height of the vessel.
(iii) wetted surface area of the vessel when it is half-filled with water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 16
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 17

Question 16.
Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along it :
(i) shorter side.
(ii) longer side.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 18

Question 17.
A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 19Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 19

Question 18.
A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embankment 2 m in width and 1.6 m in height. Find the depth of the circular tank.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 20

Question 19.
The sum of the inner and the outer curved surfaces of a hollow metallic cylinder is 1056 cm2 and the volume of material in it is 1056 cm3. Find its internal and external radii. Given that the height of the cylinder is 21 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 21
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 22

Question 20.
The difference between the outer curved surface area and the inner curved surface area of a hollow cylinder is 352 cm2. If its height is 28 cm and the volume of material in it is 704 cm3;find its external curved surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 23

Question 21.
The sum of the heights and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm2, find the volume of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 24

Question 22.
The total surface area of a solid cylinder is 616 cm2. If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 25

Question 23.
A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 26

Question 24.
Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are metled and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 28

Question 25.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 151Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 29
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 30

Question 26.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 31
The given figure shows a solid formed of a solid cube of side 40cm and a solid cylinder of radius 20 cm and height 50 cm attached to the cube as shown.
Find the volume and the total surface area of the whole solid (Take π = 3.14)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 32

Question 27.
Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3, Find the ratio between their :
(i) curved surface areas.
(ii) volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 33

Question 28.
A dosed cylindrical tank, made of thin ironsheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m2, is used in making this tank, if \(\frac{1}{15}\) of the sheet actually used was wasted in making the tank?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 34

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 35

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 153
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 36

Question 3.
The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 37

Question 4.
The radius and height of a right circular cone are in the ratio 5:12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 38

Question 5.
Two right circular cones x and y are made, x having three times the radius of y and y having half the volume of x. Calculate the ratio between the heights of x and y.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 39

Question 6.
The diameters of two cones are equal. If their slant heights are in the ratio 5:4, find the ratio of their curved surface areas.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 40

Question 7.
There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 41

Question 8.
A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 42

Question 9.
Find what length of canvas, 1.5 m in width, is required to make a conical tent 48 m in diameter and 7 m in height. Given that 10% of the canvas is used in folds and stitching. Also, find the cost of the canvas at the rate of Rs. 24 per meter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 43
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 44

Question 10.
A solid cone of height 8 cm and base radius 6 cm is melted and re-casted into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 45

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 46
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 47

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 48
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 49

Question 13.
A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result, one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 50

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 51
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 52

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 54

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 56

Question 3.
A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 57

Question 4.
How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 58

Question 5.
8 metallic sphere; each of radius 2 mm, are melted and cast into a single sphere. Calculate the radius of the new sphere.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 59

Question 6.
The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:
(i) radii
(ii) surface areas
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 59

Question 7.
If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in the volume, what is the diameter of the sphere?
Solution:
atics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 61

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 154Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 155

Question 9.
The internal and external diameters of a hollow hemi-spherical vessel are 21 cm and 28 cm respectively. Find:
(i) internal curved surface area
(ii) external curved surface area
(iii) total surface area
(iv) volume of material of the vessel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 62
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 63

Question 10.
A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 64

Question 11.
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking ∏ = 3.1, find the surface area of solid sphere formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 65

Question 12.
The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its:
(i) radius
(ii) volume
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 66
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 67

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 68

Question 2.
A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 69

Question 3.
The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively. It is melted and recast into a solid right circular cone of height 32 cm. find the diameter of the base of the cone.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 70

Question 4.
Total volume of three identical cones is the same as that of a bigger cone whose height is 9 cm and diameter 40 cm. find the radius of the base of each smaller cone, if height of each is 108 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 71

Question 5.
A solid rectangular block of metal 49 cm by 44 cm by 18 cm is melted and formed into a solid sphere. Calculate the radius of the sphere.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 72

Question 6.
A hemi-spherical bowl of internal radius 9 cm is full of liquid. This liquid is to be filled into conical shaped small containers each of diameter 3 cm and height 4 cm. How many containers are necessary to empty the bowl?
Solution:
Radius of hemispherical bowl = 9 cm
Volume = \(\frac{1}{2} \times \frac{4}{3} \pi r^{3}=\frac{2}{3} \pi 9^{3}=\frac{2}{3} \pi \times 729=486 \pi \mathrm{cm}^{2}\)
Diameter each of cylindrical bottle = 3 cm
Radius = \(\frac{3}{2}\)cm, and height = 4 cm
∴ Volume of bottle = \(\frac{1}{3} \pi \pi^{2} n=\frac{1}{3} \pi \times\left(\frac{3}{2}\right)^{2} \times 4=3 \pi\)
∴ No. of bottles = \(\frac{486 \pi}{3 \pi}=162\)

Question 7.
A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone if it is completely filled.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 73

Question 8.
A solid cone of radius 5 cm and height 8 cm is melted and made into small spheres of radius 0.5 cm. Find the number of spheres formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 74

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 75
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 76

Question 10.
A solid metallic cone, with radius 6 cm and height 10 cm, is made of some heavy metal A. In order to reduce weight, a conical hole is made in the cone as shown and it is completely filled with a lighter metal B. The conical hole has a diameter of 6 cm and depth 4 cm. Calculate the ratio of the volume of the metal A to the volume of metal B in the solid.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 77
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 78

Question 11.
A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 3 cm and height 8 cm. Find the number of cones.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 79

Question 12.
The surface area of a solid metallic sphere is 2464 cm2. It is melted and recast into solid right circular cones of radius 3.5 cm and height 7 cm. Calculate :
(i) the radius of the sphere.
(ii) the number of cones recast. (Take π = \(\frac{22}{7}\))
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 80

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 81

Question 2.
A buoy is made in the form of a hemisphere surmounted by a right circular cone whose circular base coincides with the plane surface of the hemisphere. The radius of the base of the cone is 3.5 m and its volume is two-third the volume of hemisphere. Calculate the height of the cone and the surface area of the buoy, correct to two decimal places.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 82

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 83
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 84

Question 4.
The cubical block of side 7 cm is surmounted by a hemisphere of the largest size. Find the surface area of the resulting solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 85

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of the top which is open is 5 cm. It is filled with water. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 86

Question 6.
A hemispherical bowl has negligible thickness and the length of its circumference is 198 cm. find the capacity of the bowl.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 87

Question 7.
Find the maximum volume of a cone that can be carved out of a solid hemisphere of radius r cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 88

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 89
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 90

Question 9.
A solid hemisphere of diameter 28 cm is melted and recast into a number of identical solid cones, each of diameter 14 cm and height 8 cm. Find the number of cones so formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 91

Question 10.
A cone and a hemisphere have the same base and same height. Find the ratio between their volumes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 92

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base are removed. Find the volume of the remaining solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 93

Question 2.
From a solid cylinder whose height is 16 cm and radius is 12 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume and total surface area of the remaining solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 94
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 95

Question 3.
A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 80 m, calculate the total area of canvas required. Also, find the total cost of canvas used at Rs 15 per meter if the width is 1.5 m
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 96

Question 4.
A circus tent is cylindrical to a height of 8 m surmounted by a conical part. If total height of the tent is 13 m and the diameter of its base is 24 m; calculate:
(i) total surface area of the tent
(ii) area of canvas, required to make this tent allowing 10% of the canvas used for folds and stitching.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 97
Height of the cylindrical part = H = 8 m
Height of the conical part = h = (13 – 8)m = 5 m
Diameter = 24 m → radius = r = 12 m
Slant height of the cone = l
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 98
Slant height of cone = 13 m
(i) Total surface area of the tent = 2πrh + πrl = πr(2h + l)
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 99
(ii)Area of canvas used in stitching = total area
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 100

Question 5.
A cylindrical boiler, 2 m high, is 3.5 m in diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 101

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 102
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 156
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 104

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 105
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 106

Question 8.
A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in level of the water when the solid is submerged.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 107

Question 9.
Spherical marbles of diameter 1.4 cm are dropped into beaker containing some water and are fully submerged. The diameter of the beaker is 7 cm. Find how many marbles have been dropped in it if the water rises by 5.6 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 108

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 109
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 110

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 111
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 112

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 113
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 114

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 115
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 116
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 117

Question 14.
A cylindrical can, whose base is horizontal and of radius 3.5 cm, contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate:
(i) the total surface area of the can in contact with water when the sphere is in it;
(ii) the depth of water in the can before the sphere was put into the can.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 118
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 119

Question 15.
A hollow cylinder has solid hemisphere inward at one end and on the other end it is closed with a flat circular plate. The height of water is 10 cm when flat circular surface is downward. Find the level of water, when it is inverted upside down, common diameter is 7 cm and height of the cylinder is 20 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 120

Cylinder, Cone and Sphere Surface Area and Volume Exercise 20G – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
What is the least number of solid metallic spheres, each of 6 cm diameter, that should be melted and recast to form a solid metal cone whose height is 45 cm and diameter is 12 cm?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 121

Question 2.
A largest sphere is to be carved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. (Answer correct to the nearest integer)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 122

Question 3.
A right circular cylinder having diameter 12 cm and height 15 cm is full of ice-cream. The ice-cream is to be filled in identical cones of height 12 cm and diameter 6 cm having a semi-spherical shape on the top. Find the number of cones required.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 123

Question 4.
A solid is in the form of a cone standing on a hemisphere with both their radii being equal to 8 cm and the height of cone is equal to its radius. Find in terms of π, the volume of the solid.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 124

Question 5.
The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 0.2 cm. Find the length of wire.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 125

Question 6.
Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 126

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 127
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 128

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 129
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 130

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 131
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 132

Question 10.
A cylindrical water tank of diameter 2.8m and height 4.2m is being fed by a pipe of diameter 7 cm through which water flows at the rate of 4m/s. Calculate, in minutes, the time it takes to fill the tank.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 133

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 134
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 135

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 136
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 137
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 138

Question 13.
A solid, consisting of a right circular cone, standing on a hemisphere, is placed upright, in a right circular cylinder, full of water, and touches the bottom. Find the volume of water left in the cylinder, having given that the radius of the cylinder is 3 cm and its height is 6 cm; the radius of the hemisphere is 2 cm and the height of the cone is 4 cm. Give your answer to the nearest cubic centimeter.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 139
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 140

