Raju

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 11 Mid Point Theorem

Question 1.
(a) In the figure (1) given below, D, E and F are mid-points of the sides BC, CA and AB respectively of ∆ ABC. If AB = 6 cm, BC = 4.8 cm and CA= 5.6 cm, find the perimeter of (i) the trapezium FBCE (ii) the triangle DEF.
(b) In the figure (2) given below, D and E are mid-points of the sides AB and AC respectively. If BC =
5.6 cm and∠B = 72°, compute (i) DE (ii)∠ADE.
(c) In the figure (3) given below, D and E are mid-points of AB, BC respectively and DF || BC. Prove that DBEF is a parallelogram. Calculate AC if AF = 2.6 cm.

Solution:
(a) (i) Given : AB = 6 cm, BC = 4.8 cm, and CA = 5.6 cm
Required : The perimeter of trapezium FBCA.

Question 2.
Prove that the four triangles formed by joining in pairs the mid-points of the sides c of a triangle are congruent to each other.
Solution:
Given: In ∆ ABC, D, E and r,
F are mid-points of AB, BC and CA respectively. Join DE, EF and FD.

Question 3.
If D, E and F are mid-points of sides AB, BC and CA respectively of an isosceles triangle ABC, prove that ∆DEF is also F, isosceles.
Solution:
Given : ABC is an isosceles triangle in which AB = AC

Question 4.
The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the mid-point of AD, prove that
(i) PQ || AB
(ii) PO=\(\frac { 1 }{ 2 }\)CD.
Solution:
(i) Given : ABCD is a parallelogram in which diagonals AC and BD intersect each other. At point O, P is the mid-point of AD. Join OP.

Question 5.
In the adjoining figure, ABCD is a quadrilateral in which P, Q, R and S are mid-points of AB, BC, CD and DA respectively. AC is its diagonal. Show that
(i) SR || AC and SR =\(\frac { 1 }{ 2 }\)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution:

Question 6.
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square,
Solution:

Question 7.
In the adjoining figure, AD and BE are medians of ∆ABC. If DF U BE, prove that CF = \(\frac { 1 }{ 4 }\) AC.

Solution:

Question 8.
(a) In the figure (1) given below, ABCD is a parallelogram. E and F are mid-points of the sides AB and CO respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) ∆HEB = ∆HCF
(ii) GEHF is a parallelogram.
(b) In the diagram (2) given below, ABCD is a parallelogram. E is mid-point of CD and P is a point on AC such that PC = \(\frac { 1 }{ 4 }\) AC. EP produced meets BC at F. Prove that
(i) F is mid-point of BC (ii) 2EF = BD

Solution:

Question 9.
ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
Solution:

Question 10.
(a) In the quadrilateral (1) given below, AB || DC, E and F are mid-points of AD and BD respectively. Prove that:

Solution:

Question 11.
(a) In the quadrilateral (1) given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.
(b) In the figure (2) given below, ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB respectively. Prove that:
(i) ∠EFG = 90
(ii) The line drawn through G and parallel to FE bisects DA.

Solution:

Question 12.
In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate:
(i) BG if AD = 6 cm
(ii) CF if GE = 2.3 cm
(iii) AB if BC = 2.4 cm
(iv) ED if FD = 4.4 cm.

Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 6):
Question 1.
In a ∆ABC, AB = 3 cm, BC = 4 cm and CA = 5 cm. IfD and E are mid-points of AB and BC respectively, then the length of DE is
(a) 1.5 cm
(b) 2 cm
(c) 2.5 cm
(d) 3.5 cm
Solution:
In ∆ABC, D and E are the mid-points of sides AB and BC

Question 2.
In the given figure, ABCD is a rectangle in which AB = 6 cm and AD = 8 cm. If P and Q are mid-points of the sides BC and CD respectively, then the length of PQ is
(a) 7 cm
(b) 5 cm
(c) 4 cm
(d) 3 cm
Solution:

Question 3.
D and E are mid-points of the sides AB and AC of ∆ABC and O is any point on the side BC. O is joined to A. If P and Q are mid-points of OB and OC respectively, then DEQP is
(a) a square
(b) a rectangle
(c) a rhombus
(d) a parallelogram
Solution:
D and E are mid-points of sides AB and AC respectively of AABC O is any point on BC and AO is joined P and Q are mid-points of OB and OC, EQ and DP are joined

Question 4.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectanlge if
(a) PQRS is a parallelogram
(b) PQRS is a rectangle
(c) the diagonals of PQRS are perpendicular to each other
(d) the diagonals of PQRS are equal.
Solution:
A, B, C and D are the mid-points of the sides PQ, QR, RS and SP respectively

Question 5.
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a rhombus if
(a) ABCD is a parallelogram
(b) ABCD is a rhombus
(c) the diagonals of ABCD are equal
(d) the diagonals of ABCD are perpendicular to each other.
Solution:
P, Q, R and S are the mid-points of the quadrilateral ABCD and a quadrilateral is formed by joining the mid-points in order

Question 6.
The figure formed by joining the mid points of the sides of a quadrilateral
ABCD, taken in order, is a square only if
(a) ABCD is a rhombus r
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are perpendicular to each other
(d) diagonals of ABCD are equal and perpendicular to each other.
Solution:

Chapter Test

Question 1.
ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥ QR.
Solution:

Question 2.
The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.
Solution:

Question 3.
If D, E, F are mid-points of the sides BC, CA and AB respectively of a ∆ ABC, Prove that AD and FE bisect each other.
Solution:

Question 4.
In ∆ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.
Solution:
In ∆ABC, D and E are the mid-points of

Question 5.
In the given figure, ABCD is a parallelogram and E is mid-point of AD. DL EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.

Solution:
Given In the figure

Question 6.
In the given figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively. Show that CR = \(\frac { 1 }{ 2 }\) AC.

Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 10 Triangles

Exercise 10.1

Question 1.
It is given that ∆ABC ≅ ∆RPQ. Is it true to say that BC = QR ? Why?
Solution:

Question 2.
“If two sides and an angle of one triangle are equal to two sides and an angle of another triangle, then the two triangles must be congruent.” Is the statement true? Why?
Solution:
No, it is not true statement as the angles should be included angle of there two given sides.

Question 3.
In the given figure, AB=AC and AP=AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠APC = ∠AQB.
Solution:

Question 4.
In the given figure, AB = AC, P and Q are points on BA and CA respectively such that AP = AQ. Prove that
(i) ∆APC ≅ ∆AQB
(ii) CP = BQ
(iii) ∠ACP = ∠ABQ.
Solution:

Question 5.
In the given figure, AD = BC and BD = AC. Prove that :
∠ADB = ∠BCA and ∠DAB = ∠CBA.

