# A metal bucket is

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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