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 141
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 142

Question 15.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 143
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 144

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 145
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 146
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 148

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 149
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 150

Question 18.
Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 157

Question 19.
A certain number of metallic cones, each of radius 2 cm and height 3 cm, are melted and recast into a solid sphere of radius 6 cm. Find the number of cones used.
Solution:
Let the number of cones melted be n.
Let the radius of sphere be rs = 6 cm
Radius of cone be rc = 2 cm
And, height of the cone be h = 3 cm
Volume of sphere = n (Volume of a metallic cone)
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 158

Question 20.
A conical tent is to accommodate 77 persons. Each person must have 16m3 of air to breathe. Given the radius of the tent as 7m, find the height of the tent and also its curved surface area.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Cylinder, Cone and Sphere image - 159

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Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved

Exercise 1

Solution A.1.
c) Imbibition

Solution A.2.
c) Hypertonic salt solution

Solution A.3.
b) Turgidity

Solution A.4.
d) Grow downward into the soil

Solution A.5.
b) Diffusion

Solution A.6.
c) a selectively permeable membrane in between

Solution A.7.
a) Pure water

Solution A.8.
d) water

Solution A.9.
b) Root pressure

Solution A.10.
d) it allows a solvent to pass through freely but prevents the passage of the solute

Solution B.1.
(a) Turgidity
(b) Guttation
(c) Osmosis
(d) Xylem
(e) Endosmosis
(f) Diffusion
(g) Root pressure

Solution B.2.
(a) Turgor pressure
(b) Flaccidity
(c) Bleeding

Solution B.3.
(a) the fluids inside
(b) transported inside against their concentration gradient
(c) turgor movements

Solution B.4.
(a) shrink
(b) water
(c) opposite

Solution B.5.

Column I Column II
a Xylem (iv) upward flow of water
b Phloem (iii) downward flow of sap
c Cell membrane (i) semi-permeable
d Root pressure (v) guttation
e Cell wall (ii) permeable

Solution C.1.
(a)

Plasmolysis Deplasmolysis
1. It refers to the shrinkage of the cytoplasm and withdrawal of the plasma membrane from the cell wall caused due to the withdrawal of water when placed in a hypertonic solution.

2. In Plasmolysis, the cell becomes flaccid.

1.Deplasmolysis is the recovery of a plasmolysed cell when it is placed in water, wherein the cell’s protoplasm again swells up due to the re-entry of water.

2. In deplasmolysis, the cell becomes turgid.

(b)

Turgor pressure Wall pressure
Turgor pressure is the pressure of the cell contents on the cell wall. Wall pressure is the pressure exerted by the cell wall on the cell content.

(c)

Guttation Bleeding
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure. Bleeding is the loss of cell sap through a cut stem.

(d)

Turgidity Flaccidity
1. It is the state of a cell in which the cell cannot accommodate any more water and it is fully distended. 1. It is the condition in which the cell content is shrunken and the cell is not tight.

Solution C.2.
(a)

  1. False
  2. False
  3. False
  4. True
  5. False
  6. False

(b)

  1. A plant cell placed in hypotonic solution gets turgid.
  2. Addition of salt to pickles prevents growth of bacteria because they turn flaccid.
  3. Cells that have lost their water content are said to be plasmolysed.
  4. The shrinkage of protoplasm, when a cell is kept in hypertonic solution.

Solution C.3.
The cell is said to be turgid when the plant cell wall becomes rigid and stretched by an increase in the volume of vacuoles due to the absorption of water when placed in hypotonic solution. On the other hand, the cell is said to be flaccid when the cell contents get shrunken when the cell is placed in hypertonic solution and the cell is no more tight. Flaccidity is the reverse of turgidity.

Example: Weeds can be killed in a playground by sprinkling excessive salts around their base.
Or
A plant cell when immersed in hypertonic solution like salt solution for about 30 minutes will become flaccid or limp.

Solution C.4.

(a) Common salt when sprinkled on the grass causes the Plasmolysis of grass cell ultimately leading them to death. Hence, if we sprinkle some common salt on grass growing on a lawn, it is killed at the spot.

(b) If a plant is uprooted, the leaves continue losing water by transpiration, but there is no more water absorbed the roots. This does not allow the compensation for the loss of water by transpiration; hence the leaves of the uprooted plant wilt soon.

(c) Transplantation causes stress to the seedlings. If the seedlings are transplanted in the morning, they would have to immediately bear the additional stress of excessive transpiration occurring during the hot afternoon. Transplantation in the evening helps the seedlings to adjust for a longer time during the night (cooler temperatures) because the quantity of water absorbed exceeds the loss of water through transpiration. Therefore, it is better to transplant seedling in a flower bed in the evening and not in the morning.

(d) In a hypertonic solution, the solution outside the cell has higher solute concentration than the fluids inside the cell. Therefore, water flows out from the plant cell due to exosmosis. The cytoplasm shrinks and the plasma membrane withdraws away from the cell wall and this the cell becomes flaccid. Hence a plant cell when kept in a hypertonic salt solution for about 30 minutes turns flaccid.

(e) Potato cubes contain excess of salts and sugars as compared to the water in which the cubes are placed. Hence, due to endosmosis, water from the surrounding enters the potato cubes making them firm and increasing their size.

Solution C.5.

(a) True.
Plasmolysis occurs due to outflow of water from the cell when placed in hypertonic solution due to which the cytoplasm shrinks away from the cell wall. On the other hand, deplasmolysis is the result of the re-entry of water into the plasmolysed cell when placed in hypotonic solution due to which the protoplasm again swells up pressing tight against the cell wall.

(b) False.
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure whereas bleeding is the loss of cell sap through a cut stem.

(c) False.
There is only one seed coat in a seed.

(d) False.
The leaves of the twig remain turgid since its xylem is intact and xylem is responsible for water conduction in plants.

(e) False.
Guttation occurs due to excessive root pressure. It is maximum when root pressure is maximum which occurs in the early mornings or at night. This is because during these times, transpiration is very low and water absorption is very high.

(f) False.
Dry seeds when submerged in water swell up due to imbibitions. On contact with water dry seeds imbibe water and swell up.

Solution D.1.
Examples of turgor movements in plants:

  1. In Mimosa pudica, a sensitive plant, the stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles called pulvinus. This causes the folding and drooping of leaves of the plant.
  2. The leaves of insectivorous plants close up to entrap a living prey. When the insect come in contact with the leaf, it loses it turgor hence closing the leaves of the plant.
  3. The bending movements of certain flowers towards the sun.
    (Any two)

Solution D.2.
The closing and opening of the stomata depends on the turgidity of the guard cells. Each guard cell has a thicker wall on the side facing the stoma and a thin wall on the opposite side. Guard cells contain chloroplasts. As a result of the synthesis of glucose during photosynthesis and some other chemical changes, the osmotic pressure of the contents of the guard cells increases and they absorb more water from the neighbouring cells, thus becoming turgid. Due to turgor, the guard cells become more arched outwards and the aperture between them widens, thereby opening the stoma.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 1
At night or when there is shortage of water in the leaf, the guard cells turn flaccid and their inner rigid walls become straight, thus closing the stomatal aperture.

Solution D.3.
If the concentration of mineral nutrient elements is higher inside the root-hairs than in the surrounding soil, then roots take them in from the soil by ‘active transport’. In active transport, the mineral ions are forcibly carried from the surrounding soil i.e. the region of their lower concentration into the roots i.e. the region of their higher concentration through the cell membrane by expenditure of energy. This energy is supplied by the cell in the form of ATP.

Solution D.4.
When soaked in water, the seeds swell up due to imbibition and endosmosis. During these two processes water enters the cell. Due to endosmosis, at some point, the seed coat is unable to bear the turgor pressure and hence, the seed coat bursts.

Solution D.5.
Leaves of the sensitive plant wilt and droop down on a slight touch due to turgor movement. Petiole of sensitive plant is held up by turgid pulvinus tissue. The stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles i.e. pulvinus. The cells of the lower side of pulvinus lose water and the petiole collapses. This causes the wilting and drooping of the leaves.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 2

Solution D.6.
As water is lost from the leaf surface by transpiration, more water molecules are pulled up due to the tendency of water molecules to remain joined i.e. cohesion. This produces a continuous column of water throughout the stem which is known as ‘transpiration pull’. A negative pressure or tension is produced in the xylem that pulls the water from the roots and soil. Transpirational pull is an important force which causes the ascent of sap.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 3

Solution E.1.
(a) The cell is flaccid i.e. it is plasmolysed.
(b) Plasma Membrane
(c) Plasmolysis would not occur and flaccidity would not be seen i.e. the protoplasm would not have shrunken away from the cell wall.
(d) Cell Wall is absent in animal cell.

Solution E.2.
(a) Flaccid Cell
(b) The liquid is hypertonic solution. It has higher solute concentration outside the cell than the fluids inside the cell.
(c)
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 4

Solution E.3.

(a) Osmosis

(b) Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

(c) After an hour or so, the level of sugar solution in the thistle funnel will rise and the level of water in the beaker will drop slightly.

(d) For control experiment, the beaker will contain the water. At the same time, instead of the sugar solution; the thistle funnel with the cellophane paper tied on its mouth and inverted in the beaker will also contain water.

(e)

  1. concentrated sugar solution – Cell sap (of higher concentration than that of the surrounding water) within the root hair.
  2. parchment paper – cell membrane of root hair.
  3. water in the beaker – water in soil.

(f) cellophane paper, egg membrane, animal bladder (any one)

(g)

  1. The roots of plants absorb water and minerals from surrounding soil due to osmosis.
  2. Osmosis allows plants to absorb water from the soil which helps plants to keep cells alive in roots, stems and leaves.
  3. Osmosis is also important in the opening and closing of stomata which is an important feature for the processes like transpiration and photosynthesis. (Any two)

Solution E.4.

a.
A – Cell wall
B – Cell membrane
C – Cytoplasm
D – Nucleus

b. A root hair gets turgid because of the absorption of water from the surrounding. Absorption of water by root hair is achieved by the process of osmosis. The concentration of water in the surrounding is more than that of the interior of the cell; this causes the water from the surrounding to move in because of endosmosis.

c.