Solution:

Question 6.
In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ∆ABD ≅ ∆BAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Solution:

Question 7.
In the given figure, AB = DC and AB || DC. Prove that AD = BC.
Solution:

Question 8.
In the given figure. AC = AE, AB = AD and ∠BAD = ∠CAE. Show that BC = DE.

Solution:

Question 9.
In the adjoining figure, AB = CD, CE = BF and ∠ACE = ∠DBF. Prove that
(i) ∆ACE ≅ ∆DBF
(ii) AE = DF.
Solution:

Question 10.
In the given figure, AB = AC and D is mid-point of BC. Use SSS rule of congruency to show that
(i) ∆ABD ≅ ∆ACD
(ii) AD is bisector of ∠A
(iii) AD is perpendicular to BC.
Solution:

Question 11.
Two line segments AB and CD bisect each other at O. Prove that :
(i) AC = BD
(ii) ∠CAB = ∠ABD
(iii) AD || CB
(iv) AD = CB.

Solution:

Question 12.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.2

Question 1.
In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 2.
In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.
Solution:

Question 3.
“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?
Solution:
The given statement can be true only if the corresponding (included) sides are equal otherwise not.

Question 4.
In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.
Solution:

Question 5.
In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

Solution:

Question 6.
In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Solution:

Question 7.
In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.
Solution:

Question 8.
In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that
(i) ∆ACD ≅ ∆BDC
(ii) BC = AD
(iii) ∠A = ∠B.
Solution:

Question 9.
In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that
(i) ∆EBC ≅ ∆DCB
(ii) ∆OEB ≅ ∆ODC
(iii) OB = OC.
Solution:

Question 10.
ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:

Question 11.
In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.
Solution:

Question 12.
ABCD is a rectanige. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.
Solution:

Question 13.
(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that
(i) ∆XTQ ≅ ∆XSQ
(ii) PX bisects the angle P.
(b) In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that
(i) AD = BC
(ii) AC = BD.
(c) In the figure (3) given below, BA || DF and CA II EG and BD = EC . Prove that, .
(i) BG = DF
(ii) EG = CF.


Solution:

Question 14.
In each of the following diagrams, find the values of x and y.

Solution:

Exercise 10.3

Question 1.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:

Question 2.
Show that the angles of an equilateral triangle are 60° each.
Solution:

Question 3.
Show that every equiangular triangle is equilateral.
Solution:

Question 4.
In the following diagrams, find the value of x:

Solution:

Question 5.
In the following diagrams, find the value of x:

Solution:

Question 6.
(a) In the figure (1) given below, AB = AD, BC = DC. Find ∠ ABC.
(b)In the figure (2) given below, BC = CD. Find ∠ACB.
(c) In the figure (3) given below, AB || CD and CA = CE. If ∠ACE = 74° and ∠BAE =15°, find the values of x and y.
Solution:

Question 7.
In ∆ABC, AB = AC, ∠A = (5x + 20)° and each of the base angle is \(\frac { 2 }{ 5 }\) th of ∠A. Find the measure of ∠A.
Solution:

Question 8.
(a) In the figure (1) given below, ABC is an equilateral triangle. Base BC is produced to E, such that BC’= CE. Calculate ∠ACE and ∠AEC.
(b) In the figure (2) given below, prove that ∠ BAD : ∠ ADB = 3 : 1.
(c) In the figure (3) given below, AB || CD. Find the values of x, y and ∠.

Solution:

Question 9.
In the given figure, D is mid-point of BC, DE and DF are perpendiculars to AB and AC respectively such that DE = DF. Prove that ABC is an isosceles triangle.

Solution:

Question 10.
In the given figure, AD, BE and CF arc altitudes of ∆ABC. If AD = BE = CF, prove that ABC is an equilateral triangle.
Solution:

Question 11.
In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:
(i) ∆DBC ≅ ∆ECB
(ii) ∠DCB = ∠EBC
(iii) OB = OC,where O is the point of intersection of BE and CD.
Solution:

Question 12.
ABC is an isosceles triangle in which AB = AC. P is any point in the interior of ∆ABC such that ∠ABP = ∠ACP. Prove that
(a) BP = CP
(b) AP bisects ∠BAC.
Solution:

Question 13.
In the adjoining figure, D and E are points on the side BC of ∆ABC such that BD = EC and AD = AE. Show that ∆ABD ≅ ∆ACE.
Solution:

Question 14.
(a) In the figure (i) given below, CDE is an equilateral triangle formed on a side CD of a square ABCD. Show that ∆ADE ≅ ∆BCE and hence, AEB is an isosceles triangle.

(b) In the figure (ii) given below, O is a point in the interior of a square ABCD such that OAB is an equilateral trianlge. Show that OCD is an isosceles triangle.

Question 15.
In the given figure, ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.
Solution:

Exercise 10.4

Question 1.
In ∆PQR, ∠P = 70° and ∠R = 30°. Which side of this triangle is longest? Give reason for your answer.
Solution:

Question 2.
Show that in a right angled triangle, the hypotenuse is the longest side.
Solution:

Question 3.
PQR is a right angle triangle at Q and PQ : QR = 3:2. Which is the least angle.
Solution:

Question 4.
In ∆ ABC, AB = 8 cm, BC = 5.6 cm and CA = 6.5 cm. Which is (i) the greatest angle ?
(ii) the smallest angle ?
Solution:

Question 5.
In ∆ABC, ∠A = 50°, ∠B= 60°, Arrange the sides of the triangle in ascending order.
Solution:

Question 6.
In figure given alongside, ∠B = 30°, ∠C = 40° and the bisector of ∠A meets BC at D. Show
(i) BD > AD
(ii) DC > AD
(iii) AC > DC
(iv) AB > BD

Solution:

Question 7.
(a) In the figure (1) given below, AD bisects ∠A. Arrange AB, BD and DC in the descending order of their lengths.
(b) In the figure (2) given below, ∠ ABD = 65°, ∠DAC = 22° and AD = BD. Calculate ∠ ACD and state (giving reasons) which is greater : BD or DC ?

Solution:

Question 8.
(a) In the figure (1) given below, prove that (i) CF> AF (ii) DC>DF.
(b) In the figure (2) given below, AB = AC.
Prove that AB > CD.
(c) In the figure (3) given below, AC = CD. Prove that BC < CD.

