The slant height of a frustum of a cone is 4 | vedclass

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(B) Let the radii of the circular ends of the frustum be $r_1$ and $r_2$ respectively.Given that the perimeters of the circular ends are $18 \,cm$ and

Vedclass · Accepted Answer

$48$

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(B) Let the radii of the circular ends of the frustum be $r_1$ and $r_2$ respectively.Given that the perimeters of the circular ends are $18 \,cm$ and $6 \,cm$.So,$2 \pi r_1 = 18 \implies r_1 = \frac{9}{\pi} \,cm$.And $2 \pi r_2 = 6 \implies r_2 = \frac{3}{\pi} \,cm$.The slant height $(l)$ of the frustum is given as $4 \,cm$.The formula for the curved surface area $(CSA)$ of a frustum of a cone is $CSA = \pi (r_1 + r_2) l$.Substituting the values:$CSA = \pi \left( \frac{9}{\pi} + \frac{3}{\pi} ight) 	imes 4$$CSA = \pi \left( \frac{12}{\pi} ight) 	imes 4$$CSA = 12 	imes 4 = 48 \,cm^2$.Thus,the curved surface area of the frustum is $48 \,cm^2$.

The slant height of a frustum of a cone is $4 \,cm$ and the perimeters (circumference) of its circular ends are $18 \,cm$ and $6 \,cm$. Find the curved surface area of the frustum in $cm^2$.

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