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The larger diameter of a frustum cone is double the smaller diameter which is 24 cm. The slant height of the cone is 20 cm. What is the surface area? (Use $π$ = 3.14).

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4,721.6

{cm}^{2}

4,821.6

{cm}^{2}

4,621.6

{cm}^{2}

4,521.6

{cm}^{2}

answer is D.

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Detailed Solution

Given that,
Larger diameter of a frustum cone is double the smaller diameter which is 24 cm.
Larger diameter = 2

\times

smaller diameter

\Rightarrow

\times

\Rightarrow

48 cm
Smaller diameter = 24 cm
Diameter =

2 \times radius

Radius, R = 24 cm, r = 12 cms = 20 cm
Surface area of a frustum of cone =

π

(r + R)s +

{π r}^{2}

{π R}^{2}

\Rightarrow

π

(12 + 24)20 +

{π 12}^{2}

{π 24}^{2}

\Rightarrow

π

(36)20 + 144

π

+ 576

π

\Rightarrow

3.14

\times

(720+ 144 + 576)

\Rightarrow

3.14

\times

1,440

\Rightarrow

4,521.6

{cm}^{2}

Therefore, the surface area = 4,521.6

{cm}^{2}

.
Hence, the correct option is 4.

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