## Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration

**Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration**

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### Mensuration Exercise 20A – Selina Concise Mathematics Class 7 ICSE Solutions

**Question 1.**

**The length and the breadth of a rectangular plot are 135 m and 65 m. Find, its perimeter and the cost of fencing it at the rate of ₹60 per m.**

**Solution:**

Given :

Length (l) = 135 m

Breadth (b) = 65 m

Perimeter = 2 (l + b)

= 2(135 + 65)

= 2(200) = 400 m

∴Perimeter of rectangular plot is = 400 m

Cost of fencing per m = ₹60

∴Cost of fencing 400 m = ₹60 x 400 m = ₹24000

**Question 2.**

**The length and breadth of a rectangular field are in the ratio 7 : 4. If its perimeter is 440 m, find its length and breadth. Also, find the cost of fencing it @ ₹150 per m.**

**Solution:**

Given : Perimeter = 440 m

Let the length of rectangular field = lx and breadth = 4x

2(l + b) = Perimeter

2(7x + 4x) = 440 m

2(11x) = 440 m

22x = 440 m

x = \(\frac { 440 }{ 22 }\)

x = 11 m

∴Length = 7x = 7 x 11 = 77 m

Breadth = Ax = 4 x 11 = 44 m

Cost of fencing per m = ₹150

Cost of fencing 440 m = ₹150 x 440 = ₹66,000

**Question 3.**

**The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.**

**Solution:**

**Question 4.**

**The diagonal of a square is 12\(\sqrt { 2 } \) cm. Find its perimeter.**

**Solution:**

Diagonal of square = Its side x \(\sqrt { 2 } \)

Side \(\sqrt { 2 } \) = \(\sqrt { 2 } \) \(\sqrt { 2 } \)

i.e. side = 12 cm

Perimeter of a square = 4 x Side

= 4 x 12 = 48 cm

**Question 5.**

**Find the perimeter of a rectangle whose length = 22.5 m and breadth = 16 dm.**

**Solution:**

Length = 22.5 m

Breadth = 16 dm = 1.6 m

Perimeter of rectangle = 2(l + b)

– 2(22.5 + 1.6)

– 2(24.1) = 48.2 m

**Question 6.**

**Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm**

**Solution:**

Length of a rectangle (l) = 24 cm Diagonal = 25 cm

Let breadth of the rectangle = b m

Applying Pythagoras Theorem in triangle ABC,

We get, (AC)^{2} = (AB)^{2} + (BC)^{2}

(25)^{2 }= (24)^{2} + (b)^{2}

625 = 576 + (b)^{2}

625 – 576 = b^{2}

49 = A2

\(\sqrt { 7 x 7 } \) =b

∴b = 7 cm

Now, perimeter of the rectangle

= 2(1 + b)

= 2(24 + 7)

= 2(31)

= 62 cm

**Question 7.**

**The length and breadth of rectangular piece of land are in the ratio of 5 : 3. If the total cost of fencing it at the rate of ₹48 per metre is ₹19,200, find its length and breadth.**

**Solution:**

Ratio in length and breadth of a rectangular piece of land = 5:3

Cost of fencing =₹ 19,200

and rate = ₹48 per m

∴Perimeter = \(\frac { 19200 }{ 48 }\)= 400 m 48

Let length = 5x.

Then breadth = 3x

∴Perimeter = 2(l + b)

400 = 2(5x + 3x)

400 = 2 x 8x= 16x

∴16x = 400

⇒ x = \(\frac { 400 }{ 16 }\) = 25

∴Length of the land = 5x= 5 x 25 = 125 m and breadth = 3x = 3 x 25 = 75 m

**Question 8.**

**A wire is in the shape of square of side 20 cm. If the wire is bent into a rectangle of length 24 cm, find its breadth.**

**Solution:**

Side of square = 20 cm

Perimeter of square = 4 x 20 = 80 cm

Or perimeter of rectangle = 80 cm

Length of a rectangle = 24 cm

∴ Perimeter of a rectangle = 2(l + b)

b = \(\frac { 80 }{ 2 }\) – 24

b = 40 – 24 = 16 m

**Question 9.**

**If P = perimeter of a rectangle, l= its length and b = its breadth find :**

**(i) P, if l = 38 cm and b = 27 cm**

**(ii) b, if P = 88 cm and l = 24 cm**

**(iii) l, if P = 96 m and b = 28 m**

**Solution:**

**Question 10.**

**The cost of fencing a square field at the rate of**

**Cost of fencing 440 m = ₹150 x 440 = ₹75 per meter is**

**Cost of fencing 440 m = ₹150 x 440 = ₹67,500. Find the perimeter and the side of the square field.**

**Solution:**

**Question 11.**

**The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter is equal to the perimeter of a square, find the side of the square.**

