Question

# A metallic bucket, open at the top, of height $$24cm$$ is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are $$7cm$$ and $$14cm$$ respectively. Findthe volume of water which can completely fill the bucket;the area of the metal sheet used to make the bucket.

Solution

## Let the slant height of the frustum of the cone be 'l' and the radius of the upper end be 'R' and the radius of the lower end be 'r' respectively.Therefore,l² = √(R - r)² + h²l² = √(14 - 7)² + 24²l² = √7² + 24²l² = √49 + 576l² = √625l = 25 cmSo, slant height is 25 cmVolume = 1/3πh(R² + r² + R*r)⇒ 1/3*22/7*24*(14² + 7² + 14*7)⇒ 1/3*22/7*24*(196 + 49 + 98)⇒ 1/3*22/7*24*343⇒ 181104/21⇒ 8624 cm³Curved surface area = πl(R + r)⇒ 22/7*25*(14 + 7)⇒ 22/7*25*21⇒ 11550/21⇒ 1650 cm²Area of the base (lower end) of the bucket = πr²⇒ 22/7*7*7⇒ 154 cm²Area of the metal sheet used to make the bucket = Curved surface area + Area of the base (lower end)= 1650 + 154 = 1804 cm²Mathematics

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