Mensuration ICSE Class-7th Concise Maths Selina Solutions

Mensuration ICSE Class-7th Concise Selina mathematics Solutions Chapter-20. We provide step by step Solutions of Exercise / lesson-20 Mensuration for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-20 A and Exe-20 B   to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 mathematics.

Mensuration ICSE Class-7th Concise Mathematics Selina Solutions Chapter-20


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Exe-20 A,

Exe-20 B,

 


Exercise – 20 A Mensuration for ICSE Class-7th Concise Selina mathematics

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.
Answer

Given:
Length (l) = 135 m
Breadth (b) = 65 m

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.

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 × 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.
Answer

Given : Perimeter = 440 m

Let the length of rectangular field = 7x and breadth = 4x

2(l + b) = Perimeter

2(7x + 4x)

= 440m

2(11x)

= 440m

22x = 440m

x =440/22

x = 20m

so Length = 7x = 7 x 20

= 140m

Breadth = 4x = 4 x 20 = 80m

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.
Answer

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 20 Mensuration 3

Question 4.

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

Diagonal of square = Its side x√2

Side

√2=√2√2

i.e. side = 12 cm

Perimeter of a square = 4 × side

= 4 × 12 = 48 cm

Question 5.

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

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
Answer

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

Find the perimeter of a rectangle with length = 24 cm and 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)= (24)2 + (b)2
625 = 576 + (b)2
625 – 576 = b2
49 = b2
7×7 = 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.
Answer

Ratio in length and breadth of a rectangular piece of land = 5:3
Cost of fencing = ₹19,200
and rate = ₹48 per m
∴ Perimeter =19200/48

= 400 m
Let length = 5x.
Then breadth = 3x
∴ Perimeter = 2(l + b)
400 = 2(5x + 3x)
400 = 2 × 8x = 16x
∴ 16x = 400
⇒ x = 40016 = 25
∴ Length of the land = 5x = 5 × 25 = 125 m

breadth = 3x = 3 × 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.
Answer

Side of square = 20 cm
Perimeter of square = 4 × 20 = 80 cm
Or perimeter of rectangle = 80 cm
Length of a rectangle = 24 cm
∴ Perimeter of a rectangle = 2(l + b)
b = (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
Answer

(i)

Length = 38 cm

Breadth= 27 cm

Perimeter of a triangle = 2(l + b)

= 2(38 + 27)

= 2(65)

= 130 cm

(ii)

Perimeter of a rectangle = 88 cm

Length (l) = 24 cm

Let breadth = b

P = 2(l + b)

b = (P/2)-l

b = (88/2)-24

= 44 − 24

= 20 cm

∴ Breadth of a rectangle = 20 cm

(iii)

Perimeter of a rectangle = 96 m

Breadth (b) = 28 m

Let length = l

P = 2(l + b)

l =(P/2)-b

= (96/2)-28

= 48 − 42

= 20 m

∴ Length of a rectangle = 20 m

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.
Answer

Length of the fence × its rate = ₹67,500

⇒ Length of the fence = ₹67500/75

= 900 m

∴ Perimeter of a square field

= length of its fence = 900 m

Since, perimeter of a square

= 4 × Length of its side

Length of the side of the square

= Perimeter/4

=900/4 

= 225 m

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.
Answer

Length of rectangle = 36 cm

Breadth of rectangle = 28 cm

Perimeter of the rectangle

= 2(l + b)

= 2(36 + 28)

=2(64)

= 128 cm

Given, the perimeter of the square = perimeter of rectangle = 128 cm

so Side of the square =Perimeter/4

=128/4

= 32 cm

Question 12.

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

Given, radius (r) = 21 cm and π =22/7

Circumference of the circle = 2πr

= 2×(22/7)×21 cm

= 2 × 22 × 3 cm

= 132 cm

Question 13.