**Solution:**

**Question 12.**

**The radius of a circle is 21 cm. Find the circumference (Take π = 3 \(\frac { 1 }{ 7 }\) ).**

**Solution:**

**Question 13.**

**The circumference of a circle is 440 cm. Find its radius and diameter. (Take π = \(\frac { 22 }{ 7 }\)**

**Solution:**

**Question 14.**

**The diameter of a circular field is 56 m. Find its circumference and cost of fencing it at the rate of ₹80 per m. (Take n = \(\frac { 22 }{ 7 }\))**

**Solution:**

**Question 15.**

**The radii of two circles are 20 cm and 13 cm. Find the difference between their circumferences. (Take π = \(\frac { 22 }{ 7 }\))**

**Solution:**

**Question 16.**

**The diameter of a circle is 42 cm, find its perimeter. If the perimeter of the circle is doubled, what will be the radius of the new circle. (Take π = \(\frac { 22 }{ 7 }\) )**

**Solution:**

**Question 17.**

**The perimeter of a square and the circumference of a circle are equal. If the length of each side of the square is 22 cm, find:**

**(i) perimeter of the square.**

**(ii) circumference of the circle.**

**(iii) radius of the circle.**

**Solution:**

**Question 18.**

**Find the radius of the circle whose circumference is equal to the sum of the circumferences of the circles having radii 15 cm and 8 cm.**

**Solution:**

**Question 19.**

**Find the diameter of a circle whose circumference is equal to the sum of circumference of circles with radii 10 cm, 12 cm and 18 cm.**

**Solution:**

**Question 20.**

**The circumference of a circle is eigth time the circumference of the circle with radius 12 cm. Find its diameter.**

**Solution:**

**Question 21.**

**The radii of two circles are in the ratio 3 : 5, find the ratio between their circumferences.**

**Solution:**

**Question 22.**

**The circumferences of two circles are in the ratio 5 : 7, find the ratio between their radii.**

**Solution:**

**Question 23.**

**The perimeters of two squares are in the ratio 8:15, find the ratio between the lengths of their sides.**

**Solution:**

**Question 24.**

**The lengths of the sides of two squares are in the ratio 8:15, find the ratio between their perimeters.**

**Solution:**

Let the side of first square = 8x

∴Perimeter of first square = 4 x Side = 4 x 8x = 32 x

and the side of second squares = 15x

∴Perimeter of second square = 4 x Side = 4 x 15s = 60s

Now, the ratio between their perimeter = 32x: 60x= 8: 15

**Question 25.**

**Each side of a square is 44 cm. Find its perimeter. If this perimeter is equal to the circumference of a circle, find the radius of the circle.**

**Solution:**

### Mensuration Exercise 20B – Selina Concise Mathematics Class 7 ICSE Solutions

**Question 1.**

**Find the area of a rectangle whose length and breadth are 25 cm and 16 cm.**

**Solution:**

**Question 2.**

**The diagonal of a rectangular board is 1 m and its length is 96 cm. Find the area of the board.**

**Solution:**

**Question 3.**

**The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m ^{2}, find**

**(i) its perimeter**

**(ii) cost of fencing it at the rate of ₹40 per meter.**

**Solution:**

**Question 4.**

**A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.**

**Solution:**

**Question 5.**

**The length and breadth of a rectangular piece of land are in the ratio 5 : 3. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its :**

**(i) length and breadth**

**(ii) area**

**(iii) cost of levelling at the rate of ₹60 per m ^{2}.**

**Solution:**

**Question 6.**

**Find the area of the square whose perimeter is 56 cm.**

**Solution:**

**Question 7.**

**A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m ^{2} find the area of the lawn.**

**Solution:**

**Question 8.**

**For each figure, given below, find the area of shaded region : (All measurements are in cm)**

**Solution:**

**Question 9.**

**One side of a parallelogram is 20 cm and its distance from the opposite side is 16 cm. Find the area of the parallelogram.**

**Solution:**

**Question 10.**

**The base of a parallelogram is thrice it height. If its area is 768 cm ^{2}, find the base and the height of the parallelogram.**

