WORD PROBLEMS ON SPHERE AND HEMISPHERE WORKSHEET

Problem 1 :

The diameter of a baseball is 7.4 cm. Use 3.14 for π.

a. Find the surface area to the nearest tenth of a square cm.

b. Find the volume of the baseball to the nearest tenth of a cubic cm.

Solution :

a.

Given, diameter = 7.4 cm

Radius = 7.4/2 = 3.7 cm

Surface area of baseball = 4πr²

= 4 × 3.14 × 3.7 × 3.7

S.A = 172 cm²

b.

Volume of baseball = (4/3) πr³

= (4/3) × 3.14 × (3.7)³

V = 212 cm³

Problem 2 :

The diameter of a golf ball is 4.3 cm. Use 3.14 for π.

a. Find the surface area to the nearest tenth of a square cm.

b. Find the volume of the golf ball to the nearest tenth of a cubic cm.

Solution:

a.

Given, diameter = 4.3 cm

Radius = 4.3/2 = 2.15 cm

Surface area of golf ball = 4πr²

= 4 × 3.14 × 2.15 × 2.15

S.A = 58 cm²

b.

Volume of golf ball = 4/3 πr³

= 4/3 × 3.14 × (2.15)³

 V = 42 cm³

Problem 3 :

Each tennis ball is a sphere with radius 3.4 cm. Use 3.14 for π.

a. Find the surface area of 3 tennis balls to the nearest tenth of a square cm.

b. Find the volume of 3 tennis balls to the nearest tenth of a cubic cm.

Solution :

Given, each tennis ball radius = 3.4 cm

a.

Surface area of each tennis ball = 4πr²

= 4 × 3.14 × 3.4 × 3.4

= 145 cm²

Surface area of 3 tennis balls = 3 × 145

S.A = 435 cm²

b.

Volume of each tennis ball = 4/3 × πr³

= 4/3 × 3.14 × (3.4)³

= 164 cm³

Volume of 3 tennis balls = 3 × 164

V = 492 cm³

Problem 4 :

A spherical watermelon has a 12 in. diameter. The volume of a particular spherical cantaloupe (melon) is 1/2 of the volume of the watermelon. Find the diameter of the cantaloupe to the nearest tenth of an inch.

Solution :

Given, diameter = 12 in

 Radius = 12/2 = 6 in

 The volume of watermelon = 4/3 × πr³

= 4/3 × π × (6)³

V = 288π in³

The volume of cantaloupe V₁ = V/2

= 288π/2

V₁ = 144π in³

V₁ = 4/3 × πr₁³

144π = 4/3 × πr₁³

r₁³ = 108

r₁ = 4.76 in

Diameter = 2r₁

= 2 × 4.76

= 9.52 in

So, the diameter of the cantaloupe is 9.5 in.

Problem 5 :

For the sun to collapse to a “black hole”, it would have to be compressed into a sphere with radius of 1 kilometer.

a. What would be its surface area?

b. What would be its volume?

Solution :

a.

Given, radius = 1 km

Surface area of sphere = 4πr²

= 4 × 3.14 × (1)²

S.A = 12.56 km

b.

Volume of sphere = 4/3 × πr³

= 4/3 × 3.14 × (1)³

V = 4.19 km³

Problem 6 :

Find the volume of a hemisphere with radius 12.

Solution :

Given, radius = 12

Volume of hemisphere = 2/3 × πr³

 = 2/3 × 3.14 × (12)³

V = 3617.28