Class 8 Maths Solution | Lesson 8 Area and Volume | Curriculum Development Centre (CDC)

Lesson 8 Area and Volume

Exercise 8.2

1. Find the area of a circle with the following measurements: (π = 3.14)

(a) Radius = 3 cm

Here,

Radius of circle (r) = 3 cm

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (3)² = 3.14 × 9 = 28.26 cm²

(b) Radius = 8 ft

Here,

Radius of circle (r) = 8 ft

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (8)² = 3.14 × 64 = 200.96 ft²

(c) Radius = 16 m

Here,

Radius of circle (r) = 16 m

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (16)² = 3.14 × 256 = 803.84 m²

(d) Diameter = 5 cm

Here,

Radius of circle (r) = Diameter / 2 = 5 / 2 = 2.5 cm

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (2.5)² = 3.14 × 6.25 = 19.625 cm²

(e) Diameter = 12 inch

Here,

Radius of circle (r) = Diameter / 2 = 12 / 2 = 6 inch

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (6)² = 3.14 × 36 = 113.04 inch²

(f) Diameter = 18 m

Here,

Radius of circle (r) = Diameter / 2 = 18 / 2 = 9 m

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (9)² = 3.14 × 81 = 254.34 m²

(g) Diameter = 20 km

Here,

Radius of circle (r) = Diameter / 2 = 20 / 2 = 10 km

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (10)² = 3.14 × 100 = 314 km²

(h) Diameter = 15 mm

Here,

Radius of circle (r) = Diameter / 2 = 15 / 2 = 7.5 mm

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (7.5)² = 3.14 × 56.25 = 176.625 mm²

(i) Diameter = 22 cm

Here,

Radius of circle (r) = Diameter / 2 = 22 / 2 = 11 cm

Area of circle (A) = ?