Cell wall Cell membrane
The cell wall of a root hair is freely permeable and allows both salt and water to pass through. The cell membrane of a root hair is semi-permeable and does not allow large dissolved salt molecules to pass through.

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 5

Solution E.5.

(a) Water is hypotonic to the potato cells, due to which endosmosis occurs and water enters the potato cells. The protoplasm swells up pressing tight against the cell wall. The cells are fully distended i.e. turgid. This causes the firmness and increase in the size of the potato cubes when placed in water.

(b) Sugar solution is hypertonic to the potato cells, due to which exosmosis occurs and water flows out of the potato cells. The potato cell loses its distended appearance, the cytoplasm shrinks and the plasma membrane withdraws from the cell wall. The cells become limp or flaccid. This causes the softness and decrease in size of the potato cubes when placed in sugar solution.

(c) The process being investigated is osmosis. Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

Solution E.6.
(a) It is the diagrammatic cross-section of a part of a root.

(b)

  1. Root hair
  2. Epidermis
  3. Cortex
  4. Endodermis
  5. Phloem
  6. Xylem

(c) Cortex (label 3) is the ground tissue and is active in the uptake of water and minerals. It also helps in storage of photosynthetic products.
Phloem (label 5) helps in transporting the prepared food from leaves to different parts of the plant.

Solution E.7.
(a) The process of water absorption by plant roots through osmosis is being studied here.

(b) A root-hair contains cell sap which contains higher concentration of salts as compared to outside soil water. This difference sets off osmosis and outside water diffuses into the root-hair. From the cell bearing root-hair, water passes into adjoining cells one after another to finally the xylem vessels.

(c) The surface of water was covered with oil to prevent any loss of water by evaporation.

Solution E.8.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 6

More Resources for Selina Concise Class 10 ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression

Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 11 Geometric Progression

Geometric Progression Exercise 11A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find, which of the following sequence form a G.P. :
(i) 8, 24, 72, 216, ……
(ii) \(\frac{1}{8}, \frac{1}{24}, \frac{1}{72}, \frac{1}{216}\), ……..
(iii) 9, 12, 16, 24, ……
Solution 1(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 1

Solution 1(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 2

Solution 1(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 3

Question 2.
Find the 9th term of the series :
1, 4, 16, 64 ……..
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 4

Question 3.
Find the seventh term of the G.P. :
1, \(\sqrt{3}\), 3, \(3 \sqrt{3}\) …..
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 5

Question 4.
Find the 8th term of the sequence :
\(\frac{3}{4}, 1 \frac{1}{2}\) 3, …….
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 6

Question 5.
Find the 10th term of the G.P. :
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 7

Question 6.
Find the nth term of the series :
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 8

Question 7.
Find the next three terms of the sequence :
\(\sqrt{5}\), 5, \(5 \sqrt{5}\), ……
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 9

Question 8.
Find the sixth term of the series :
22, 23, 24, ……….
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 10

Question 9.
Find the seventh term of the G.P. :
[late]\sqrt{3}+1,1, \frac{\sqrt{3}-1}{2}[/latex], ……………..
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 11

Question 10.
Find the G.P. whose first term is 64 and next term is 32.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 12

Question 11.
Find the next three terms of the series:
\(\frac{2}{27}, \frac{2}{9}, \frac{2}{3}\), ………….
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 13

Question 12.
Find the next two terms of the series
2 – 6 + 18 – 54 …………
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 14

Geometric Progression Exercise 11B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Which term of the G.P. :
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 15
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 16

Question 2.
The fifth term of a G.P. is 81 and its second term is 24. Find the geometric progression.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 17

Question 3.
Fourth and seventh terms of a G.P. are \(\frac{1}{18} \text { and }-\frac{1}{486}\) respectively. Find the GP.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 18

Question 4.
If the first and the third terms of a G.P. are 2 and 8 respectively, find its second term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 19

Question 5.
The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 20

Question 6.
Find the geometric progression with 4th term = 54 and 7th term = 1458.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 21

Question 7.
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 22

Question 8.
The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ratio and number of terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 23

Question 9.
If the 4th and 9th terms of a G.P. are 54 and 13122 respectively, find the GP. Also, find its general term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 24

Question 10.
The fifth, eight and eleventh terms of a geometric progression are p, q and r respectively. Show that : q2 = pr.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 25

Geometric Progression Exercise 11C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the seventh term from the end of the series : \(\sqrt{2}\) , 2, \(2 \sqrt{2}\), ………. 32.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 26

Question 2.
Find the third term from the end of the GP.
\(\frac{2}{27}, \frac{2}{9}, \frac{2}{3}\), ………….. 162
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 27

Question 3.
For the \(\frac{1}{27}, \frac{1}{9}, \frac{1}{3}\), ………… 81;
find the product of fourth term from the beginning and the fourth term from the end.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 28

Question 4.
If for a G.P., pth, qth and rth terms are a, b and c respectively ; prove that :
(q – r) log a + (r – p) log b + (p – q) log c = 0
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 29

Question 5.
If a, b and c in G.P., prove that : log an, log bn and log cn are in A.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 30

Question 6.
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 31

Question 7.
If a, b and c are in A.P. a, x, b are in G.P. whereas b, y and c are also in G.P. Show that : x2, b2, y2 are in A.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 32

Question 8.
If a, b, c are in G.P. and a, x, b, y, c are in A.P., prove that :
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 33
Solution 8(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 34

Solution 8(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 35

Question 9.
If a, b and c are in A.P. and also in G.P., show that: a = b = c.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 36

Question 10.
The first term of a G.P. is a and its nth term is b, where n is an even number.If the product of first n numbers of this G.P. is P ; prove that : p2 – (ab)n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 37

Question 11.
If a, b, c and d are consecutive terms of a G.P. ; prove that :
(a2 + b2), (b2 + c2) and (c2 + d2) are in GP.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 38

Question 12.
If a, b, c and d are consecutive terms of a G.P. To prove:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 39
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 40

Geometric Progression Exercise 11D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the sum of G.P. :
(i) 1 + 3 + 9 + 27 + ……….. to 12 terms.
(ii) 0.3 + 0.03 + 0.003 + 0.0003 + …… to 8 terms.
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 41
Solution 1(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 42

Solution 1(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 43

Solution 1(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 44

Solution 1(iv).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 45

Solution 1(v).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 46

Solution 1(vi).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 47

Question 2.
How many terms of the geometric progression 1+4 + 16 + 64 + ……… must be added to get sum equal to 5461?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 48

Question 3.
The first term of a G.P. is 27 and its 8th term is \(\frac{1}{81}\). Find the sum of its first 10 terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 49

Question 4.
A boy spends ₹ 10 on first day, ₹ 20 on second day, ₹ 40 on third day and so on. Find how much, in all, will he spend in 12 days?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 50

Question 5.
The 4th and the 7th terms of a G.P. are \(\frac{1}{27} \text { and } \frac{1}{729}\) respectively. Find the sum of n terms of this G.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 51

Question 6.
A geometric progression has common ratio = 3 and last term = 486. If the sum of its terms is 728 ; find its first term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 52

Question 7.
Find the sum of G.P. : 3, 6, 12, ……………. 1536.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 53

Question 8.
How many terms of the series 2 + 6 + 18 + ………….. must be taken to make the sum equal to 728 ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 54

Question 9.
In a G.P., the ratio between the sum of first three terms and that of the first six terms is 125 : 152.
Find its common ratio.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 54

Question 10.
Find how many terms of G.P. \(\frac{2}{9}-\frac{1}{3}+\frac{1}{2}\) ………. must be added to get the sum equal to \(\frac{55}{72}\)?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 56

Question 11.
If the sum 1 + 2 + 22 + ………. + 2n-1 is 255, find the value of n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 57

Question 12.
Find the geometric mean between :
(i) \(\frac{4}{9} \text { and } \frac{9}{4}\)
(ii) 14 and \(\frac{7}{32}\)
(iii) 2a and 8a3
Solution 12(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 59

Solution 12(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 58

Solution 12(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 60

Question 13.
The sum of three numbers in G.P. is \(\frac{39}{10}\) and their product is 1. Find the numbers.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 61

Question 14.
The first term of a G.P. is -3 and the square of the second term is equal to its 4th term. Find its 7th term.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 62

Question 15.
Find the 5th term of the G.P. \(\frac{5}{2}\), 1, …..
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 63

Question 16.
The first two terms of a G.P. are 125 and 25 respectively. Find the 5th and the 6th terms of the G.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 64

Question 17.
Find the sum of the sequence –\(\frac{1}{3}\), 1, – 3, 9, …………. upto 8 terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 65

Question 18.
The first term of a G.P. in 27. If the 8thterm be \(\frac{1}{81}\), what will be the sum of 10 terms ?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 66

Question 19.
Find a G.P. for which the sum of first two terms is -4 and the fifth term is 4 times the third term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 67

Additional Questions

Question 1.
Find the sum of n terms of the series :
(i) 4 + 44 + 444 + ………
(ii) 0.8 + 0.88 + 0.888 + …………..
Solution 1(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 68

Solution 1(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 69

Question 2.
Find the sum of infinite terms of each of the following geometric progression:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 70
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 71
Solution 2(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 72

Solution 2(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 73

Solution 2(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 74

Solution 2(iv).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 75

Solution 2(v).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 76

Question 3.
The second term of a G.P. is 9 and sum of its infinite terms is 48. Find its first three terms.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 77

Question 4.
Find three geometric means between \(\frac{1}{3}\) and 432.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 78

Question 5.
Find :
(i) two geometric means between 2 and 16
(ii) four geometric means between 3 and 96.
(iii) five geometric means between \(3 \frac{5}{9}\) and \(40 \frac{1}{2}\)
Solution 5(i).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 79

Solution 5(ii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 80.