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

(i) 

Circumference of the circle = 440 cm

Radius =(C/2π)

=(440×7)/(2×22) cm

=3088/44 cm

(ii) 

Diameter = 2 × radius

= 2 × 70 = 140 cm

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 })
Answer

Given, Diameter of a circular field = 56 m

so Radius =562 = 28 m

Circumference of the circle = 2πr

= 2×(22/7)×28 m

= 2 × 22 × 4 m

= 176 m

Cost of fencing of 176 m is
= 176 m × ₹80 per m
= ₹1,40,780

Question 15.

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

Radius of 1st circle = 20 cm

Circumference of the circle = 2πr

= 2×(22/7)×20

= 8807

= 122.8 cm

The radius of 2nd circle = 13 cm

Circumference of the circle = 2πr

= 2×(22/7)×13

= 572/7

= 81.7

∴ The difference of circumference of two circles
= 122.8 − 81.7 cm

= 41.1 cm

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 } )
Answer

Given, the Diameter of a circle = 42 cm

∴ Radius of circle = 42/2

= 21 cm

The perimeter of the circle

= 2πr

=2×(22/7)×21

= 132 cm

If the perimeter of the circle doubled
= 2 × 132 =264 cm

Radius = (C/2π)=(264/2×(2/27))

= (264×7)/(2×22)

= 42 cm

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.
Answer

(i) 

Side of square = 22 cm

Perimeter of square = 4 × side

= 4 × 22

= 88 cm

(ii) 

Circumference of circle

Given, Perimeter of square = Circumference of circle

= 88 cm

(iii)

Circumference of circle

= 88 cm

∴ Radius =(C/2π)

=(88×7)/(2×22)

=616/44

=14 cm

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.
Answer

For circle with radius = 15 cm

Circumference of circle = 2πr

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 3

Sum of the circumferences of these two circles

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 4

= 23 cm

∴ Required radius = 23 cm

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.
Answer

The radius of the circle = R cm

so 2πR = 2π × 10 + 2π × 12 + 2π × 18

On dividing each term by 2π, we get:

R = 10 + 12 + 18

= 40 cm

so Radius of the circle obtained = 40 cm

And, its diameter = 2 × Radius

= 2 × 40 cm

= 80 cm

Question 20.

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

Radius of the given circle = 12 cm

Circumference of the given circle = 2πr

= 2×(22/7)×12=(528/7) cm

Circumference of the required circle

= 5×(528/7)

=2640/7 cm

If the radius of the required circle = R cm
Its circumference = 2πR

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 6

∴ Required radius = 60 cm

Question 21.

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

The ratio of the radius of the circles = 3: 5

The radius of the first circle = 3x
and radius of the second circle = 5x

∴ Circumference of the first circle = 2πr

= 2π × 3x = 6πx

and circumference of the second circle = 2πr

= 2π × 5x = 10x

∴ The ratio between their circumference
= 6πx : 10πx
= 16 : 10
= 3 : 5

Question 22.

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

The ratio of the circumference of the circle = 5: 7

Let circumference of the first ratio = 5x

∴ 2πr = 5x

⇒ r =5x/2π

and the circumference of the second ratio = 7x

∴ 2πr = 7x

⇒ r = 7x/2π

The ratio between their radius =(5x/2π):(7x/2π)

= 5: 7

Question 23.

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

The perimeter of the first square = 8x

∴ Side of the first square =Perimeter/4

=8x/4

and the perimeter of the second square = 7x

∴ Side of the second square

=Perimeter/4

= 15x/4

Now, the ratio between the sides of the square =(8x/4):(15x/4)

= 8 : 15

Question 24.

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

Let the side of first square = 8x
∴ Perimeter of first square = 4 × Side = 4 × 8x

= 32x
and the side of second squares = 15x
∴ Perimeter of second square = 4 × Side

= 4 × 15x

= 60x
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.
Answer

The side of a square = 44 cm

so Its perimeter = 4 × Side

= 4 × 44 = 176 cm

Since, Its is given that, Circumference of a circle = Perimeter of a square

∴ Circumference of a circle = 176 cm

Let, the radius of the circle = r

⇒ 2πr = 176 cm

r = (176×7)/(2×22) = 28 cm

so The radius of the circle

= 28 cm


Concise Selina Solutions Exercise – 20 B Mensuration for ICSE Class-7th 

Question 1.

