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Find the surface area of a frustum of a cone, whose larger and smaller radius is 10 and 2 mm. The slant height of the cone is 15 mm.

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274

π

{mm}^{2}

264

π

{mm}^{2}

284

π

{mm}^{2}

254

π

{mm}^{2}

answer is C.

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Detailed Solution

It is given larger and smaller radius is 10 and 2 mm and slant height of the cone is 15 mm.
Surface area of a frustum of cone =

π

(r + R) s +

π

r^{2}

+

π

R^{2}

\Rightarrow (2 + 10) \times 15 + 2 + 10

\Rightarrow π (12) 15 + 4 π + 100 π

\Rightarrow π (180 + 4 + 100)

\Rightarrow 284 π

mm^{2}

Therefore, the surface area of a frustum of cone is

284 πm m^{2}

.
Hence, the correct option is 3.

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