Solution 5(iii).
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 81

Question 6.
The sum of three numbers in G.P. is \(\frac{39}{10}\) and their product is 1. Find the numbers.
Solution:
Sum of three numbers in G.P. = \(\frac{39}{10}\) and their product = 1
Let number be \(\frac{a}{r}\), a, ar, then
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 82
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 86

Question 7.
Find the numbers in G.P. whose sum is 52 and the sum of whose product in pairs is 624.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 84

Question 8.
The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Geometric Progression - 85

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line

Equation of a Line Exercise 14A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find, which of the following points lie on the line x – 2y + 5 = 0:
(i) (1, 3) (ii) (0, 5)
(iii) (-5, 0) (iv) (5, 5)
(v) (2, -1.5) (vi) (-2, -1.5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 1

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 3

Question 3.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 5

Question 4.
For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?
Solution:
The given equation of the line is 9x + 4y = 3.
Put x = 3 and y = -k, we have:
9(3) + 4(-k) = 3
27 – 4k = 3
4k = 27 – 3 = 24
k = 6

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 7

Question 6.
Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 8

Question 7.
(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 9

Question 8.
(i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.
(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.
Solution:
(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.
Substituting x = -3 and y = 2 in the given equation, we have:
a(-3) + 3(2) + 6 = 0
-3a + 12 = 0
3a = 12
a = 4
(ii) Given, the line y = mx + 8 contains the point (-4, 4).
Substituting x = -4 and y = 4 in the given equation, we have:
4 = -4m + 8
4m = 4
m = 1

Question 9.
The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x – 5y + 15 = 0?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 10

Question 10.
The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2. Does the line x – 2y = 0 contain Q?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 11

Question 11.
Find the point of intersection of the lines:
4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1)x – 2y = 4; find the value of k.
Solution:
Consider the given equations:
4x + 3y = 1 ….(1)
3x – y + 9 = 0 ….(2)
Multiplying (2) with 3, we have:
9x – 3y = -27 ….(3)
Adding (1) and (3), we get,
13x = -26
x = -2
From (2), y = 3x + 9 = -6 + 9 = 3
Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).
The point (-2, 3) lies on the line (2k – 1)x – 2y = 4.
(2k – 1)(-2) – 2(3) = 4
-4k + 2 – 6 = 4
-4k = 8
k = -2

Question 12.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
We know that two or more lines are said to be concurrent if they intersect at a single point.
We first find the point of intersection of the first two lines.
2x + 5y = 1 ….(1)
x – 3y = 6 ….(2)
Multiplying (2) by 2, we get,
2x – 6y = 12 ….(3)
Subtracting (3) from (1), we get,
11y = -11
y = -1
From (2), x = 6 + 3y = 6 – 3 = 3
So, the point of intersection of the first two lines is (3, -1).
If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.
Substituting x = 3 and y = -1, we have:
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 – 5 = 0 = R.H.S.
Thus, (3, -1) also lie on the third line.
Hence, the given lines are concurrent.

Equation of a Line Exercise 14B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 13

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 14
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 15

Question 3.
Find the slope of the line passing through the following pairs of points:
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b) and (b, -a)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 16

Question 4.
Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 17

Question 5.
Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 18

Question 6.
The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 19

Question 7.
The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 20

Question 8.
Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 21

Question 9.
Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 22

Question 10.
(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 23
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 24

Question 11.
Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 25

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 27

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 29

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 30
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 31

Question 15.
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:
(i) the slope of the altitude of AB,
(ii) the slope of the median AD, and
(iii) the slope of the line parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 32

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 33
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 34

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 35
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 36

Question 18.
The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 37

Question 19.
The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 38

Question 20.
Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC?
Justify your conclusion by calculating the slopes of AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 39

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 41

Equation of a Line Exercise 14C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the equation of a line whose:
y-intercept = 2 and slope = 3.
Solution:
Given, y-intercept = c = 2 and slope = m = 3.
Substituting the values of c and m in the equation y = mx + c, we get,
y = 3x + 2, which is the required equation.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 42
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 43

Question 3.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 45

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 141
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 46

Question 5.
Find the equation of the line passing through:
(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 47

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:
(i) the gradient of PQ;
(ii) the equation of PQ;
(iii) the co-ordinates of the point where PQ intersects the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 48

Question 7.
The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find:
(i) the equation of AB;
(ii) the co-ordinates of the point where the line AB intersects the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 49

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 51

Question 9.
In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 52

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 53
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 54

Question 11.
Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 55

Question 12.
In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.
Also, find the equation of the line through vertex B and parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 56

Question 13.
A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 57

Question 14.
Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 58

Question 15.
Find the equation of the line, whose:
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 59

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 142
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 60

Question 17.
Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 61

Question 18.
Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 62

Question 19.
Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 63

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 65

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 67

Question 22.
A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC, Find:
(i) the co-ordinates of the centroid of triangle ABC.
(ii) the equation of a line, through the centroid and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 68

Question 23.
A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP: CP = 2: 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 69

Equation of a Line Exercise 14D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the slope and y-intercept of the line:
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 70

Question 2.
The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 71

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 72
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 73

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 74
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 75

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 76
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 77

Question 6.
(i) Lines 2x – by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 78

Question 7.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 79

Question 8.
The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 80

Question 9.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 81

Question 10.
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 82

Question 11.
The line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y =5. Find the value of b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 83

Question 12.
Find the equation of the line through (-5, 7) and parallel to:
(i) x-axis (ii) y-axis
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 84

Question 13.
(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 85

Question 14.
Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 86

Question 15.
Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 87

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 88
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 89

Question 17.
B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 90

Question 18.
A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 91

Question 19.
A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 92

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 93

Question 21.
The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 94

Question 22.
The point P is the foot of perpendicular from A (-5, 7) to the line whose equation is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP
(ii) the co-ordinates of P
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 95

Question 23.
The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.
If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 96

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 97
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 98

Question 25.
Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 99

Equation of a Line Exercise 14E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.
Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 100

Question 2.
The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y + 4 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 101
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 102
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 103
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 104
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 105
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 106
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 107
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 108
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 109
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 110

Question 3.
A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Solution:

Question 5.
Solution:

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:

Question 9.
A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 111

Question 11.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 112
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 113

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of triangle OAB through vertex O.
(ii) the equation of altitude of triangle OAB through vertex B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 114

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does the line 3x = y + 1 bisect the line segment joining the two given points?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 115

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 143Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 116

Question 15.
Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 117

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 118

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 119
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 120

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 121

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 122

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
(i) the co-ordinates of A and B.
(ii) the equation of line through P and perpendicular to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 123

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 124

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 125

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 126
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 127

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 146
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 147

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 157
Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find:
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ if (-2, p) lies on it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 149

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 150
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 158
Question 28.
The equation of a line 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 151
Question 29.
ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find:
(i) Co-ordinates of A
(ii) Equation of diagonal BD
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 152

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 153
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 154
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 155

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 156
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 128

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 129

Question 33.
A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid- t point of the line segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 130

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 131
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 132

Question 35.
A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 133

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 134

Question 37.
Find the equation of the line through the Points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 135

Question 38.
Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2), find :
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 136

Question 39.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 137
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 138

Question 40.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 139

Question 41.

i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).
Since B lies on the Y-axis, let the co-ordinates of B be (0, y).
Let m = 1 and n = 2
Using Section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 140
⇒ Slope of line perpendicular to AB = m = -2
P = (4, -1)
Thus, the required equation is
y – y1 = m(x – x1)
⇒ y – (-1) = -2(x – 4)
⇒ y + 1 = -2x + 8
⇒ 2x + y = 7

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Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 9 Electrical Power and Household Circuits. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Physics Chapter 9 Electrical Power and Household Circuits

Solution 1.

The electric power is generated at 11 KV, 50Hz at the power generating station.

Solution 2.

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 1
At a power generating station, the electric power is generated at 11 kV. From here, the alternating voltage is transmitted to the grid sub-station and stepped up to 132 kV using a step-up transformer. It is then transmitted to the main sub-station where the voltage is stepped down to 33 kV using a step-down transformer and is then transmitted to the intermediate sub-station. At the intermediate sub-station, the voltage is stepped down to 11 kV using a step-down transformer and is transmitted to the city sub-station, where the voltage is further stepped down to 220 V and is supplied to our houses.

Solution 3.

Electric power from the generating station is transmitted at 11 kV because voltage higher than this causes insulation difficulties, while the voltage lower than this involves high current and loss of energy in form of heat (I2Rt).

Solution 4.

At 220 V of voltage and 50 Hz of frequency, the a.c. is supplied to our houses.

Solution 5.

(a) Step-up transformer
(b) Step-down transformer

Solution 6.

(a) The three connecting wires used in a household circuit are:

  1. Live (or phase) wire (L),
  2. Neutral wire (N), and
  3. Earth wire (E).

(b) Among them neutral and earth wires are at the same potential.
(c) The switch is connected in the live wire.

Solution 7.

Before the electric line is connected to the meter in a house, a fuse of rating (≈ 50 A) is connected in the live wire at the pole or just before the meter. This fuse is called the pole fuse.
Its current rating is ≈ 50 A.

Solution 8.

  1. After the company fuse, the cable is connected to a kWh meter and from this meter; connections are made to the distribution board through a main fuse and a main switch.
  2. Main fuse is connected in the live wire and in case of high current it gets burnt and cut the connections to save appliances.
  3. Main switch is connected in the live and neutral wires. It is used to cut the connections of the live as well as the neutral wires simultaneously from the main supply.

Solution 9.

The electric meter in a house measures the electrical energy consumed in kWh.
Its value in S.I. unit is 1kWh = 3.6 x 106J.

Solution 10.

The main fuse in a house circuit is connected on the distribution board, in live wire before the main switch.

Solution 12.

Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 2
Advantages of ring system over tree system

  1. In a ring system the wiring is cheaper than tree system.
  2. In ring system the sockets and plugs of same size can be used while in a tree system sockets and plugs are of different size.
  3. In ring system, each appliance has a separate fuse due to which if there is a fault and the fuse of one appliance burns it does not affect other appliances; while in a tree system when fuse in one distribution line blows, it disconnects all the appliances connected to that distribution circuit.

Solution 13.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 3

Solution 14.

All the electrical appliances in a building should be connected in parallel at the mains, each with a separate switch and a separate fuse connected in the live wire so that the switching on or off in a room has no effect on other lamps in the same building.

Solution 16.

In set A, the bulbs are connected in series. Thus, when the fuse of one bulb blows off, the circuit gets broken and current does not flow through the other bulbs also.
In set B, the bulbs are connected in parallel. Thus, each bulb gets connected to its voltage rating (= 220 V) and even when the fuse of one bulb blows off, others remain unaffected and continue to glow.