We know that,

Area of circle (A) = πr² = 3.14 × (11)² = 3.14 × 121 = 379.94 cm²

2. Find the area of a circle with the following circumference: (π = 3.14)

(a) Circumference = 34.54 cm

Here,

Circumference of circle (C) = 34.54 cm

Radius of circle (r) = C / 2π = 34.54 / (2 × 3.14) = 34.54 / 6.28 = 5.5 cm

Area of circle (A) = πr² = 3.14 × (5.5)² = 3.14 × 30.25 = 95.585 cm²

(b) Circumference = 65.94 m

Here,

Circumference of circle (C) = 65.94 m

Radius of circle (r) = C / 2π = 65.94 / (2 × 3.14) = 65.94 / 6.28 = 10.5 m

Area of circle (A) = πr² = 3.14 × (10.5)² = 3.14 × 110.25 = 346.185 m²

(c) Circumference = 18.84 inch

Here,

Circumference of circle (C) = 18.84 inch

Radius of circle (r) = C / 2π = 18.84 / (2 × 3.14) = 18.84 / 6.28 = 3 inch

Area of circle (A) = πr² = 3.14 × (3)² = 3.14 × 9 = 28.26 inch²

(d) Circumference = 113.04 m

Here,

Circumference of circle (C) = 113.04 m

Radius of circle (r) = C / 2π = 113.04 / (2 × 3.14) = 113.04 / 6.28 = 18 m

Area of circle (A) = πr² = 3.14 × (18)² = 3.14 × 324 = 1017.36 m²

(e) Circumference = 376.80 ft

Here,

Circumference of circle (C) = 376.80 ft

Radius of circle (r) = C / 2π = 376.80 ÷ (2 × 3.14) = 376.80 / 6.28 = 60 ft

Area of circle (A) = πr² = 3.14 × (60)² = 3.14 × 3600 = 11304 ft²

(f) Circumference = 157 m

Here,

Circumference of circle (C) = 157 m

Radius of circle (r) = C / 2π = 157 / (2 × 3.14) = 157 / 6.28 = 25 m

Area of circle (A) = πr² = 3.14 × (25)² = 3.14 × 625 = 1962.5 m²

3. Find radius of a circle having the following area: (π = 22/7)

(a) Area = 154 cm²

Here,

Area of circle (A) = 154 cm²

We know that, A = πr²

r² = A / π = 154 / (22/7) = 154 × (7/22) = 49

∴ Radius of circle (r) = √49 = 7 cm

(b) Area = 346.5 ft²

Here,

Area of circle (A) = 346.5 ft²

We know that, A = πr²

r² = A / π = 346.5 / (22/7) = 346.5 × (7/22) = 110.25

∴ Radius of circle (r) = √110.25 = 10.5 ft

(c) Area = 616 m²

Here,

Area of circle (A) = 616 m²

We know that, A = πr²

r² = A / π = 616 / (22/7) = 616 × (7/22) = 196

∴ Radius of circle (r) = √196 = 14 m

(d) Area = 1386 m²

Here,

Area of circle (A) = 1386 m²

We know that, A = πr²

r² = A / π = 1386 / (22/7) = 1386 × (7/22) = 441

∴ Radius of circle (r) = √441 = 21 m

(e) Area = 38.5 km²

Here,

Area of circle (A) = 38.5 km²

We know that, A = πr²

r² = A / π = 38.5 / (22/7) = 38.5 × (7/22) = 12.25

∴ Radius of circle (r) = √12.25 = 3.5 km

(f) Area = 3850 ft²

Here,

Area of circle (A) = 3850 ft²

We know that, A = πr²

r² = A / π = 3850 / (22/7) = 3850 × (7/22) = 1225

∴ Radius of circle (r) = √1225 = 35 ft

4. Find the area of the shaded part of the figures below: (π = 3.14)

(a)

Here,

Diameter of the larger circle = 12 cm

Radius of the larger circle (R) = 12 / 2 = 6 cm

Diameter of the smaller circle = 6 cm

Radius of the smaller circle (r) = 6 / 2 = 3 cm

Area of the larger circle = πR² = 3.14 × (6)² = 3.14 × 36 = 113.04 cm²

Area of the smaller circle = πr² = 3.14 × (3)² = 3.14 × 9 = 28.26 cm²

∴ Shaded area = Area of larger circle - Area of smaller circle = 113.04 - 28.26 = 84.78 cm²

(b)

Here,

Side of the square = 4 cm

Area of the square = (Side)² = (4)² = 16 cm²

Area of the circle = πr² = 3.14 × (4)² = 3.14 × 16 = 50.24 cm²

∴ Shaded area = Area of circle - Area of square = 50.24 - 16 = 34.24 cm²

(c)

Here,

Radius of the circle (r) = 4 in

Sides of the rectangle, l = 6 in and b = 5 in

We know that,

Area of the circle = πr² = 3.14 × (4)² = 3.14 × 16 = 50.24 in²

Area of the rectangle = l × b = 6 × 5 = 30 in²

∴ Shaded area = Area of circle - Area of rectangle = 50.24 - 30 = 20.24 in²

(d)

Here,

Radius of the circle (r) = 5 cm

Sides of the rectangle, l = 8 cm and b = 6 cm

Area of the circle = πr² = 3.14 × (5)² = 3.14 × 25 = 78.5 cm²

Area of the rectangle = Length × Breadth = 8 × 6 = 48 cm²

∴ Shaded area = Area of circle - Area of rectangle = 78.5 - 48 = 30.5 cm²

(e)

Here,

Radius of the circle (r) = 6 ft

Side of the square = 12 ft

Area of the circle = πr² = 3.14 × (6)² = 3.14 × 36 = 113.04 ft²

Area of the square = (Side)² = (12)² = 144 ft²

∴ Shaded area = Area of square - Area of circle = 144 - 113.04 = 30.96 ft²

(f)

Here,

Side of the square = 28 cm

Radius of the circle (r) = 28 / 2 = 14 cm

Area of the square = (Side)² = (28)² = 784 cm²

Area of the four quadrants = Area of one circle = πr² = 3.14 × (14)² = 3.14 × 196 = 615.44 cm²

∴ Shaded area = Area of square - Area of four quadrants = 784 - 615.44 = 168.56 cm²

5. (a) If the radius of a circular room is 14 m then, how much area does it cover?

Here,

Radius of the circular room (r) = 14 m

Area of the circular room = πr² = (22/7) × (14)² = (22/7) × 196 = 616 m²

(b) If the radius of a circular fun park is 21 m, then, how much area does it cover?