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

Length of rectangle = 25 cm

Breadth of rectangle = 16 cm

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

Area of rectangle = l × b or AB × BC

= 25 × 16 cm2 = 400 cm2

so Area of rectangle = 400 cm2

Question 2.

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

Length of diagonal (AB) = 96 cm

Diagonal (AC) = 1 m = 100 cm

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

In right-angled triangle ABC,

By Applying Pythagoras Theorem,

(AC)= (AB)2 + (BC)2

= (100)2 = (96)2 + BC2

10000 = 9216

= BC2

10000 − 9216

= BC2

√784 = BC

∴ BC = 28 cm

Area of the rectangular board

= l × b or AB × BC

= 96 × 28

= 2688 cm2

Question 3.

The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m2, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹40 per meter.
Answer

Ratio in the sides of a rectangle = 4 : 3

Area = 1728 m2

Let length = 4x, and breadth = 3x

∴ Area = l × b

1728 = 4x × 3x

⇒ 12x2 = 1728

⇒ x2=(1728/12)

⇒ x2 = 144 = (12)2

∴ x = 12

∴ Length = 4x = 4 × 12 = 48 m

Breadth = 3x = 3 × 12

= 36 m

(i) 

Now perimeter = 2(l + b)

= 2(48 + 36) m

= 2 × 84

= 168 m

(ii) 

Rate of fencing = ₹40 per metre

Total cost = 168 × 40

= ₹6720

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.
Answer

Length of floor (l) = 40 m

breadth of floor (b) = 15 m

∴ Area of floor = l × b = 40 × 15 = 600 m2

Length of one tile = 60 cm

=6/10 m

and breadth = 50 cm

= 5/10 m

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 8

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 m2.
Answer

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

Cost of fencing = ₹9600

And rate = ₹24 per m

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 9

= 400 m

Let length = 5x

Then breadth = 3x

∴ Perimeter = 2(l + b)

400 = 2 × (5x + 3x)

∴ 16x = 400

x = 400/16

=25

(i)

∴ Length of land = 5x = 5 × 25

= 125 m

and breadth = 3x = 3 × 25

= 75 m

(ii) 

Area = l × b

= 125 × 75

= 9375 m2

(iii) 

Cost of levelling at rate ₹60 per m2

= ₹60 × 9375 m2

= ₹5,62,500

Question 6.

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

Perimeter of square = 56 cm

⇒ 4 × Side = 56 cm

⇒ Side = 56/4cm

⇒ Side = 14 cm

∴ Area of square = (Side)2

= (14)2

= 14 × 14 cm2

= 196 cm2

Question 7.

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m2 find the area of the lawn.
Answer

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m2 find the area of the lawn.

Area of path

= 165 m2

Width of path = 2.5 m

Let the side of square lawn = x m

so Outer side = x + 2 × 2.5

= (x + 5) m

∴ Area of path = (x + 5)2 − x2

⇒ x2 + 10x + 25 − x2

= 165

⇒ 10x = 165 − 25

= 140

⇒ x = 140/10

= 14 m

∴ Side of lawn = 14 m

and area of lawn = (14)2 m2

= 196 m2

Question 8.

For each figure, given below, find the area of shaded region : (All measurements are in cm)
Selina Concise Mathematics class 7 ICSE Solutions - Mensuration-8
Answer

(i)

Outer length = 20 cm

and breadth = 16 cm

∴ Outer area = l × b

= 20 × 16 cm

= 320 cm2

Inner length = 15 cm

and Inner breadth = 10 cm

∴ Inner area = 15 × 10

= 150 cm2

∴ Area of shaded region = Area of whole region − Area of unshaded region

= 320 − 150 cm2

= 170 cm2

(ii)