Solution 1 (MCQ).

The main fuse is connected in live wire.
Hint: The main fuse is connected in live wire so that if the current exceeds its rating, the fuse melts and breaks the circuit; thus, preventing the excessive current from flowing into the circuit.

Solution 2 (MCQ).

Electrical appliances in a house are connected in parallel.
Hint: On connecting the electrical appliances in parallel, each appliance works independently without being affected whether the other appliance is switched on or off.

Solution 3 (MCQ).

Energy
Hint: The electric meter in a house records the amount of electrical energy consumed in a house.

Exercise 9(B)

Solution 1.

An electric fuse is a safety device, which is used to limit the current in an electric circuit. The use of fuse safeguards the circuit and appliances connected in that circuit from being damaged.
An alloy of lead and tin is used as a material of fuse because it has low melting point and high resistivity.

Solution 2.

‘Fuse’ is used to protect electric circuits from overloading and short circuiting. It works on heating effect of current.

Solution 3.

(a) A fuse is a short piece of wire of material of high resistance and low melting point.
(b) A fuse wire is made of an alloy of lead and tin. If the current in a circuit rises too high, the fuse wiremelts
(c) A fuse is connected in series with the live wire.
(d) Higher the current rating, Thicker is the fuse wire.

Solution 4.

The fuse wire is fitted in a porcelain casing because porcelain is an insulator of electricity.

Solution 5.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 4

Solution 6.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 5

Solution 7.

The fuse wire is always connected in the live wire of the circuit because if the fuse is put in the neutral wire, then due to excessive flow of current when the fuse burns, current stops flowing in the circuit, but the appliance remains connected to the high potential point of the supply through the live wire. Now if a person touches the appliance, he may get a shock as the person will come in contact with the live wire through the appliance.

Solution 8.

The 20 A fuse wire will be thicker so that its resistance be low.

Solution 9.

It means that the line to which this fuse is connected has a current carrying capacity of 5 A.

Solution 10.

The safe limit of current which can flow through the electrical appliance is I = P/V = 5000/200 = 25 A; which is greater than 8 A. So, such fuse cannot be used.

Solution 11.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 16

Solution 12.

A switch is an on-off device for current in a circuit (or in an appliance). The switch should always be connected in the live wire so that the appliance could be connected to the high potential point through the live wire. In this position the circuit is complete as the neutral wire provides the return path for the current. When the appliance does not work i.e., in off position of the switch, the circuit is incomplete and no current reaches the appliance.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 6
On the other hand, if switch is connected in the neutral wire, then in ‘off’ position, no current passes through the bulb. But the appliance remains connected to the high potential terminal through the live wire.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 7
Thus, if the switch is connected in the neutral wire, it can be quite deceptive and even dangerous for the user.
Precaution while handling a switch: A switch should not be touched with wet hands.

Solution 13.

A switch should not be touched with wet hands. If water reaches the live wire, it forms a conducting layer between the hand and the live wire of the switch through which the current passes to the hand and the person may get a fatal shock.

Solution 14.

Let a switch S1 be fitted at the bottom and a switch S2 at the top of the staircase. Fig. (a) shows the off position of the bulb.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 8
The bulb can now be switched on independently by either the switch S1 or the switch S2. If the switch S1 is operated, the connection ‘ab’ is changed to ‘bc’, which completes the circuit and the bulb lights up [Fig. (b)].
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 9

Similarly, on operating the switch S2, the connection ‘bc’ changes to ‘ba’, which again completes the circuit [Fig. (c)].
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 10
Similarly if the bulb is in on position as shown in Fig. (b) or (c), one can switch off the bulb either from the switch S1 or the switch S2.

Solution 15.

All electrical appliances are provided with a cable having a plug at one end to connect the appliance to the electric supply.
In this three way pin plug, the top pin is for earthing (E), the live pin (L) in on the left and the neutral pin (N) is on the right.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 11

Solution 16.

The three pins in the plug are labelled as
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 12
Here E signifies the earth pin,
L is for live wire, and
N is for neutral wire.

  1. The earth pin is made long so that the earth connection is made first. This ensures the safety of the user because if the appliance is defective, the fuse will blow off. The earth pin is thicker so that even by mistake it cannot be inserted into the hole for the live or neutral connection of the socket.
  2. The pins are splitted at the end to provide spring action so that they fit in the socket holes tightly.

Solution 17.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 13

Solution 18.

(a) 1 – Earth, 2 – Neutral, 3 – Live
(b) Terminal 1 is connected to the outer metallic case of the appliance.
(c) The fuse is connected to live wire joined to 3 so that in case of excessive flow of current fuse melts first and breaks down the circuit to protect appliances.

Solution 19.

Local earthing is made near kWh meter. In this process a 2 – 3 metre deep hole is dug in the ground. A copper rod placed inside a hollow insulating pipe, is put in the hole. A thick copper plate of dimensionsis 50 cm x 50 cm welded to the lower end of the copper rod and it is buried in the ground. The plate is surrounded by a mixture of charcoal and salt to make a good earth connection.
To keep the ground damp, water is poured through the pipe from time to time. This forms a conducting layer between the plate and the ground. The upper end of the copper rod is joined to the earth connection at the kWh meter.

Solution 20.

If the live wire of a faulty appliance comes in to direct contact with the metallic case due to some reason then the appliance acquires the high potential of live wire. This may results in shock if any person touches the body of appliance. But if the appliance is earthed then as soon as the live wire comes in to contact with the metallic case, high current flows through the case to the earth. The fuse connected to the appliance will also blows off, so the appliance get disconnected.

Solution 21.

(a) The fuse must be connected in the live wire only. If the fuse is in the neutral wire, then although the fuse burns due to the flow of heavy current, but the appliance remains at the supply voltage so that on touching the appliance current flows through the appliance to the person touching it.
(b) Metallic case of the appliance should be earthed.

Solution 22.

The paint provides an insulating layer on the metal body of the appliance. To make earth connection therefore, the paint must be removed from the body part where connection is to be made.

Solution 23.

  1. According to new international convention
  2. Live wire is brown in colour.
  3. Neutral is light blue and
  4. Earth wire is yellow or green in colour.

Solution 25.

(a) The three wires are: Live wire, Earth wire and Neutral wire.
(b) The heating element of geyser should be connected to live wire and neutral wire.
(c) The metal case should be connected to earth wire.
(d) The switch and fuse should be connected to live wire.

Solution 26.

One may get an electric shock from an electrical gadget in the following two cases:

  1. If the fuse is put in the neutral wire instead of live wire and due to fault, if an excessive current flows in the circuit, the fuse burns, current stops flowing in the circuit but the appliance remains connected to the high potential point of the supply through the live wire. In this situation, if a person touches the faulty appliance, he may get an electric shock as the person will come in contact with the live wire through the appliance.
    Preventive measure: The fuse must always be connected in the live wire.
  2. When the live wire of a faulty appliance comes in direct contact with its metallic case due to break of insulation after constant use (or otherwise), the appliance acquires the high potential of the live wire. A person touching it will get a shock because current flows through his body to earth.
    Preventive measure: Proper ‘earthing’ of the electric appliance should be done.

Solution 27.

Power circuit carries high power and costly devices. If there is some unwanted power signal (noise) in the wire it can damage the device. To reduce this effect earth is necessary.
Lighting circuit carries low power (current).So, we ignore the earth terminal.

Solution 28.

A high tension wire has a low resistance and large surface area.

Solution 29.

To carry larger current, the resistance of the wire should be low, so its area of cross section should be large. Therefore 15 A current rated wire will be thicker.

Solution 30.

(a) Switches 2 and 3.
(b) The lamps are connected in series.

Solution 31.
(a)
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 14
(b)

Wire no. Wire name Colour (Old convention) Colour (New convention)
1 Neutral wire Black Light blue
2 Earth wire Green Green or yellow
3 Live wire Red Brown

(c) The bulbs are joined in parallel.

Solution 1 (MCQ).

5 A
Hint: The electric wiring for light and fan circuit uses a thin fuse of low current rating (= 5 A) because the line wire has a current carrying capacity of 5 A.

Solution 2 (MCQ).

A switch must be connected in live wire.
Explanation: A switch must be connected in live wire, so that when it is in ‘off’ position, the circuit is incomplete and no current reaches the appliance through the live wire.
Selina Concise Physics Class 10 ICSE Solutions Electrical Power and Household Circuits img 15

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Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

Selina Concise Biology Class 10 ICSE Solutions The Reproductive System

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Selina ICSE Solutions for Class 10 Biology Chapter 11 The Reproductive System

Exercise 1

Solution A.1.
(d) Epididymis → vas deferens → urethra

Solution A.2.
(d) 28 days

Solution A.3.
(d) About seven days

Solution B.1.
(a) Scrotum
(b) Seminiferous Tubules
(c) GraafianFollicle
(d) Seminal vesicle
(e) Epididymis

Solution B.2.
(a) Testosterone
(b) Ureter
(c) Ovum
(d) After birth

Solution B.3.
(a) Testes → Sperms → Sperm duct → Semen → Penis
(b) Menarche Puberty → Reproductive age → Menstruals → Menopause
(c) Graafian follicle → Ostium → Fallopian tube → Uterus

Solution B.4.
Seminiferous tubule → Epididymis → Vas deferens → Penis

Solution C.1.
Semen is the mixture of sperms and secretions from seminal vesicles, prostate gland and Cowper’s (bulbo-urethral gland).

Solution C.2.
(a) Inguinal canal: It is the canal which allows the descent of testes along with their ducts, blood vessels and nerves into the abdomen.
(b) Prostate gland: It is a bilobed structure which surrounds the urethra and pours an alkaline secretion into the semen.
(c) Testis: Testis is a male reproductive organ. There a pair of testes present in a scrotal sac descended outside the body cavity. Testes produce sperms which are the male gametes.
(d) Ovary: Ovary is a female reproductive organ.It produces ova i.e. female gametes.
(e) Oviduct: A pair of oviduct is present on either side of the uterus. Oviduct carries the released ovum from the ovary to the uterus.