Solution:

Question 9.
(a) In the figure (i) given below, ∠B < ∠A and ∠C < ∠D. Show that AD < BC. (b) In the figure (ii) given below, D is any point on the side BC of ∆ABC. If AB > AC, show that AB > AD.
Solution:

Question 10.
(i) Is it possible to construct a triangle with lengths of its sides as 4 cm, 3 cm and 7 cm? Give reason for your answer,
(ii) Is it possible to construct a triangle with lengths of its sides as 9 cm, 7 cm and 17 cm? Give reason for your answer.
(iii) Is it possible to construct a triangle with lengths of its sides as 8 cm, 7 cm and 4 cm? Give reason for your answer.
Solution:
(i) Length of sides of a triangle are 4 cm, 3 cm and 7 cm
We know that sum of any two sides of a triangle is greatar than its third side But 4 + 3 = 7 cm
Which is not possible
Hence to construction of a triangle with sides 4 cm, 3 cm and 7 cm is not possible.
(ii) Length of sides of a triangle are 9 cm, 7 cm and 17 cm
We know that sum of any two sides of a triangle is greater than its third side Now 9 + 7 = 16 < 17 ∴ It is not possible to construct a triangle with these sides.
(iii) Length of sides of a triangle are 8 cm, 7 cm and 4 cm We know that sum of any two sides of a triangle is greater than its third side Now 7 + 4 = 11 > 8
Yes, It is possible to construct a triangle with these sides.

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 18):
Question 1.
Which of the following is not a criterion for congruency of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Solution:
Criteria of congruency of two triangles ‘SSA’ is not the criterion. (c)

Question 2.
In the adjoining figure, AB = FC, EF=BD and ∠AFE = ∠CBD. Then the rule by which ∆AFE = ∆CBD is
(a) SAS
(b) ASA
(c) SSS
(d) AAS

Solution:

Question 3.
In the adjoining figure, AB ⊥ BE and FE ⊥ BE. If AB = FE and BC = DE, then
(a) ∆ABD ≅ ∆EFC
(b) ∆ABD ≅ ∆FEC
(c) ∆ABD ≅ ∆ECF
(d) ∆ABD ≅ ∆CEF
Solution:
In the figure given,

Question 4.
In the adjoining figure, AB=AC and AD is median of ∆ABC, then AADC is equal to
(a) 60°
(b) 120°
(c) 90°
(d) 75°
Solution:

Question 5.
In the adjoining figure, O is mid point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to
(a) ∠OCA
(b) ∠ODB
(c) ∠OBD
(d) ∠BOD
Solution:

Question 6.
In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to
(a) ∠BAD
(b) ∠ABC
(c) ∠ABD
(d) ∠BDA

Solution:

Question 7.
In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. If OB = 4 cm, then BD is
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:

Question 8.
In ∆ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Solution:

Question 9.
In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 80°
(b) 40°
(c) 50°
(d) 100°
Solution:

Question 10.
In ∆PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 4 cm
(b) 5 cm
(c) 2 cm
(d) 2.5 cm
Solution:

Question 11.
In ∆ABC and APQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) isosceles but not congruent
(b) isosceles and congruent
(c) congruent but isosceles
(d) neither congruent nor isosceles
Solution:

Question 12.
Two sides of a triangle are of lenghts 5 cm and 1.5 cm. The length of the third side of the triangle can not be
(a) 3.6 cm
(b) 4.1 cm
(c) 3.8 cm
(d) 3.4 cm
Solution:

Question 13.
If a, b, c are the lengths of the sides of a trianlge, then
(a) a – b > c
(b) c > a + b
(c) c = a + b
(d) c < A + B
Solution:
a, b, c are the lengths of the sides of a trianlge than a + b> c or c < a + b
(Sum of any two sides is greater than its third side) (d)

Question 14.
It is not possible to construct a triangle when the lengths of its sides are
(a) 6 cm, 7 cm, 8 cm
(b) 4 cm, 6 cm, 6 cm
(c) 5.3 cm, 2.2 cm, 3.1 cm
(d) 9.3 cm, 5.2 cm, 7.4 cm
Solution:
We know that sum of any two sides of a triangle is greater than its third side 2.2 + 3.1 = 5.3 ⇒ 5.3 = 5.3 is not possible (c)

Question 15.
In ∆PQR, if ∠R> ∠Q, then
(a) QR > PR
(b) PQ > PR
(c) PQ < PR
(d) QR < PR
Solution:
In ∆PQR, ∠R> ∠Q
∴ PQ > PR (b)

Question 16.
If triangle PQR is right angled at Q, then
(a) PR = PQ
(b) PR < PQ
(c) PR < QR
(d) PR > PQ
Solution:

Question 17.
If triangle ABC is obtuse angled and ∠C is obtuse, then
(a) AB > BC
(b) AB = BC
(c) AB < BC
(d) AC > AB
Solution:

Question P.Q.
A triangle can be constructed when the lengths of its three sides are
(a) 7 cm, 3 cm, 4 cm
(b) 3.6 cm, 11.5 cm, 6.9 cm
(c) 5.2 cm, 7.6 cm, 4.7 cm
(d) 33 mm, 8.5 cm, 49 mm
Solution:
We know that in a triangle, if sum of any two sides is greater than its third side, it is possible to construct it 5.2 cm, 7.6 cm, 4.7 cm is only possible. (c)

Question P.Q.
A unique triangle cannot be constructed if its
(a) three angles are given
(b) two angles and one side is given
(c) three sides are given
(d) two sides and the included angle is given
Solution:
A unique triangle cannot be constructed if its three angle are given, (a)

Question 18.
If the lengths of two sides of an isosceles are 4 cm and 10 cm, then the length of the third side is
(a) 4 cm
(b) 10 cm
(c) 7 cm
(d) 14 cm
Solution:
Lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then length of the third side is 10 cm
(Sum of any two sides of a triangle is greater than its third side and 4 cm is not possible as 4 + 4 > 10 cm.

Chapter Test

Question 1.
In triangles ABC and DEF, ∠A = ∠D, ∠B = ∠E and AB = EF. Will the two triangles be congruent? Give reasons for your answer.
Solution:

Question 2.
In the given figure, ABCD is a square. P, Q and R are points on the sides AB, BC and CD respectively such that AP= BQ = CR and ∠PQR = 90°. Prove that
(a) ∆PBQ ≅ ∆QCR
(b) PQ = QR
(c) ∠PRQ = 45°

Solution:

Question 3.
In the given figure, AD = BC and BD = AC. Prove that ∠ADB = ∠BCA.
Solution:

Question 4.
In the given figure, OA ⊥ OD, OC X OB, OD = OA and OB = OC. Prove that AB = CD.

Solution:

Question 5.
In the given figure, PQ || BA and RS CA. If BP = RC, prove that:
(i) ∆BSR ≅ ∆PQC
(ii) BS = PQ
(iii) RS = CQ.

Solution:

Question 6.
In the given figure, AB = AC, D is a point in the interior of ∆ABC such that ∠DBC = ∠DCB. Prove that AD bisects ∠BAC of ∆ABC.
Solution:

Question 7.
In the adjoining figure, AB || DC. CE and DE bisects ∠BCD and ∠ADC respectively. Prove that AB = AD + BC.
Solution:

Question 8.
In ∆ABC, D is a point on BC such that AD is the bisector of ∠BAC. CE is drawn parallel to DA to meet BD produced at E. Prove that ∆CAE is isosceles
Solution:

Question 9.
In the figure (ii) given below, ABC is a right angled triangle at B, ADEC and BCFG are squares. Prove that AF = BE.