Outer length = 30 cm

and Outer breadth = 20 cm

∴ Outer area = l × b

= 30 × 20 

= 600 cm2

Inner length = 12 cm and inner breadth = 12 cm

Inner area = l × b = 12 × 12

= 144 cm2

Area of shaded portion = Area of outer figure − Area of inner figure

= 600 − 144

= 456 cm2

(iii)

Area of shaded portion = Area of outer region − Area of unshaded region

= 40 × 40 − 32 × 15

= 1600 − 480 cm2

= 1120 cm2

(iv)

Area of shaded region = Area of outer region − Area of inner region

= 40 × 40 − 15 × 15

= 1600 − 225

= 1375 cm2

(v)

Area of shaded portion

= 2 × 20 + 2 × 8 + 2 ×(12 + 2)

= 40 + 16 + 28 cm2

= 84 cm2

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.
Answer

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

Area of parallelogram = Base × Height

= AB × DE

= 20 cm × 16 cm = 320 cm2

so Area of parallelogram = 320 cm2

Question 10.

The base of a parallelogram is thrice it height. If its area is 768 cm2, find the base and the height of the parallelogram.
Answer

Area of the parallelogram = 768 cm2

Let the height of the parallelogram = x

Then base = 3x

∴ Area = Base × Height

⇒ 768 = 3x × x

⇒ 768 = 3x2

⇒ x2 =768/3

= 256 cm

so x = 16×16

= 16 cm

so Base = 3 × 16

= 48 cm

and height = x = 16 cm

Question 11.

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

Given, diagonal (d1) = 30 cm

Other diagonal (d2) = 24 cm

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

If AC and BD are the diagonals of a rhombus its

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 13

= 15 × 24 = 360 cm2

∴ Area of rhombus = 360 cm2

Question 12.

If the area of a rhombus is 112 cm2 and one of its diagonals is 14 cm, find its other diagonal.
Answer

Area of rhombus = 112 cm2

One diagonal = 14 cm

Let second diagonal = x

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 14

⇒ x = 16

∴ Second diagonal = 16 cm

Question 13.

One side of a parallelogram is 18 cm and its area is 153 cm2. Find the distance of the given side from its opposite side.
Answer

Area of parallelogram ABCD = 153 cm2

Side (Base) AB = 18 cm

One side of a parallelogram is 18 cm and its area is 153 cm2. Find the distance of the given side from its opposite side.

∴ Distance DL between AB and DC …..(altitude)

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 16

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.
Answer

In parallelogram ABCD

AB = DC = 15 cm

BC = AD = 10 cm

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.

Distance between longer sides AB and DC is 6 cm

i.e., perpendicular DL = 6 cm

DM ⊥ BC

Area of parallelogram = Base × Altitude

= AB × DL = 15 × 6

= 90 cm2

Again let DM = x cm

∴ Area of parallelogram ABCD

= BC × DM

= 10 × x = 10x cm2

∴ 10x cm2 = 90 cm2

⇒ x = 90/10

= 9 cm

Question 15.

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

Area of rhombus = 84 cm2

Perimeter = 56 cm

∴ Its side =564 = 14 cm

∴ Height =(Area of rhombus/Base)

=84/14

= 6 cm

Question 16.

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

Base of triangle = 30 cm

Height of triangle = 18 cm

∴ Area =((1/2)base×height)

= (1/2)×30×18

= 270 cm2

Question 17.

Find the height of a triangle whose base is 18 cm and area is 270 cm2.
Answer

Base of triangle = 18 cm

Area of triangle = 270 cm2

∴ Height = (Area×2/Base)

= (270×2)/18

=540/18

= 30 cm

Question 18.

The area of a right-angled triangle is 160 cm2. If its one leg is 16 cm long, find the length of the other leg.
Answer

Area of the right-angled triangle = 160 cm2

Let base (one side) = 16 cm

The area of a right-angled triangle is 160 cm2. If its one leg is 16 cm long, find the length of the other leg.