Solution C.3.
Secondary sexual characters in males:

  1. Beard and moustache
  2. Stronger muscular built
  3. Deeper voice

Secondary sexual characters in females:

  1. Breasts in females
  2. Large hips
  3. High pitched voice

Solution C.4.
The accessory reproductive organs include all those structures which help in the transfer and meeting of two kinds of sex cells leading to fertilization and growth and development of egg up to the birth of the baby.
For example: uterus in females, penis in males.

Solution C.5.

Primary Reproductive Organs

Accessory Reproductive Organs
The primary reproductive organs produce sex cells.

The accessory reproductive organs help in the transfer and meeting of two kinds of sex cells leading to fertilization.

The primary reproductive organs do not help in the development of baby.

The accessory organs help in the growth and development of egg up to the birth of baby.
Example: Testes in males and ovaries in females.

Example: penis in males, Uterus, vagina in female.

Solution C.6.
Hymen is a thin membrane which partially covers the opening of the vagina in young females.

Solution C.7.
(a) Hernia: It is an abnormal condition which is caused when the intestine due to the pressure in abdomen bulges into the scrotum through the inguinal canal.
(b) Ovulation: It is the release of the mature ovum by the rupture of the Graafian follicle.
(c) Puberty: It is the period during which immature reproductive system in boys and girls matures and becomes capable of reproduction.

Solution C.8.
Changes in human male:

  1. Development of Beard and moustache
  2. Voice becomes deeper

Changes in human female:

  1. Development of Breasts in females
  2. Development of high pitched voice

Solution C.9.
(a) Menarche is the onset of menstruation in young females at about 13 years of age whereas menopause is the permanent stoppage of menstruation at about 45 years of age.

(b) Cowper’s gland opens into urethra in human males and its secretion serves as a lubricant whereas the prostate gland surrounds the urethra in males and its alkaline secretion neutralizes acid in female’s vagina.

(c) Hymen is a thin membrane that partially covers the opening of vagina in young females whereas clitoris is a small erectile structure located in the uppermost angle of vulva in front of the urethral opening.

(d) Uterus is a hollow, pear shaped muscular organ located in the pelvic cavity. It is the site of implantation for the embryo after fertilisation whereas the vagina is the muscular tube extending from the cervix to the outside. At the time of sexual intercourse, the vagina receives the male penis and provides entry for the sperms.

(e) Efferent ducts join to form the epididymis whereas the epididymis is continued by the side of the testes to give rise to the sperm duct or vas deferens.

Solution D.1.

  • Testes are responsible for the production of male gametes i.e. sperms. The normal body temperature does not allow the maturation of the sperms. Being suspended outside the body cavity, the temperature in the scrotal sac is 2 to 3oC which is the suitable temperature for the maturation of the sperms.
  • When it is too hot, the skin of the scrotum loosens so that the testes hang down away from the body. When it is too cold, the skin contracts in a folded manner and draws the testes closer to the body for warmth.
  • In an abnormal condition, in the embryonic stage, the testes do not descend into the scrotum. It can lead to sterility or incapability to produce sperms.

Solution D.2.
Testosterone is the male reproductive hormone produced by the interstitial cells or the Leydig cells. These cells are located in the testes. They serve as a packing tissue between the coils of the seminiferous tubules. Therefore, it can be said that the testes produce the male hormone testosterone.

Solution D.3.
otal reproductive period = 45 – 13 = 32 years
Total eggs produced = 32 x 12 = 384 eggs approximately

Solution E.1.
(a) Excretory system and Female Reproductive system

(b)

  1. Kidney
  2. Ureter
  3. Fallopian Tube
  4. Infundibulum
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Cervix
  9. Vagina
  10. Vulva

(c)

  • Function of Fallopian Tube (part 3): The fallopian tubes carry the ovum released from the ovary to the uterus.
  • Function of Infundibulum (part 4): Infundibulum is the funnel shaped distal end of the ovary which picks up the released ovum and pushes it further on its passage into the fallopian tube.
  • Function of Ovary (part 5): Ovary produces female gametes i.e. ova.
  • Function of Uterus (part 6): Uterus allows the growth and development of the embryo.

Solution E.2.
(a)

  1. Fallopian Tube
  2. Infundibulum
  3. Ureter
  4. Vagina
  5. Ovary
  6. Uterus
  7. Urinary Bladder
  8. Urethra

(b) Oestrogen secreted by the corpus luteum secrets oestrogen. Oestrogen stimulates the thickening of the endometrial wall of the uterus. The uterine wall becomes thickened and is supplied with a lot of blood to receive the fertilized egg.
(c) If fertilization fails to take place, the endometrial lining of the uterus starts shedding on the 28th day of the menstrual cycle. Finally it is discharged out along with the unfertilised ovum as the menstrual flow.

Solution E.3.
a.

  1. Seminal vesicles
  2. Prostate gland
  3. Bulbo-urethral gland
  4. Epididymis
  5. Testis
  6. Scrotum
  7. Urinary bladder
  8. Vas deferens
  9. Erectile tissue
  10. Penis
  11. Urethra

b. Functions of

  1. Seminal vesicles
    They produce the fluid which serves as the transporting medium for sperms.
  2. Prostate gland
    It produces an alkaline secretion which mixes with the semen and helps neutralise the vaginal acids.
  3. Bulbo-urethral gland
    It produces a secretion which serves as a lubricant for the semen to pass through the urethra.
  4. Testis
    It produces the male gamete sperm and the male sex hormone testosterone.
  5. Vas deferens
    They carry the sperms from the epididymis to the urethra.
  6. Urethra
    It serves as an outlet for delivering the sperms into the vagina.

Exercise 2

Solution A.1.
(c) fallopian tube

Solution A.2.
(a) energy

Solution A.3.
(c) 280 days

Solution B.1.
(a) Amniotic fluid
(b) Uterus
(c) Amniotic membrane
(d) Inguinal canal

Solution B.2.
(a) Sperm
(b) Follicle

Solution B.3.
(a) Ovulation → fertilization → implantation → gestation → child birth
(b) Sperm → sperm duct → urethra → coitus → vagina → ovum

Solution B.4.
(a) Menarche
(b) Ovulation
(c) Menstruation
(d) Fertilization
(e) Implantation

Solution B.5.

Column I

Column II
(a) Acrosome

(v) spermatozoa

(b) Gestation

(vii) Time taken by a fertilized egg till the delivery of baby
(c) Menopause

(vi) complete stoppage of menstrual cycle

(d) Foetus

(i) An embryo which looks like human baby
(e) Oogenesis

(iii) ovum producing cells

(f) Ovulation

(ii) Luteinizing hormone

 

Solution C.1.
(a)

  1. False
  2. False
  3. False
  4. False

(b)

  1. Fertilization occurs in the fallopian tube.
  2. Vagina is also known as the birth canal.
  3. Nutrition and oxygen diffuse from the mother’s blood into the foetus’s blood through placenta.
  4. Gestation period in humans is about 280 days.

Solution C.2.

Structure

Function
1. Corpus luteum

1. secretes progesterone & other hormones to prepare the uterine wall for the receival of the embryo.

2. Testes

2. produces male gametes in mass
3. Placental disc

3. supplies oxygen and nutrients to embryo

4. Oxytocin

4. increases the force in uterine contractions during child birth
5. Umbilical cord

5. connects placenta with foetus

6. Fallopian tube

6. The site of fertilization for the sperm and ovum

Solution C.3.

(a) Foetus:

  1. It is contained in the uterus.
  2. In foetus, limbs have appeared and resembles the humans unlike the embryo which is a growing or dividing zygote.

(b) Hyaluronidase:

  1. Enzyme
  2. It is an enzyme secreted by the sperm that allows the sperm to penetrate the egg.

(c) Morula:
It is the stage in the development of human embryo which consists of a spherical mass of cells. Blastocyst

(d) Amniotic fluid:

  1. Between amnion and embryo
  2. It protects the embryo from physical damage, keeps the pressure all around embryo and prevents sticking of foetus to amnion.

(e) Gestation:
Gestation is the full term of the development of an embryo in the uterus. 280 days in humans.

(f) Placenta:

  1. Placenta is formed by two sets of minute finger like processes called the villi. One set of villi is from the uterine wall and the other set is from the allantois.
  2. Oxygen and amino acids.
  3. Progesterone and oestrogen.

(g) Implantation:

  1. Blastocyst
  2. It occurs in about 5-7 days after ovulation.

Solution D.1.

(a) Sperm is the male gamete produced by the testes. Semen on the other hand is the mixture of sperms and alkaline secretions from the seminal vesicle, prostate gland and Cowper’s gland.

(b) Implantation is the fixing of embryo in the wall of uterus. The state that implantation produces is known as pregnancy.

(c) Follicle is the cellular sac containing a maturing egg. Corpus luteum on the other hand is the remnant of the follicle the release of ovum during ovulation.

(d) Amnion is a sac which develops around the embryo whereas allantois is an extension from the embryo which forms villi of placenta.

(e) Sterility is the incapability to produce sperms whereas impotency is the inability to copulate.

(f) Prostate gland pours alkaline secretions into the semen to neutralize the acid in female’s vagina whereas the secretion of Cowper’s gland serves as a lubricant.

(g) Identical twins are produced from one ovum i.e. one developing zygote splits and grows into two foetuses whereas fraternal twins are produced when two ova get fertilized at a time.

Solution D.2.

  1. After fertilization zygote is formed inside the fallopian tube.
  2. The zygote then divides repeatedly to form a spherical mass of cells known as ‘Morula’.
  3. The morula then develops into a hollow sphere of cells with a surrounding cellular layer and an inner cell mass projecting from it centrally. This stage is known as the ‘blastocyst’. It implants itself into the uterine wall.
    Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -1
  4. From the blastocyst arises an embryo which is around 3 weeks old. It is a tiny organism that hardly resembles human being.
  5. By the end of 5 weeks, the embryo is with a develoed heart and blood vessels.
  6. By the end of 8 weeks, limbs are developed. This stage is known as ‘foetus’.
  7. At the end of nearly 40 weeks i.e. end of gestation period, the infant is born.