Solution:

Question 10.
In the given figure, BD = AD = AC. If ∠ABD = 36°, find the value of x .

Solution:

Question 11.
In the adjoining figure, TR = TS, ∠1 = 2∠2 and ∠4 = 2∠3. Prove that RB = SA.

Solution:
Given: In the figure , RST is a triangle

Question 12.
(a) In the figure (1) given below, find the value of x.
(b) In the figure (2) given below, AB = AC and DE || BC. Calculate
(i)x
(ii) y
(iii) ∠BAC
(c) In the figure (1) given below, calculate the size of each lettered angle.

Solution:

Question 13.
(a) In the figure (1) given below, AD = BD = DC and ∠ACD = 35°. Show that
(i) AC > DC (ii) AB > AD.
(b) In the figure (2) given below, prove that
(i) x + y = 90° (ii) z = 90° (iii) AB = BC

Solution:

Question 14.
In the given figure, ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.
Solution:

Question 15.
In the given figure, AP ⊥ l and PR > PQ. Show that AR > AQ.
Solution:

Question 16.
If O is any point in the interior of a triangle ABC, show that
OA + OB + OC > \(\frac { 1 }{ 2 }\)
(AB + BC + CA).

Solution:

Question P.Q.
Construct a triangle ABC given that base BC = 5.5 cm, ∠ B = 75° and height = 4.2 cm.
Solution:

Question P.Q.
Construct a triangle ABC in which BC = 6.5 cm, ∠ B = 75° and ∠ A = 45°. Also construct median of A ABC passing through B.
Solution:

Question P.Q.
Construct triangle ABC given that AB – AC = 2.4 cm, BC = 6.5 cm. and ∠ B = 45°.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 9 Logarithms

Exercise 9.1

Question 1.
Convert the following to logarithmic form:
(i) 5² = 25
(ii) a⁵ =64
(iii) 7^x =100
(iv) 9° = 1
(v) 6¹ = 6
(vi) 3^-2 = \(\frac { 1 }{ 9 }\)
(vii) 10^-2 = 0.01
(viii) (81)^{\(\frac { 3 }{ 4 }\)} = 27
Solution:

Question 2.
Convert the following into exponential form:
(i) log₂ 32 = 5
(ii) log₃ 81=4
(iii) log₃\(\frac { 1 }{ 3 }\)= -1
(iv) log₃ 4= \(\frac { 2 }{ 3 }\)
(v) log₈ 32= \(\frac { 5 }{ 3 }\)
(vi) log₁₀ (0.001) = -3
(Vii) log₂ 0.25 = -2
(viii) log_a (\(\frac { 1 }{ a }\)) =-1
Solution:

Question 3.
By converting to exponential form, find the values of:
(i) log₂ 16
(ii) log₅ 125
(iii) log₄ 8
(iv) log₉ 27
(v) log₁₀(.01)
(vi) log₇ \(\frac { 1 }{ 7 }\)
(vii) log₅ 256
(Viii) log₂ 0.25
Solution:

Question 4.
Solve the following equations for x.

Solution:

Question 5.
Given log₁₀a = b, express 10^2b-3 in terms of a.
Solution:

Question 6.
Given log₁₀ x= a, log₁₀ y = b and log₁₀ z =c,
(i) write down 10^2a-3 in terms of x.
(ii) write down 10^3b-1 in terms of y.
(iii) if log₁₀ P = 2a + \(\frac { b }{ 2 }\)– 3c, express P in terms of x, y and z.
Solution:

Question 7.
If log₁₀x = a and log₁₀y = b, find the value of xy.
Solution:

Question 8.
Given log₁₀ a = m and log₁₀ b = n, express \(\frac { { a }^{ 3 } }{ { b }^{ 2 } }\) in terms of m and n.
Solution:

Question 9.
Given log₁₀a= 2a and log₁₀y = –\(\frac { b }{ 2 }\)
(i) write 10^a in terms of x.
(ii) write 10^2b+1 in terms of y.
(iii) if log₁₀P= 3a -2b, express P in terms of x and y .
Solution:

Question 10.
If log₂ y = x and log₃ z = x, find 72^x in terms of y and z.
Solution:

Question 11.
If log₂ x = a and log₅y = a, write 100^2a-1 in terms of x and y.
Solution:

Exercise 9.2

Question 1.
Simplify the following :

Solution:

Question 2.
Evaluate the following:


Solution:

Question 3.
Express each of the following as a single logarithm:

Solution:

Question 4.
Prove the following :
(i) log₁₀ 4 ÷ log₁₀ 2 = l0g₃ 9
(ii) log₁₀ 25 + log₁₀ 4 = log₅ 25
Solution:

Question 5.
If x = 100)^a , y = (10000)^b and z = (10)^c, express

Solution:

Question 6.
If a = log₁₀x, find the following in terms of a :
(i) x
(ii) log₁₀\(\sqrt [ 5 ]{ { x }^{ 2 } }\)
(iii) log₁₀5x
Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If x = log₁₀ 12, y = log₄ 2 x log₁₀ 9 and z = log₁₀ 0.4, find the values of
(i)x-y-z
(ii) 7^x-y-z
Solution:

Question 10.
If log V + log3 = log π + log4 + 3 log r, find V in terns of other quantities.
Solution:

Question 11.
Given 3 (log 5 – log3) – (log 5-2 log 6) = 2 – log n , find n.
Solution:

Question 12.
Given that log₁₀y + 2 log₁₀x= 2, express y in terms of x.
Solution:

Question 13.
Express log₁₀2+1 in the from log₁₀x.
Solution:

Question 14.

Solution:

Question 15.
Given that log m = x + y and log n = x-y, express the value of log m²n in terms of x and y.
Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.
Solve for x:

Solution:

Question 19.
Given 2 log₁₀x+1= log₁₀250, find
(i) x
(ii) log₁₀2x
Solution:

Question 20.

Solution:

Question 21.
Prove the following :
(i) 3^{_{log 4}} = 4^{_{log 3}}
(ii) 27^{_{log 2}} = 8^{_{log 3}}
Solution:

Question 22.
Solve the following equations :
(i) log (2x + 3) = log 7
(ii) log (x +1) + log (x – 1) = log 24
(iii) log (10x + 5) – log (x – 4) = 2
(iv) log₁₀5 + log₁₀(5x+1) = log₁₀(x + 5) + 1
(v) log (4y – 3) = log (2y + 1) – log3
(vi) log₁₀(x + 2) + log₁₀(x – 2) = log₁₀3 + 31og₁₀4.
(vii) log(3x + 2) + log(3x – 2) = 5 log 2.
Solution:

Question 23.
Solve for x :
log₃ (x + 1) – 1 = 3 + log₃ (x – 1)
Solution:

Question 24.

Solution:

Question 25.

Solution:

Question 26.