∴ Altitude (second side)

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 19

= 20 cm

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.
Answer

In right-angled Δ ABC,

Base BC = 12 cm

and hypotenuse AC = 13 cm

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

Applying Pythagoras Theorem,

(AC)2 = (AB)2 + (BC)2

(13)= (AB)2 + (12)2

169 = (AB)2 + 144

(AB)= 169 − 144

(AB)2 = 25

∴ AB =√25

= √5×5 = 5 cm

Now, area of Δ ABC = (1/2)base×altitude

= (1/2)×12×5

= 30 cm2

Question 20.

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

Side of the equilateral triangle = 16 cm

Find the area of an equilateral triangle whose each side is 16 cm. (Take

∴ Area = √3/4(a)²

=(√3/4)×16×16

= 1.73 × 4 × 16

= 110.72 cm2

Question 21.

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

Let a = 21 cm, b = 17 cm and c = 10 cm

∴ a + b + c

= 21 cm + 17 cm + 10 cm = 48 cm

s = (a + b + c)/2

= (48/2)

= 24 cm

Area of the triangle

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 22

= 2 × 2 × 3 × 7

= 84 cm2

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.
Answer

In isosceles Δ ABC

Base BC = 16 cm

and AB = AC = 10 cm

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

Let AD ⊥ BC and BD = (1/2)BC=16/2

∴ BD = 8 cm

In right Δ ABD

AB= AD2 + BD2 …………..(Pythagoras Theorem)

(10)= AD2 + (8)2

100 = AD+ 64

100 − 64 = AD2

36 = AD2

AD = √36=√6×6

∴ AD = 6 cm

Now, the area of triangle =(Base×Altitude)/2

= 16×6/2

= 48 cm2

Question 23.

Find the base of a triangle whose area is 360 cm2and height is 24 cm.
Answer

Area of triangle = 360 cm2

and height (h) = 24 cm

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 24

= 30 cm

Question 24.

The legs of a right-angled triangle are in the ratio 4 :3 and its area is 4056 cm2. Find the length of its legs.
Answer

Area of right-angled triangle = 4056 cm2

Legs of a right-angled triangled are in the ratio i.e. 4 : 3

Let one leg (Base) = 3x

Then second leg (altitude) = 4x

The legs of a right-angled triangle are in the ratio 4 : 3 and its area is 4056 cm2. Find the length of its legs.

The legs of a right-angled triangle are in the ratio 4 : 3 and its area is 4056 cm2. Find the length of its legs.

∴ x = 26 cm

∴ One leg (base) = 3x = 3 × 26

= 78 cm

and second leg (altitude) 4x = 4 × 26

= 104 cm

Question 25.

The area of an equilateral triangle is (64 x \sqrt { 3 }  ) cm2– Find the length of each side of the triangle.
Answer

Area of equilateral triangle = 64√3 cm2

Let each side = a

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 27

a = (16)2

∴ a = 16 cm

∴ Each side = 16 cm

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.
Answer

Perimeter of the triangle = 168 cm

Sum of ratio of sides = 15 + 13 + 14 = 42

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 28

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 29

= 2 × 2 × 2 × 2 × 2 × 2 × 3 × 7

= 1344 cm2

Question 27.

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

Diameter of circle (d) = 20 cm

SO. Circumference = dπ = 20 × 3.14 = 62.8 cm

Radius (r) = d/2

= 10 cm

∴ Area of a circle

= πr2

= 3.14 × 10 × 10

= 314 cm2

Question 28.

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

c be the circumference and d be the diameter of the circle.

∴ c = d + 18

⇒ dπ = d + 18

⇒ dπ − d = 18

d(π − 1) = 18

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 30

Question 29.

The ratio between the radii of two circles is 5 : 7. Find the ratio between their :
(i) circumference
(ii) areas
Answer

(i) 

The ratio of the radius of the circles = 5 : 7
Let radius of first circle = 5x
and radius of second circle = 7x

∴ Circumference of first circle = 2πr

= 2π × 5x

= 10πx

and circumference of second circle

= 2π × 7x

= 14πx

∴ Ratio between their circumference

= 10πx : 14πx

= 10 : 14 = 5 : 7

(ii) Area of first circle = πr2

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 31

= 550 : 1078 …………(Dividing by 22)

= 25 : 49

Question 30.