**Solution:**

**Question 11.**

**Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.**

**Solution:**

**Question 12.**

**If the area of a rhombus is 112 cm ^{2} and one of its diagonals is 14 cm, find its other diagonal.**

**Solution:**

**Question 13.**

**One side of a parallelogram is 18 cm and its area is 153 cm ^{2}. Find the distance of the given side from its opposite side.**

**Solution:**

**Question 14.**

**The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.**

**Solution:**

**Question 15.**

**The area of a rhombus is 84 cm2 and its perimeter is 56 cm. Find its height.**

**Solution:**

**Question 16.**

**Find the area of a triangle whose base is 30 cm and height is 18 cm.**

**Solution:**

**Question 17.**

**Find the height of a triangle whose base is 18 cm and area is 270 cm ^{2}.**

**Solution:**

**Question 18.**

**The area of a right-angled triangle is 160 cm ^{2}. If its one leg is 16 cm long, find the length of the other leg.**

**Solution:**

**Question 19.**

**Find the area of a right-angled triangle whose hypotenuse is 13 cm long and one of its legs is 12 cm long.**

**Solution:**

**Question 20.**

**Find the area of an equilateral triangle whose each side is 16 cm. (Take \(\sqrt { 3 } \)= 1.73)**

**Solution:**

**Question 21.**

**The sides of a triangle are 21 cm, 17 cm and 10 cm. Find its area.**

**Solution:**

**Question 22.**

**Find the area of an isosceles triangle whose base is 16 cm and length of each of the equal sides is 10 cm.**

**Solution:**

**Question 23.**

**Find the base of a triangle whose area is 360 cm ^{2}and height is 24 cm.**

**Solution:**

**Question 24.**

T**he legs of a right-angled triangle are in the ratio 4 :3 and its area is 4056 cm ^{2}. Find the length of its legs.**

**Solution:**

**Question 25.**

**The area of an equilateral triangle is (64 x \(\sqrt { 3 } \) ) cm ^{2}– Find the length of each side of the triangle.**

**Solution:**

**Question 26.**

**The sides of a triangle are in the ratio 15 : 13 : 14 and its perimeter is 168 cm. Find the area of the triangle.**

**Solution:**

**Question 27.**

**The diameter of a circle is 20 cm. Taking π = 3.14, find the circumference and its area.**

**Solution:**

**Question 28.**

**The circumference of a circle exceeds its diameter by 18 cm. Find the radius of the circle.**

**Solution:**

**Question 29.**

**The ratio between the radii of two circles is 5 : 7. Find the ratio between their :**

**(i) circumference**

**(ii) areas**

**Solution:**

**Question 30.**

**The ratio between the areas of two circles is 16 : 9. Find the ratio between their :**

**(i) radii**

**(ii) diameters**

**(iii) circumference**

**Solution:**

**Question 31.**

**A circular racing track has inner circumference 528 m and outer circumference 616 m. Find the width of the track.**

**Solution:**

**Question 32.**

**The inner circumference of a circular track is 264 m and the width of the track is 7 m. Find:**

**(i) the radius of the inner track.**

**(ii) the radius of the outer circumference.**

**(iii) the length of the outer circumference.**

**(iv) the cost of fencing the outer circumference at the rate of ₹50 per m.**

**Solution:**

**Question 33.**

**The diameter of every wheel of a car is 63 cm. How much distance will the car move during 2000 revolutions of its wheel.**

**Solution:**

**Question 34.**

**The diameter of the wheel of a car is 70 cm. How many revolutions will it make to travel one kilometre?**

**Solution:**

**Question 35.**

**A metal wire, when bent in the form of a square of largest area, encloses an area of 484 cm ^{2}. Find the length of the wire. If the same wire is bent to a largest circle, find:**

**(i) radius of the circle formed.**

**(ii) area of the circle.**

**Solution:**

**Question 36.**

**A wire is along the boundary of a circle with radius 28 cm. If the same wire is bent in the form of a square, find the area of the square formed.**

**Solution:**

**Question 37.**

**The length and the breadth of a rectangular paper are 35 cm and 22 cm. Find the area of the largest circle which can be cut out of this paper.**

**Solution:**

**Question 38.**

**From each comer of a rectangular paper (30 cm x 20 cm) a quadrant of a circle of radius 7 cm is cut. Find the area of the remaining paper i.e., shaded portion.**

**Solution:**