Solution:

Question 27.
If p = log₁₀20 and q = log₁₀25, find the value of x if 2 log₁₀ (x +1) = 2p – q.
Solution:

Question 28.
Show that:

Solution:

Question 29.
Prove the following identities:

Solution:

Question 30.

Solution:

Question 31.
Solve for x :

Solution:

Multiple Choice Questions

correct Solution from the given four options (1 to 7):
Question 1.
If log√3 27 = x, then the value of x is
(a) 3
(b) 4
(c) 6
(d) 9
Solution:

Question 2.
If log₅ (0.04) = x, then the vlaue of x is
(a) 2
(b) 4
(c) -4
(d) -2
Solution:

Question 3.
If log_0.5 64 = x, then the value of x is
(a) -4
(b) -6
(c) 4
(d) 6
Solution:

Question 4.
If log_{10\(\sqrt [ 3 ]{ 5 }\)} x = -3, then the value of x is

Solution:

Question 5.
If log (3x + 1) = 2, then the value of x is

Solution:

Question 6.
The value of 2 + log₁₀ (0.01) is
(a)4
(b)3
(c)1
(d)0
Solution:

Question 7.


Solution:

Chapter Test

Question 1.

Solution:

Question 2.
Find the value of log√3 3√3 – log₅ (0.04)
Solution:

Question 3.
Prove the following:

Solution:

Question 4.
If log (m + n) = log m + log n, show that n = \(\frac { m }{ m-1 }\)
Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.
Solve the following equations for x:

Solution:

Question 8.
Solve for x and y:

Solution:

Question 9.
If a = 1 + log_xyz, 6 = 1+ log_y zx and c=1 + log_zxy, then show that ab + bc + ca = abc.
Solution:

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 7 Quadratic Equations

Question 1.
Solve the following (1 to 12) equations:
(i) x² – 11x + 30 = 0
(ii) 4x² -25 = 0
Solution:

Question 2.
(i) 2x² – 5x = 0
(ii) x² – 2x = 48
Solution:

Question 3.
(i) 6 + x = x²
(ii) 2x² + 3x + 1= 0
Solution:

Question 4.
(i) 3x² = 2x + 8
(ii) 4x² + 15 = 16x
Solution:

Question 5.
(i) x (2x + 5) = 25
(ii) (x +3) (x – 3) = 40
Solution:

Question 6.
(i) (2x + 3) (x – 4) = 6
(ii) (3x + 1) (2x + 3) = 3
Solution:

Question 7.
(i) 4x² + 4x + 1 = 0
(ii) (x – 4)² + 5²= 132
Solution:

Question 8.
(i) 21x² = 4(2x + 1)
(ii) \(\frac { 2 }{ 3 }\) x2 – \(\frac { 1 }{ 3 }\) x – 1 = 0
Solution:

Question 9.
(i) 6x + 29 = \(\frac { 5 }{ x }\)
(ii) x + \(\frac { 1 }{ x }\) = 2 \(\frac { 1 }{ 2 }\)
Solution:

Question 10.

Solution:

Question 11.

Solution:

Question 12.

Solution:

Multiple Choice Questions

Choose the correct Solution from the given four options (1 to 5):
Question 1.
Which of the following is not a quadratic equation :
(a) 2x² = 3x – 5
(b) (2x- 1) (x- 1) = 2x² – 7x + 2
(c) (2x – 1) (x + 2) = (x – 1) (x + 1)
(d) (x+ 1), = x, + 2x+2
Solution:
(2x – 1) (x – 1) = 2x² – 7x + 2 is not a quadratic equation. (b)

Question 2.
If 2 is a root of the quadratic equation 2x² – kx + 1 = 0, then the value of k is

Solution:

Question 3.
If -3 is a root of the quadratic equation kx² + 2x – 3 = 0, then the value of k is
(a) 1
(b) -1
(c) \(\frac { 1 }{ 9 }\)
(d) \(\frac { 1 }{ -9 }\)
Solution:

Question 4.
Which of the following quadratic equations has -1 as a root?
(a) x² + 5x + 6 = 0
(b) 2x² – 3x + 1 = 0
(c) 2x² + x – 3 = 0
(d) 2x² – x – 3 = 0
Solution:

Question 5.
The root of the quadratic equation x² – 3x – 4 = 0 are
(a) -4, 1
(b) 4, -1
(c) 4, 1
(d) -4, -1
Solution:

Chapter Test

Solve the following (1 to 3) equations :
Question 1.
(i) x(2x+ 5) = 3
(ii) 3x² – 4x – 4 = 0
Solution:

Question 2.
(i) 4x² – 2x + \(\frac { 1 }{ 4 }\) = 0
(ii) 2x² + 7x + 6 = 0
Solution:

Question 3.

Solution:

 

ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 6 Problems on Simultaneous Linear Equations

Question 1.
The sum of two numbers is 50 and their difference is 16. Find the numbers.
Solution:

Question 2.
The sum of two numbers is 2. If their difference is 20, find the numbers.
Solution:

Question 3.
The sum of two numbers is 43. If the larger is doubled and the smaller is tripled, the difference is 36. Find the two numbers.
Solution:

Question 4.
The cost of 5 kg of sugar and 7 kg of rice is Rs. 153, and the cost of 7 kg of sugar and 5 kg of rice is Rs. 147. Find the cost of 6 kg of sugar and 10 kg of rice.
Solution:

Question 5.
The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. 70 per kg and the other costing Rs. 84 per kg. They estimated that 36 kg of sweets were needed. If the total money spent on sweets was Rs. 2800, find how much sweets of each kind they purchased.
Solution:

Question 6.
If from twice the greater of two numbers 16 is subtracted, the result is half the other number. If from half the greater number 1 is subtracted, the result is still half the other number. What are the numbers.
Solution:

Question 7.
There are 38 coins in a collection of 20 paise coins and 25 paise coins. If the total value of the collection is Rs. 8.50, how many of each are there ?
Solution:

Question 8.
A man has certain notes of denominations Rs. 20 and Rs. 5 which amount to Rs. 380. If the number of notes of each kind is interchanged, they amount to Rs. 60 less as before. Find the number of notes of each denomination.
Solution:

Question 9.
The ratio of two numbers is \(\frac { 2 }{ 3 }\). If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.
Solution:

Question 10.
If 1 is added to the numerator of a fraction, it becomes \(\frac { 1 }{ 5 }\) ; if 1 is taken from the denominator, it becomes \(\frac { 1 }{ 7 }\), find the fraction.
Solution:

Question 11.
If the numerator of a certain fraction is increased by 2 and the denominator by 1, the fraction becomes equal to \(\frac { 5 }{ 8 }\) and if the numerator and denominator are each diminished by 1, the fraction becomes equal to \(\frac { 1 }{ 2 }\) , find the fraction.
Solution:

