The radii of the circular bases of a frustum of a right cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum.

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Solution

Soln:

Let slant height of the frustum cone be ‘L’

Given height of frustum cone = 12 cm

Radii of a frustum cone are 12 cm and 3 cm

$r_{1} = 12 c m; r_{2} = 3 c m$

$L = \sqrt{(r_{1} - r_{2})^{2} + h^{2}} = \sqrt{(12 - 3)^{2} + 12^{2}} = \sqrt{81 + 144} = 15 c m$

Total surface area of cone

$= π (r_{1} + r_{2}) L + π r_{1}^{2} + π r_{2}^{2}$

$= π (12 + 3) 15 + π 12^{2} + π 3^{2} = 378 c m^{2}$

Volume of frustum cone = 13Π(r21+r22+r1r2)×h

$= \frac{1}{3} (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2}) h$

$= \frac{1}{3} (12^{2} + 3^{2} + 12 \times 3) \times 12 = 756 c m^{3}$

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