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Find the surface area of a frustum of a cone, whose larger and smaller radius is 12 and 5 m. The slant height of the cone is 20 m.

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519

π

m^{2}

509

π

m^{2}

529

π

m^{2}

539

π

m^{2}

answer is B.

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Detailed Solution

It is given larger and smaller radius is 12 and 5 m and slant height of the cone is 20 m.
Surface area of a frustum of cone =

π

(r + R)s +

π r^{2}

π R^{2}

\Rightarrow π (5 + 12) 20 + π 5^{2} + π 1 2^{2}

\Rightarrow π (17) 20 + 25 π + 144 π

\Rightarrow π (340 + 25 + 144)

\Rightarrow 509 π m^{2}

Therefore, surface area of a frustum of cone

509 π m^{2}

.
Hence, the correct option is 2.

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