Question 12.
Find the fraction which becomes \(\frac { 1 }{ 2 }\) when the denominator is increased by 4 and is equal to \(\frac { 1 }{ 8 }\) , when the numerator is diminished by 5.
Solution:

Question 13.
In a two digit number the sum of the digits is 7. If the number with the order of the digits reversed is 28 greater than twice the unit’s digit of the original number, find the number.
Solution:

Question 14.
A number of two digits exceeds four times the sum of its digits by 6 and it is increased by 9 on reversing the digits. Find the number.
Solution:

Question 15.
When a two digit number is divided by the sum of its digits the quotient is 8. If the ten’s digit is diminished by three times the unit’s digit the remainder is 1. What is the number ?
Solution:

Question 16.
The result of dividing a number of two digits by the number with digits reversed is 1 \(\frac { 3 }{ 4 }\) . If the sum of digits is 12, find the number.
Solution:

Question 17.
The result of dividing a number of two digits by the number with the digits reversed is \(\frac { 5 }{ 6 }\) . If the difference of digits is 1, find the number.
Solution:

Question 18.
A number of three digits has the hundred digit 4 times the unit digit and the sum of three digits is 14. If the three digits are written in the reverse order, the value of the number is decreased by 594. Find the number.
Solution:

Question 19.
Four years ago Marina was three times old as her daughter. Six years from now the mother will be twice as old as her daughter. Find their present ages.
Solution:

Question 20.
On selling a tea set at 5% loss and a lemon set at 15% gain, a shopkeeper gains Rs. 70. If he sells the tea set at 5% gain and lemon set at 10% gain, he gains Rs. 130. Find the cost price of the lemon set.
Solution:

Question 21.
A person invested some money at 12% simple interest and some other amount at 10% simple interest. He received yearly interest of Rs, 1300. If he had interchanged the amounts, he would have received Rs. 40 more as yearly interest. How much did he invest at different rates ?
Solution:

Question 22.
A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting Rs. 1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got Rs. 20 more. Find the cost price of the table and the list price of the chair.
Solution:

Question 23.
A and B have some money with them. A said to B, “if you give me Rs. 100, my money will become 75% of the money left with you.” B said to A” instead if you give me Rs. 100, your money will become 40% of my money, How much money did A and B have originally ?
Solution:

Question 24.
The students of a class are made to stand in (complete) rows. If one student is extra in a row, there would be 2 rows less, and if one student is less in a row, there would be 3 rows more. Find the number of students in the class.
Solution:

Question 25.
A jeweller has bars of 18-carat gold and 12- carat gold. How much of each must be melted together to obtain a bar of 16-carat gold weighing 120 grams ? (Pure gold is 24 carat)
Solution:

Question 26.
A and B together can do a piece of work in 15 days. If A’s one day work is 1 \(\frac { 1 }{ 2 }\) times the one day’s work of B, find in how many days can each do the work.
Solution:

Question 27.
men and 5 women can do a piece of work in 4 days, while one man and one woman can finish it in 12 days. How long would it take for 1 man to do the work ?
Solution:

Question 28.
A train covered a certain distance at a uniform speed. If the train had been 30 km/hr faster, it would have taken 2 hours less than the scheduled time. If the train were slower by 15 km/hr, it would have taken 2 hours more than the scheduled time. Find the length of the journey.
Solution:

Question 29.
A boat takes 2 hours to go 40 km down the stream and it returns in 4 hours. Find the speed of the boat in still water and the speed of the stream.
Solution:

Question 30.
A boat sails a distance of 44 km in 4 hours with the current. It takes 4 hours 48 minutes longer to cover the same distance against the current. Find the speed of the boat in still water and the speed of the current.
Solution:

Question 31.
An aeroplane flies 1680 km with a head wind in 3.5 hours. On the return trip with same wind blowing, the plane takes 3 hours. Find the plane’s air speed and the wind speed.
Solution:

Question 32.
A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When Bhawana takes food for 20 days, she has to pay Rs. 2600 as hostel charges; whereas when Divya takes food for 26 days, she pays Rs. 3020 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:

Multiple Choice Questions

Choose the correct answer from the given four options (1 to 8):
Question 1.
Sum of digits of a two digit number is 8. If the number obtained by reversing the digits is 18 more than the original number, then the original number is
(a) 35
(b) 53
(c) 26
(d) 62
Solution:

Question 2.
The sum of two natural numbers is 25 and their difference is 7. The numbers are
(a) 17 and 8
(b) 16 and 9
(c) 18 and 7
(d) 15 and 10
Solution:

Question 3.
The sum of two natural numbers is 240 and their ratio is 3 : 5. Then the greater number is
(a) 180
(b) 160
(c) 150
(d) 90
Solution:

Question 4.
The sum of the digits of a two digit number is 9. If 27 is added to it, the digits of the number get reversed. The number
(a) 27
(b) 72
(c) 63
(d) 36
Solution:

Question 5.
The sum of the digits of a two digit number is 12. If the number is decreased by 18, its digits get reversed. The number is
(a) 48
(b) 84
(c) 57
(d) 75
Solution:

Question 6.
Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of the money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 75
(d) 25 and 25
Solution:

Question 7.
The age of a woman is four times the age of her daughter. Five years hence, the age of the woman will be three times the age of her daughter. The present age of the daughter is
(a) 40 years
(b) 20 years
(c) 15 years
(d) 10 years
Solution:

Question 8.
Father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present age in years of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Solution:

Chapter Test

Question 1.
A 700 gm dry fruit packcosts Rs. 216. It contains some almonds and the rest cashew kernel. If almonds cost Rs. 288 ‘per kg and cashew kernel.cost Rs. 336 per kg, what are the quantities of the two dry fruits separately ?
Solution:

Question 2.
Drawing pencils cost 80 paise each and coloured pencils cost Rs. 1.10 each. If altogether two dozen pencils cost Rs. 21.60, how many coloured pencils are there ?
Solution:

Question 3.
Shikha works in a factory. In one week she earned Rs. 390 for working 47 hours, of which 7 hours were overtime. The next week she earned Rs. 416 for working 50 hours, of which 8 hours were overtime. What is Shikha’s hourly earning rate ?
Solution:

Question 4.
The sum of the digits of a two digit number is 7. If the digits are reversed, the new number increased by 3 equals 4 times the original number. Find the number.
Solution:

Question 5.
Three years hence a man’s age will be three times his son’s age and 7 years ago he was seven times as old as his son. How old are they now ?
Solution:

Question 6.
Rectangles are drawn on line segments of fixed lengths. When the breadths are 6 m and 5 m respectively the sum of the areas of the rectangles is 83 m². But if the breadths are 5 m and 4 m respectively the sum of the areas is 68 m². Find the sum of the areas of the squares drawn on the line segments.
Solution:

Question 7.
If the length and the breadth of a room are increased by 1 metre each, the area is increased by 21 square metres. If the length is decreased by 1 metre and the breadth is increased by 2 metres, the area is increased by 14 square metres. Find the perimeter of the room.
Solution:

Question 8.
The lenghts (in metres) of the sides of a triangle are 2x + \(\frac { y }{ 2 }\), \(\frac { 5x }{ 3 }\) + y + \(\frac { 1 }{ 2 }\) and \(\frac { 2 }{ 3 }\)x + 2y + \(\frac { 5 }{ 2 }\). If the triangle is equilateral, find its perimeter.
Solution:

Question 9.
On Diwali eve, two candles, one of which is 3 cm longer than the other are lighted. The longer one is lighted at 530 p.m. and the shorter at 7 p.m. At 930 p.m. they both are of the same length. The longer one burns out at 1130 p.m. and the shorter one at 11 p.m. How long was each candle originally ?
Solution:

We hope that ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 6 Problems on Simultaneous Linear Equations

are helpful  if you have doubts, comment below we will get back you as soon as possible

 

Check the below ICSE Class 10 Geography Goyal Brothers Solutions Chapter 7 Water Resources with Questions and Answers Pdf free download. Students can also read Water Resources Class 10 ICSE Solutions.

ICSE Class 10 Geography Goyal Brothers Solutions Chapter 7 Water Resources

Question 1.
What is irrigation?
Answer:
Irrigation refers to the process of watering of agricultural plants through artificial means from wells, tanks, tube wells, canals etc.

Question 2.
Why is irrigation necessary for a country like India?
Question
Irrigation necessary for a country like India due to the following reasons:

• Irrigation is required as rainfall is erratic, sporadic and unevenly distributed.
• Irrigation is required to grow crops like rice and jute require more water than supplied by rain.
• Irrigation is required to grow crops throughout the year as rainfall is seasonal.
• Irrigation is required to increase agricultural output by providing the right amount of water at the right time.

Question 3.
State the different methods of irrigation.
Answer:
The different methods of Irrigation given below:

Question 4.
Of the two types of canal irrigation, which do you think is better? Why?
Answer:
Two types of canal Irrigation are:

1. Inundation canals
2. Perennial canals

Perennial canals are preferred over inundation canals as these (perennial) over canals are taken out from the perennial rivers by constructing berrages to regulate the flow of water. Most of the cause in India belong to this category.

Question 5.
What is tank Irriagtion?
Answer:
Tank Irrigation is useful in the areas which are dependent on rainfall for their water supply.
Tanks are constructed in the areas of natural depression by building earthen embankments or masonry walls to collect water. This water is used in dry season.

• In India, about 12% of the total irrigated area is done by the tanks.
• Tanks are used mostly in peninsular India, where the underlying hard rocks do not allow the water to seep through.

Question 6.
What are the advantages and disadvantages of tank irrigation?
Answer:
Advantages of Tank Irrigation:

• Tank Irrigation is cheaper than other type of irrigation as the rainwater is collected in natural pits and depressions winch are easily available in peninsular India
• As the water is collected and used in dry season, it uses the water table of the surrounding region.
• The tank water is used for domestic purposes in villages.

Disadvantages:

• Tanks get silted up soon and regular desalting is required to make these suitable for irrigation.
• When the monsoons fail they go dry and therefore, are not dependable source of irrigation.

Question 7.
Where are tanks most widely used and why?
Answer:
Tanks are most widely used in India (more in South India than in North India, in states like Orissa, Andhra Pradesh, Tamil Nadu and Karnataka) because of the following reasons:

• Most of the area of peninsular India is uneven with many natural depressions where the tanks can be built.
• The Deccan Plateau consists of underlying hard rocks which are impervious or non-porous.
  These rocks do not allow the water to seep through.
• Peninsular India receives monsoon type of climate.
  Hence, these tanks collect water during rainy season and help in raising the water table.

Question 8.
Of the two types of wells, which is the best and why?
Answer:
There two types of wells:

1. Surface Wells
2. Tube Wells

Tube Wells is the best than Surface Wells due to the following reasons:

• Tube Wells are very useful during drought conditions when Surface Wells dry up.
• They can irrigate a larger area (about 400 hectares) as compared to Surface Wells.
• They occupy less area as compared to Surface Wells.

Question 9.
Name two states where tank irrigation and well irrigation is practised.
Answer:
Well Irrigation is used in India in states like Uttar Pradesh, Goa, Punjab, Haryana, Bihar, Rajasthan, Gujarat, Maharashtra, Madhya Pradesh, Andhra Pradesh, Karnataka and Tamil Nadu.
Well Irrigation is used in India in states like Uttar Pradesh and Bihar.

Question 10.
Name two methods of drawing water from the well.
Answer:
Two methods of drawing water from the well are:

1. Tube Well Irrigation
2. Surface Well Irrigation

Question 11.
Despite big river projects, irrigation through groundwater has still been growing in popularity. Why is this ?
Answer:
Despite big river projects, irrigation through groundwater has still been growing in popularity due to the following reasons:

• The distribution of groundwater is very uneven in India.
• Coastal regions are usually rich in groundwater owing to the largely alluvial terrain.
• Due to highly variable nature of climate, groundwater has become a popular alternative for irrigation and domestic water use across India.

Question 12.
For an Indian farmer state two advantages that well irrigation has over canal irrigation.
Answer:
Advantages of Well Irrigation over Canal Irrigation are given following:

• Well is an independent source of irrigation and can be used as and when the necessity arises. Canal irrigation, on the other hand, is controlled by other agencies and cannot be used at will.
• There is a limit of to the extent of Canal irrigation, while a Well can be dug at any convenient place.
• The farmer has to pay regularly for Canal irrigation which is not the case in Well irrigation.

Question 13.
What is the difference between surface water resource and groundwater resource? State two advantages of each.
Answer:
Surface Water Resource:
Surface water is available on the surface of the earth in the form of rivers, lakes, ponds and canals.

Advantages of Surface Water:

• A good part of its lost through the process of evaporation and plant transpiration.
• In contrast the rivers of the peninsular India are seasonal. They carry no or very little water in the dry summer season.

Groundwater Resource:
The part of rainwater that seeps through the cracks and crevices into the ground is called groundwater.

Advantages of Groundwater:

• Due to the highly variable nature of climate, groundwater has become a popular alternative for irrigation and domestic water sue across India.
• In addition to being accessible, groundwater quality is generally excellent in most area and present a relatively safe source of drinking water for Indians in rural and urban areas.

Question 14.
Name two important canal systems each in Uttar Pradesh and Punjab.
Answer:
Uttar Pradesh – Upper Ganga Canal and Agra Canal.
Punjab – Nangal Canal and Sirhind Canal.

Question 15.
Irrigation in India faces many problems. State any two such problems.
Answer:
Irrigation in India faces following problems:

• Uneven and uncertain distribution of rainfall.
• The ever increasing population in India leads to the demand of even more intensive agriculture. This needs more and enhanced irrigation facilities to get maximum yield from the same land area.

