Question

if the radii of the circular ends of a conical bucket , which is 16cm high, are 20 cm and 8 cm . find the capacity and total surface area of the bucket ( use py = 22/7)

Answer 1

Solution:-

Radius of the bigger end of the frustum (bucket) of cone = R = 20 cm 
Radius of the smaller end of the frustum (bucket) of the cone = r = 8 cm
Height = 16 cm
Volume = 1/3πh[R² + r² + R*r]
= 1/3*22/7*16[20² + 8² + 20*8]
= 352/21[400 + 64 + 160]
= (352*624)/21
= 219648/21
= 10459.43 cu cm
Now,
Slant height of the cone = l = √(R - r)² + h²
l = √(20 - 8)² + 16²
l = √12² + 16²
l = √144 + 256
l = √400
l = 20 cm
Slant height is 20 cm
Now,
Surface area = π[R² + r² + (R + r)*l]
= 22/7[20² + 8² + (20 + 8)*16]
= 22/7[400 + 64 + 448]
= 22/7*912
= 20064/7
= 2866.29 sq cm
Answer

Answer 2

Solution:-

Radius of the bigger end of the frustum (bucket) of cone = R = 20 cm

Radius of the smaller end of the frustum (bucket) of the cone = r = 8 cm

Height = 16 cm

Volume = 1/3πh[R² + r² + R*r]

= 1/3*22/7*16[20² + 8² + 20*8]

= 352/21[400 + 64 + 160]

= (352*624)/21

= 219648/21

= 10459.43 cu cm

Now,

Slant height of the cone = l = √(R - r)² + h²

l = √(20 - 8)² + 16²

l = √12² + 16²

l = √144 + 256

l = √400

l = 20 cm

Slant height is 20 cm

Now,

Surface area = π[R² + r² + (R + r)*l]

= 22/7[20² + 8² + (20 + 8)*16]

= 22/7[400 + 64 + 448]

= 22/7*912

= 20064/7

= 2866.29 sq cm