Solution D.3.
(a) Amnion:

  1. Amnion contains the amniotic fluid which surrounds the embryo.
  2. This fluid protects the embryo from physical damage.
  3. It maintains even pressure all around the embryo.
  4. It also prevents sticking of foetus to amnion.

(b) Placenta:

  1. The placenta allows the diffusion of oxygen and nutrients such as glucose, vitamins and amino acids from mother to foetus.
  2. Similarly, it also allows the diffusion of carbon dioxide, urea and waste products from foetus to mother.
  3. Placenta also acts as an endocrine tissue. It secretes oestrogen and progesterone.

Solution E.1.
a. A – ovum
B – sperm

b. Sperms are produced in the testis.
The ovum is produced in the ovary.

c. The reproductive cells unite in the fallopian tubes of the female reproductive system.

d. Ovary – Oestrogen and progesterone
Testis – Testosterone

e. Accessory glands:

  • Seminal vesicle – Seminal fluid
  • Prostate gland – Alkaline secretion
  • Bulbo-urethral gland – Lubricant

Solution E.2.
(a)

  1. umbilical cord,
  2. placenta,
  3. amnion,
  4. mouth of uterus,
  5. muscular wall of uterus

(b) Gestation
(c) 280 days
(d) Placenta provides the foetus with oxygen and nutrients. In addition, the placenta also removes carbon dioxide and waste products of the foetus.
(e) Progesterone

Solution E.3.

Selina Concise Biology Class 10 ICSE Solutions The Reproductive System image -2

Solution E.4.
(a) A – Muscular wall of uterus,
B – Oviduct,
C – Ovary,
D – Cervix
(b) If part B will get blocked, ovum released from the ovary will not get fertilized by the sperm and hence pregnancy will be prevented.

Solution E.5.

  1. Prostate gland
  2. Bulbo-urethral gland
  3. Urethra
  4. Vas deferens
  5. Testis

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Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression

Arithmetic Progression Exercise 10A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 2

Question 2.
The nth term of sequence is (2n – 3), find its fifteenth term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 3

Question 3.
If the pth term of an A.P. is (2p + 3), find the A.P.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 4

Question 4.
Find the 24th term of the sequence:
12, 10, 8, 6,……
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 5

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 6
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 7

Question 6.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 8
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 9

Question 7.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 10
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 11

Question 8.
Is 402 a term of the sequence :
8, 13, 18, 23,………….?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 12

Question 9.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 13
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 14

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 15
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 16
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 17

Question 11.
Which term of the A.P. 1 + 4 + 7 + 10 + ………. is 52?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 18

Question 12.
If 5th and 6th terms of an A.P are respectively 6 and 5. Find the 11th term of the A.P
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 19

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 85
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 20

Question 14.
Find the 10th term from the end of the A.P. 4, 9, 14,…….., 254
Solution:
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Question 15.
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Solution:
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Question 16.
Find the 31st term of an A.P whose 10th term is 38 and 10th term is 74.
Solution:
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Question 17.
Which term of the services :
21, 18, 15, …………. is – 81?
Can any term of this series be zero? If yes find the number of term.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 24

Question 18.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31st term.
Solution:
For a given A.P.,
Number of terms, n = 60
First term, a = 7
Last term, l = 125
⇒ t60 = 125
⇒ a + 59d = 125
⇒ 7 + 59d = 125
⇒ 59d = 118
⇒ d = 2
Hence, t31 = a + 30d = 7 + 30(2) = 7 + 60 = 67

Question 19.
The sum of the 4th and the 8th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
t4 + t8 = 24 (given)
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….(i)
And,
t6 + t10 = 34 (given)
⇒ (a + 5d) + (a + 9d) = 34
⇒ 2a + 14d = 34
⇒ a + 7d = 17 ….(ii)
Subtracting (i) from (ii), we get
2d = 5

Question 20.
If the third term of an A.P. is 5 and the seventh terms is 9, find the 17th term.
Solution:

Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t3 = 5 (given)
⇒ a + 2d = 5 ….(i)
And,
t7 = 9 (given)
⇒ a + 6d = 9 ….(ii)
Subtracting (i) from (ii), we get
4d = 4
⇒ d = 1
⇒ a + 2(1) = 5
⇒ a = 3
Hence, 17th term = t17 = a + 16d = 3 + 16(1) = 19

Arithmetic Progression Exercise 10B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Solution:
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Question 2.
How many two-digit numbers are divisible by 3?
Solution:
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Question 3.
Which term of A.P. 5, 15, 25 ………… will be 130 more than its 31st term?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 27

Question 4.
Find the value of p, if x, 2x + p and 3x + 6 are in A.P
Solution:
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Question 5.
If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?
Solution:
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Question 6.
How many three-digit numbers are divisible by 87?
Solution:
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Question 7.
For what value of n, the nth term of A.P 63, 65, 67, …….. and nth term of A.P. 3, 10, 17,…….. are equal to each other?
Solution:
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Question 8.
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
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Question 9.
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Solution:
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Question 10.
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Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 34

Question 11.
If a, b and c are in A.P show that:
(i) 4a, 4b and 4c are in A.P
(ii) a + 4, b + 4 and c + 4 are in A.P.
Solution:
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Question 12.
An A.P consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution:
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Question 13.
4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd time by I. Find the first term and the common difference.
Solution:
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Question 14.
The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P
Solution:
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Question 15.
In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n – r)
Solution:
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Question 16.
Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13th term?
Solution:
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Arithmetic Progression Exercise 10C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the sum of the first 22 terms of the A.P.: 8, 3, -2, ………..
Solution:
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Question 2.
How many terms of the A.P. :
24, 21, 18, ……… must be taken so that their sum is 78?
Solution:
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Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
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Question 4(i).
Find the sum of all odd natural numbers less than 50
Solution:
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Question 4(ii).
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 46

Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 47

Question 6.
The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
Solution:
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Question 7.
The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
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Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 50

Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
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Question 10.
In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.
Solution:
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Question 11.
If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Solution:
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Question 12.
Find the sum of all multiples of 7 between 300 and 700.
Solution:
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Question 13.
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Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 55

Question 14.
The fourth term of an A.P. is 11 and the term exceeds twice the fourth term by 5 the A.P and the sum of first 50 terms
Solution:
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Arithmetic Progression Exercise 10D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
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Question 2.
The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.
Solution:
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Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 59.

Question 4.
Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Solution:
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Question 5.
Find five numbers in A.P. whose sum is \(12 \frac{1}{2}\) and the ratio of the first to the last terms is 2: 3.
Solution:
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Question 6.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Solution:
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Question 7.
The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers.
Solution:
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Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
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Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 65

Question 10.
The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Solution:
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Question 11.
\(\frac{1}{a}, \frac{1}{b} \text { and } \frac{1}{c}\) are in A.P. Show that : be, ca and ab are also in A.P.
Solution:
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Question 12.
Insert four A.M.s between 14 and -1.
Solution:
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Question 13.
Insert five A.M.s between -12 and 8.
Solution:
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Question 14.
Insert six A.M.s between 15 and -15.
Solution:
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Arithmetic Progression Exercise 10E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km hr-1 The second car goes at a speed of 8 km h-1 in the first hour and thereafter increasing the speed by 0.5 km h-1 each succeeding hour. After how many hours will the two cars meet?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 71

Question 2.
A sum of ₹ 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is ₹ 20 less than its preceding prize; find the value of each of the prizes.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 72

Question 3.
An article can be bought by paying ₹ 28,000 at once or by making 12 monthly instalments. If the first instalment paid is ₹ 3,000 and every other instalment is ₹ 100 less than the previous one, find :
(i) amount of instalment paid in the 9th month
(ii) total amount paid in the instalment scheme.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 73

Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7th year.
Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10th year.
(iii) the total production in 7 years.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 74

Question 5.
Mrs. Gupta repays her total loan of ₹ 1.18,000 by paying instalments every month. If the instalment for the first month is ₹ 1,000 and it increases by ₹ 100 every month, what amount will she pay as the 30th instalment of loan? What amount of loan she still has to pay after the 30th instalment?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 75

Question 6.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 76

Arithmetic Progression Exercise 10F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The 6th term of an A.P. is 16 and the 14th term is 32. Determine the 36th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t6 = 16 (given)
⇒ a + 5d = 16 ….(i)
And,
t14 = 32 (given)
⇒ a + 13d = 32 ….(ii)
Subtracting (i) from (ii), we get
8d = 16
⇒ d = 2
⇒ a + 5(2) = 16
⇒ a = 6
Hence, 36th term = t36 = a + 35d = 6 + 35(2) = 76

Question 2.
If the third and the 9th terms of an A.P. term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
For an A.P.,a
t3 = 4
⇒ a + 2d = 4 … (i)
t9 = -8
⇒ a + 8d = -8 …. (ii)
Subtracting (i) from (ii), we get
6d = -12
⇒ d = -2
Substituting d = -2 in (i), we get
a = 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
⇒ General term = tn = 8 + (n – 1)(-2)
Let pth term of this A.P. be 0.
⇒ 8 + (0 – 1) (-2) = 0
⇒ 8 – 2p + 2 = 0
⇒ 10 – 2p = 0
⇒ 2p = 10
⇒ p = 5
Thus, 5th term of this A.P. is 0.

Question 3.
An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term of the A.P.
Solution:
For a given A.P.,
Number of terms, n = 50
3rd term, t3 = 12
⇒ a + 2d = 12 ….(i)
Last term, l = 106
⇒ t50 = 106
⇒ a + 49d = 106 ….(ii)
Subtracting (i) from (ii), we get
47d = 94
⇒ d = 2
⇒ a + 2(2) = 12
⇒ a = 8
Hence, t29 = a + 28d = 8 + 28(2) = 8 + 56 = 64

Question 4.
Find the arithmetic mean of :
(i) -5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)2 and (m – n)2
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 77

Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 + ………
Solution:
Here,
First term, a = 4
Common difference, d = 6 – 4 = 2
n = 10
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 78

Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Here,
First term, a = 3
Last term, l = 57
n = 20
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 79

Question 7.
How many terms of the series 18 + 15 + 12 + ……. when added together will give 45 ?
Solution:
Here, we find that
15 – 18 = 12 – 15 = -3
Thus, the given series is an A.P. with first term 18 and common difference -3.
Let the number of term to be added be ‘n’.
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 80
⇒ 90 = n[36 – 3n + 3]
⇒ 90 = n[39 – 3n]
⇒ 90 = 3n[13 – n]
⇒ 30 = 13n – n2
⇒ n2 – 13n + 30 = 0
⇒ n2 – 10n – 3n + 30 = 0
⇒ n(n – 10) – 3(n – 10) = 0
⇒ (n – 10)(n – 3) = 0
⇒ n – 10 = 0 or n – 3 = 0
⇒ n = 10 or n = 3
Thus, required number of term to be added is 3 or 10.