The ratio between the areas of two circles is 16 : 9. Find the ratio between their :
(i) radii
(ii) diameters
(iii) circumference
Answer

(i)

Let radius of first circle = r1

and radius of second circle = r2

Given that ratio of the areas of circles = 16 : 9

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 32

(ii)

the diameter of first circle = d1

and diameter of second circle = d2

Since, we know that diameter = 2 × radius

∴ d= 2 × r= 2 × 4x

= 8x

and d2 = 2 × r2 = 2 × 3x

= 6x

Now, the ratio between the diameter of two circles = d1 : d2

= 8x : 6x = 4 : 3

(iii)

Now, consider the ratio of circumference of the circles

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 33

so The ratio between the circumference of two circles = 4 : 3

Question 31.

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

Outer circumference = 616 m

Radius (R) = C/2π= (616×7)/(2×22)m
= 98 m

Inner circumference = 528 m

∴ Inner radius (r) =(528×7)/(2×2)m = 84 m

∴ Width of track = R − r

= 98 − 84

= 14 m

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.
Answer

Inner circumference of the circular track = 264 m

(i) ∴ Inner radius (r) = C/2π

= (264×7)/(2×22)

=(1848/44) = 42 cm

(ii) Width of the track = 7 m

∴ Outer radius (R) = 42 + 7 = 49 m

(iii) Outer circumference = 2πR

= 2×(22/7)×49

= 308 m

(iv) Rate of fencing = ₹50 per metre

∴ Total cost of fencing outer circumference = ₹50 × 308

= ₹15,400

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.
Answer

Diameter of car wheel (d) = 63 cm

∴ Circumference = πd = (22/7)×63 = 198 cm

Distance covered in 2000 revolutions

= 2000 × 198 cm

= (2000×198)/100 = 3960 m

= 3.96 km

Question 34

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

Diameter of car wheel (d) = 70 cm

∴ Circumference = πd = (22/7)×70cm

= 220 cm = 220/100 m

No. of revolutions in 1 km

Exercise - 20 A Mensuration for ICSE Class-7th Concise Selina mathematics img 34

Question 35.

A metal wire, when bent in the form of a square of largest area, encloses an area of 484 cm2. 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.
Answer

Area of the square made wire = 484 cm2

∴ Length (side) = √Area=√484 = 22 cm

(i) Perimeter of wire = 4 × Side

= 4 × 22 = 88 cm

∴ Circumference of circular wire = 88 cm

∴ Radius (r) = C/2π

=(88×7)/(2×22)cm = 14 cm

(ii) ∴ Area of the circle = πr2

= (22/7)×14×14 = 616 cm2

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.
Answer

Radius of circular wire = 28 cm

∴ Circumference = 2πr

= 2×(22/7)×28cm = 176 cm

∴ Perimeter of the square formed by this wire

= 176 cm

∴ Side (a) =176/4

= 44 cm

Area of square so formed = a2

= (44)2 cm2

= 1936 cm2

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.
Answer

Length of rectangular paper (l) = 35 cm

Breadth of rectangular paper (b) = 22 cm

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.

∴ Area = 35 × 22

= 770 cm2

The largest circle which can be cut from the rectangular paper will have a radius of 17.5 cm

∴ Area of a circle = πr2

= (22/7)×17.5×17.5

= 962.50 cm2

 

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.
Selina Concise Mathematics class 7 ICSE Solutions - Mensuration-38b
Answer

Length of paper (l) = 30 cm

and breadth (b) = 20 cm

∴ Area of rectangular paper = l × b

= 30 × 20

= 600 cm

Radius of each quadrant at the corner = 7 cm

Area of 4 quadrants =4×(1/4)πr²

= πr= 227×7×7 = 154 cm2

∴ Area of remaining paper
= 600 − 154 = 446 cm2

 

— End of Mensuration ICSE Class-7th Solutions :–

 

Return to – Concise Selina Maths Solutions for ICSE Class -7 


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