# A metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used. Also find the volume of water the bucket can hold. (π = 3.14)

Given.

Height of bucket is 40 cm

Height of cylinder is 10 cm

Radii of both circular ends of frustum are 60 cm, 20 cm

Formula used/Theory.

CSA of cylinder = 2πrh

CSA of frustum = π(r + R) × l

Volume of cylinder = πr2h

Volume of frustum = h[R2 + r2 + Rr]

As the bucket is always open from mouth

Metallic sheet require will be sum of CSA of Frustum and CSA of cylinder and Area of base circle

In frustum

L2 = height2 +

Height of frustum = height of bucket – height of cylinder

= 40 – 10 = 30 cm

L2 = 302 + [60 – 20]2

L2 = 900 cm2 + 1600 cm2 = 2500 cm2

L = √(2500 cm2) = 50cm

CSA of Frustum = π(60 + 20)50 cm

= 3.14 × 80cm × 50cm

= 12560cm2

CSA of cylinder = 2πrh

= 2 × 3.14 × 20cm × 10cm

= 1256 cm2

Area of base = πr2

= 3.14 × 20 × 20

= 1256 cm2

Area of metallic sheet = 12560 cm2 + 1256 cm2 + 1256 cm2

= 15072 cm2

Volume of Frustum = h[R2 + r2 + Rr]

= × 30cm × [(60 cm)2 + (20 cm)2 + 60 cm × 20 cm]

= 31.4 cm × [3600cm2 + 400cm2 + 1200cm2]

= 31.4 cm × 5200 cm2

= 163280 cm3

1 cm3 = litres

163280 cm3 = litres

= 163.280 litres

Area of metallic sheet used in making bucket is 15072 cm2

Volume of bucket is 163.280 litres

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