Question 16.
Differentiate between inundation canal and perennial canal.
Answer:

Inundation Canals	Perennial Canals
(i) These provide irrigation mostly during the rainy season.	These flow throughout the year.
(ii) These are constructed by putting some form of barrage across the river.	These are drawn from the rivers without making any kind of barrage or dam.
(iii) These irrigate large areas.	These irrigate small areas.

Question 17.
Give two reasons why inundation canals being converted to perennial canals?
Answer:
Inundation canals are being converted to Perennial Canals because of the following reasons :

• Inundation Canals are taken out from the rivers when they are in flood and there is excess of water. Hence if they are converted into perennial canal they will have water throughout the year.
• Only lower level areas can be irrigated by such canals as they have no regulating systems.

Question 18.
Give two reasons why tube well irrigation is more important in northern India than in southern India.
Answer:
Tube well irrigation is important in northern India than in southern India because of soft land, even land, flat land, agricultural prosperous land for tube well construction.

Question 19.
Give two reasons why tank irrigation is more popular in southern India.
Answer:
Tank irrigation is widely prevalent in Tamil Nadu, Andhra Pradesh, Karnataka and Odisha. The tank irrigation is more important in the Deccan Plateau because:

• The terrain of Deccan Plateau is undulating and is made up of hard rocks which makes it difficult to dig canals and wells.
• There is little percolation of rain water due to hard rock structure and groundwater is not available in large quantity.

Question 20.
Name two states which suffer periodically from droughts. Give a reason for your answer.
Answer:
Rajasthan and Gujarat suffer periodically from droughts The reasons for drought are:

• Location –
  These regions are drought prone. These areas receive an annual rainfall up to 60 cm and hence are severely affected where the monsoons aren’t enough.
• Inadequate rainfall-
  Inadequate or failure of monsoons is a major reasons for droughts in India Inadequate rainfall causes crop failure which lead to famines.

Question 21.
State the objectives of National Water Policy.
Answer:
Objectives of National Water Policy are:

• The main objective of the policy is to provide surplus water to the deficit areas. (Meghalaya).
• The policy aims at reducing the runoff, soil erosion and silting of river beds.
• Recharging of groundwater is also a priority of the National Water Policy.
• The policy aims to reduce water pollution and to improve the quality of water of the rivers. There is an emphasis an recycling of water too.
• There is increasing demand of water for irrigation industries and domestic uses. The policy aims to take suitable measures for conservation of water resources, by minimising losses at the storage.

Question 22.
What is meant by the term Rainwater Harvesting?
Answer:
“The activity of collecting rainwater directly or recharging it into the ground to improve groundwater storage in the aquifer” is called Rainwater harvesting.
Or
It is a technique of increasing the recharge of groundwater by capturing and storing rainwater.

Question 23.
Why is it necessary to conserve water?
Answer:
We need to conserve water for the following reasons:

• The overexploitation of underground water often results in the lowering of water table.
• The loss of vegetation causes drought and reduction of rainfall and of the water table.
• Irrigation utilises more than 90 percent of the total fresh water.
• The water demand for industrial use will increase more than two times of water demand for domestic use by 2025.
• The increase in population with the progress of time results in water scarcity.
• Out water resources like the underground water, rivers, lakes etc. are polluted and their water can hardly be used without adequate treatement.

Question 24.
How is rainwater harvesting done? Explain.
Answer:
Rainwater can be collected over our rooftops and within our house premises and can be channelized through small PVC pipes into the underground pits wells or borewells to be used by hand pump or from wells.

Question 25.
State the advantages of rooftop rainwater harvesting.
Answer:
The advantages of rainwater harvesting are:

• It reduces ground water pollution.
• It reduces soil erosion.
• It meets the increasing demand for water and supplements household requirement of water.
• It reduces surface runoff.
• It avoids flooding of roads.
• It raises the groundwater table by adding to groundwater reserves.
• It conserves water.
• It improves the quality of groundwater.

Question 26.
Name the state where rainwater harvesting is commonly practised.
Answer:
Rajasthan and Karnataka.

Question 27.
Give geographical reasons –
(a) River Damodar is called “river of sorrow”.
(b) Irrigation by canals is more suitable in northern India as compared to the south.
(c) Tank irrigation is used peninsular India.
Answer:
(a) Its main purpose was to control floods. So as river Damodar is called “River of Sorrow” owing to its devastating floods.

(b) Canal Irrigation is popular in Northern because it has perennial rivers. Soft land, even land, flat land, agricultural prosperous land for canal to be constructed from the rivers to the field.

(c) Tank irrigation is preferred over other means of irrigation in Peninsular India because of the following reasons:

• In the uneven rocky plateau of Peninsular India, where rainfall is highly seasonal, tank irrigation is useful.
• In the large stretches of the Indian plateau, due to hard rocky terrain, wells and canals are difficult to construct, so only tank irrigation is feasible.
• Tank irrigation is highly significant in stroing the abundant rainwater that would otherwise flow out and go waste.

Question 28.
Answer in one word –
(a) A plastic tube which has small holes through which water keeps on dripping.
(b) An ancient irrigation system in North East part of India.
(c) Tubewells are used extensively in UP and Bihar and are very popular.
Answer:
(a) Drip Irrigation.
(b) Bamboo Irrigation.
(c) Well Irrigation

Question 29.
State the advantages and disadvantages of Sprinkler Irrigation.
Answer:
Advantages of Sprinkler Irrigation:

• This method is best for conserving water as there is no wastage of water.
• It is best suited for arid and semi-arid regions.
• Water management is easier than surface irrigation system.
• Easy mechanization and automation.
• High application efficiency.
• This type of irrigation does not involve any loss of water by seepage or evaporation.

Disadvantages of Sprinkler Irrigation:

• High initial cost.
• High operating cost.
• Water application efficiency under sprinkler irrigation is strongly affected by direction of wind.
• Saline water may cause problem.
• Water must be free from sand, debris and salt.
• Some crop are particularly sensitive and may suffer leaf scorch because of the salt deposited on the leaves after the water evaporates.

Question 30.
Name the modern methods of irrigation.
Answer:
Modem methods of irrigation includes furrow irrigation, spray irrigation, drip irrigation.

Question 31.
Study the picture below and answer the questions that follows:
(a) Name the type of Irrigation.
Answer:
Drip Irrigation.

(b) What are the advantages of this type of irrigation.
Answer:
Advantages of Drip Irrigation :

• Fertilizer and nutrient loss is minimized if managed localized and reduced leaching.
• Water application efficiency is high if managed correctly.
• Field levelling is not necessary.
• Fields with irregular shapes are easily accomodated.
• Moisture within the root zone can be maintained at field capacity.

Goyal Brothers Prakashan Class 10 ICSE Geography Solutions