The circumference of lower and upper bases of the frustum cone is 3.5 and 1.5 ft. The slant height is 5 ft. What is the curved surface area ?

12.5

f t^{2}

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72.5

f t^{2}

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11.5

f t^{2}

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10.5

f t^{2}

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Solution

The correct option is C 12.5 $f t^{2}$
Curved surface area of the frustum cone
$= \frac{1}{2} (2 π R + 2 π r) l$
= $\frac{1}{2} (3.5 + 1.5) \times 5$
= $5 \times 2.5$
= 12.5 $f t^{2}$

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