Here,

Radius of the circular fun park (r) = 21 m

Area of the circular fun park = πr² = (22/7) × (21)² = (22/7) × 441 = 1386 m²

(c) A cow is tied with a rope of 7 ft. length. Find the grazing area of that cow.

Here,

Length of rope = Radius of grazing area (r) = 7 ft

Grazing area = πr² = (22/7) × (7)² = (22/7) × 49 = 154 ft²

6. (a) If a diameter of the base of a cylindrical bowl is 9 cm,, find the area of the base of the bowl.

Here,

Diameter of the base of the bowl = 9 cm

Radius (r) = 9 / 2 = 4.5 cm

∴ Area of the base = πr² = 3.14 × (4.5)² = 3.14 × 20.25 = 63.585 cm²

(b) If the diameter of the base of a cylindrical pipe is 30 cm, find the area of the base of pipe.

Here,

Diameter of the base of the pipe = 60 cm

Radius (r) = 60 / 2 = 30 cm

∴ Area of the base = πr² = 3.14 × (30)² = 3.14 × 900 = 2826 cm²

7. (a) If the area of base of cylindrical tank is 154 sq. ft. Find the radius and the circumference of it.

Here,

Area of the base of the cylindrical tank = 154 ft²

We know, Area = πr²

r² = Area / π = 154 / (22/7) = 154 × (7/22) = 49

Radius (r) = √49 = 7 ft

Circumference = 2πr = 2 × (22/7) × 7 = 44 ft

(b) A circular playground of area 153.86 m2 is plastered. What is the diameter of the plastered part of the playground? Also find its circumference.

Here,

Area of the circular playground = 153.86 m²

We know, Area = πr²

r² = Area / π = 153.86 / 3.14 = 49

Radius (r) = √49 = 7 m

Diameter = 2r = 2 × 7 = 14 m

Circumference = 2πr = 2 × 3.14 × 7 = 43.96 m

8. (a) Sharmila has drawn a circle of radius 5 cm. Similarly, Prakash also has drawn a circle of radius 7 cm. Now, whose area of circle is more and by how much?

Here,

Radius of Sharmila's circle (r₁) = 5 cm

Radius of Prakash's circle (r₂) = 7 cm

Area of Sharmila's circle = πr₁² = 3.14 × (5)² = 3.14 × 25 = 78.5 cm²

Area of Prakash's circle = πr₂² = 3.14 × (7)² = 3.14 × 49 = 153.86 cm²

∴ Prakash's circle is larger by = 153.86 - 78.5 = 75.36 cm²

(b) Salman dug a well of 14 m diameter. Similarly, Pramila also dug another well with a diameter of 18 m. Now whose well occupies more land and how much?

Here,

Diameter of Salman's well = 14 m

Diameter of Pramila's well = 18 m

Radius of Salman's well (r₁) = 14 / 2 = 7 m

Radius of Pramila's well (r₂) = 18 / 2 = 9 m

Area of Salman's well = πr₁² = 3.14 × (7)² = 3.14 × 49 = 153.86 m²

Area of Pramila's well = πr₂² = 3.14 × (9)² = 3.14 × 81 = 254.34 m²

∴ Pramila's well occupies more land by = 254.34 - 153.86 = 100.48 m²

9. (a)

Here,

Area of the circular pond = 616 m²

We know,

Area = πr²

or, r² = Area / π = 616 / (22/7) = 616 × (7/22) = 196

∴ Radius (r) = √196 = 14 m

(b)

Circumference = 2πr = 2 × (22/7) × 14 = 88 m

(c)

Cost of fencing per meter = Rs. 250

∴ Total cost = Circumference × Cost per meter = 88 × 250 = Rs. 22,000

10. (a)

Here,

Total distance covered = 3520 m

Number of rotations = 4

∴ Length of the circular track = Total distance / Number of rotations = 3520 / 4 = 880 m

(b)

We know, Circumference = 2πr

880 = 2 × 3.14 × r

or, r = 880 / (2 × 3.14) = 140 m

∴ Diameter = 2r = 2 × 140 = 280 m

(c)

Area of the track = πr² = 3.14 × (140)² = 3.14 × 19600 = 61544 m²

(d)

Cost of fencing per meter = Rs. 600

∴ Total cost = Circumference × Cost per meter = 880 × 600 = Rs. 528,000