The slant hieght o the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm Find the curved surface of the frustum.

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Solution

Soln:

Given slant height of frustum of cone (l) = 4 cm
Let ratio of the top and bottom circles be $r_{1}$ and $r_{2}$

Given perimeters of its ends as 18 cm and 6 cm

$=> 2 π r_{1} = 18; 2 π r_{2} = 6$

$=> π r_{1} = 9; π r_{2} = 3$

Curved surface area of frustum cone $= π (r_{1} + r_{2}) l = (π r_{1} + π r_{2}) l = (9 + 3) \times 4 = 12 \times 4 = 48 c m^{2}$

∴ Curved surface area of the frustum cone $= 48 c m^{2}$

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