Question 8.
The nth term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
tn = 8 – 5n
Replacing n by (n + 1), we get
tn+1 = 8 – 5(n + 1) = 8 – 5n – 5 = 3 – 5n
Now,
tn+1 – tn = (3 – 5n) – (8 – 5n) = -5
Since, (tn+1 – t2) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.

Question 9.
The the general term (nth term) and 23rd term of the sequence 3, 1, -1, -3, ……
Solution:
The given sequence is 1, -1, -3, …..
Now,
1 – 3 = -1 – 1 = -3 – (-1) = -2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.
The general term (nth term) of an A.P. is given by
tn = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 – 2n + 2
= 5 – 2n
Hence, 23rd term = t23 = 5 – 2(23) = 5 – 46 = -41

Question 10.
Which term of the sequence 3, 8, 13, …….. is 78 ?
Solution:
The given sequence is 3, 8, 13, …..
Now,
8 – 3 = 13 – 8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.
Let the nth term of the given A.P. be 78.
⇒ 78 = 3 + (n – 1)(5)
⇒ 75 = 5n – 5
⇒ 5n = 80
⇒ n = 16
Thus, the 16th term of the given sequence is 78.

Question 11.
Is -150 a term of 11, 8, 5, 2, ……… ?
Solution:
The given sequence is 11, 8, 5, 2, …..
Now,
8 – 11 = 5 – 8 = 2 – 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.
The general term of an A.P. is given by
tn = a + (n – 1)d
⇒ -150 = 11 + (n – 1)(-5)
⇒ -161 = -5n + 5
⇒ 5n = 166
⇒ n =\(\frac{166}{5}\)
The number of terms cannot be a fraction.
So, clearly, -150 is not a term of the given sequence.

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
The two-digit numbers divisible by 3 are as follows: 12, 15, 18, 21, …….. 99
Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = nth term= 99
The general term of an A.P. is given by
tn = a + (n – 1)d
⇒ 99 – 12 + (n – 1)(3)
⇒ 99 – 12 + 3n-3
⇒ 90 – 3n
⇒ n = 30
Thus, 30 two-digit numbers are divisible by 3.

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Numbers between 10 and 250 which are multiple of 4 are as follows: 12, 16, 20, 24,……, 248
Clearly, this forms an A.P. with first term a = 12,
common difference d= 4 and last term l = 248
l – a + (n – 1)d
⇒ 248 – 12 + (n – 1) × 4
⇒ 236 – (n – 1) × 4
⇒ n – 1 = 59
⇒ n = 60
Thus, 60 multiples of 4 lie between 10 and 250.

Question 14.
The sum of the 4th term and the 8th term of an A.P. is 24 and the sum of 6th term and the 10th term is 44. Find the first three terms of the A.P.
Solution:
Given, t4 + t8 = 24
(a + 3d) + (a + 7d) = 24
= 2a+ 10d = 24
> a + 5d = 12 ….(i)
And,
t62 + t10 = 44
= (a + 5d) + (a + 9d) = 44
= 2a+ 14d = 44
= a + 7 = 22 …(ii)
Subtracting (i) from (ii), we get
2d = 10
= d = 5
Substituting value of din (i), we get
a + 5 × 5 = 12
= a + 25 = 12
= a = -13 = 1st term
a + d = -13 + 5 = -8 = 2nd term
a + 2d = -13 + 2 × 5 = -13 + 10= -3 = 3rd term
Hence, the first three terms of an A.P. are – 13,- 8 and -5.

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14th term is 140. Find the 20th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Given,
S14= 1050
\(\frac{14}{2}[2 a+(14-1) d]=1050\)
⇒ 7[2a + 13d] = 1050
⇒ 2a + 13d = 150
⇒ a + 6.5d = 75 ….(i)
And, t14 = 140
⇒ a + 13d = 140 ….(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20th term = t20 = 10 + 19d = 10 + 19(10) = 200

Question 16.
The 25th term of an A.P. exceeds its 9th term by 16. Find its common difference.
Solution:
nth term of an A.P. is given by tn= a + (n – 1) d.
⇒ t25 = a + (25 – 1)d = a + 24d and
t9 = a + (9 – 1)d = a + 8d
According to the condition in the question, we get
t25 = t9 + 16
⇒ a + 24d = a + 8d + 16
⇒ 16d = 16
⇒ d = 1

Question 17.
For an A.P., show that:
(m + n)th term + (m – n)th term = 2 × mthterm
Solution:
Let a and d be the first term and common difference respectively.
⇒(m + n)th term = a + (m + n – 1)d …. (i) and
(m – n)th term = a + (m – n – 1)d …. (ii)
From (i) + (ii), we get
(m + n)th term + (m – n)th term
= a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a + md – nd – d
= 2a + 2md – 2d
= 2a + (m – 1)2d
= 2[ a + (m – 1)d]
= 2 × mth term
Hence proved.

Question 18.
If the nth term of the A.P. 58, 60, 62,…. is equal to the nth term of the A.P. -2, 5, 12, …., find the value of n.
Solution:
In the first A.P. 58, 60, 62,….
a = 58 and d = 2
tn = a + (n – 1)d
⇒ tn = 58 + (n – 1)2 …. (i)
In the first A.P. -2, 5, 12, ….
a = -2 and d = 7
tn = a + (n – 1)d
⇒ tn = -2 + (n – 1)7 …. (ii)
Given that the nth term of first A.P is equal to the nth term of the second A.P.
⇒58 + (n – 1)2 = -2 + (n – 1)7 … from (i) and (ii)
⇒58 + 2n – 2 = -2 + 7n – 7
⇒ 65 = 5n
⇒ n = 15

Question 19.
Which term of the A.P. 105, 101, 97 … is the first negative term?
Solution:
Here a = 105 and d = 101 – 105 = -4
Let an be the first negative term.
⇒ a2n < 0
⇒ a + (n – 1)d < 0
⇒ 105 + (n – 1)(-4)

Question 20.
How many three digit numbers are divisible by 7?
Solution:
The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and tn = 994.
We know that nth term of A.P is given by
tn = a + (n – 1)d.
⇒ 994 = 105 + (n – 1)7
⇒ 889 = 7n – 7
⇒ 896 = 7n
⇒ n = 128
∴ There are 128 three digit numbers which are divisible by 7.

Question 21.
Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
Solution:
Let the three parts of 216 in A.P be (a – d), a, (a + d).
⇒a – d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒ a(a – d ) = 5040
⇒ 72(72 – d) = 5040
⇒ 72 – d = 70
⇒ d = 2
∴ a – d = 72 – 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.

Question 22.
Can 2n2 – 7 be the nth term of an A.P? Explain.
Solution:
We have 2n2 – 7,
Substitute n = 1, 2, 3, … , we get
2(1)2 – 7, 2(2)2 – 7, 2(3)2 – 7, 2(4)2 – 7, ….
-5, 1, 11, ….
Difference between the first and second term = 1 – (-5) = 6
And Difference between the second and third term = 11 – 1 = 10
Here, the common difference is not same.
Therefore the nth term of an A.P can’t be 2n2 – 7.

Question 23.
Find the sum of the A.P., 14, 21, 28, …, 168.
Solution:
Here a = 14 , d = 7 and tn = 168
tn = a + (n – 1)d
⇒ 168 = 14 + (n – 1)7
⇒ 154 = 7n – 7
⇒ 154 = 7n – 7
⇒ 161 = 7n
⇒ n = 23
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 81
Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.

Question 24.
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31st term of this A.P.
Solution:
Here a = 20 and S7 = 2100
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 82
To find: t31 =?
tn = a + (n – 1)d
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 83
Therefore the 31st term of the given A.P. is 2820.

Question 25.
Find the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58.
Solution:
First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
58, …., -8, -10, -12.
Here a = 58 , d = -2
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 84
Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.

More Resources for Selina Concise Class 10 ICSE Solutions

ICSE Solutions Selina ICSE Solutions

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)

Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 1
Prove— Q is equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 2
Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.

Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 3
Prove— P is equidistant from AC and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 4

Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 5
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 6

Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 7
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than \(\frac{1}{2} \mathrm{BC}\) draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 12
Hence, D is equidistant from B and C.

Question 5.
In each of the given figures: PA = PH and QA = QB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 4
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 9
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.

Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 10
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 11
Hence, T is equidistant from PQ and QR.

Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 12
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 13

Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 14

Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 15
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M

Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 16
PB=3.6 cm

Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 17
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 18
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 20

Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 21
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.

Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 22
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 23
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.

Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 24
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.

Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 25
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 26
Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.

Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 27
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 28
Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.

Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 29
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.

Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 30
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.

Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 31
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.

Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 32
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.

Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 33
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.

Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 34
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 35
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.

Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 36
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.

Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 31
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.

Question 7.
Describe the locus of the door handle as the door opens.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 38
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.

Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 39
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.

Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 40
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.

Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 41
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.

Question 12.
Describe the locus of a point P so that:
AB2 = AP2 + BP2, where A and B are two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 42
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB2 = AP2 + BP2.

Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 43
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 44
The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.

Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.

Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 45
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 46
vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 47
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 48
Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 49
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.

Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 50
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 51
Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.

Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 52

 

Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 53
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.

Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 54
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm

Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 55
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.

Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 56
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.

Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 57

 

Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 58
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.

Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 59
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.

Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 60
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 61
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.

Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(i) from AB and AD;
(ii) from the vertices A and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 62

Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 63+
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm

Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 64
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.

Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 65

Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 66
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.

Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 67
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.

Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 68
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm

Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:

  1. Draw BC = 6.5 cm using a ruler.
  2. With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
  3. With Q as center and same radius, cut the previous arc at P.
  4. Join BP and extend it.
  5. With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
  6. Join AC to obtain ΔABC.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 69

ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 70
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 71
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)

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