# 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.

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Given:

The slant height of the frustum of a cone is $4 \mathrm{~cm}$ and the perimeters of its circular ends are $18 \mathrm{~cm}$ and $6 \mathrm{~cm}$.

To do:

We have to find the curved surface of the frustum.

Solution:

The perimeter of the top of frustum $= 18\ cm$

Let the top radius be $r_1$

This implies,

$2 \pi r_1=18$

$r_{1}=\frac{18 \times 7}{2 \times 22}$

$=\frac{63}{22} \mathrm{~cm}$

Perimeter of the bottom $=6 \mathrm{~cm}$

Let the bottom radius be $r_2$.

$2 \pi r_2=6$ $r_{2}=\frac{6 \times 7}{2 \times 22}$

$=\frac{21}{22} \mathrm{~cm}$

Slant height of the frustum $l=4 \mathrm{~cm}$

Curved surface area of the frustum $=\pi(r_{1}+r_{2}) l$

$=\frac{22}{7}(\frac{63}{22}+\frac{21}{22}) \times 4$

$=\frac{22}{7} \times \frac{84}{22} \times 4$

$=48 \mathrm{~cm}^{2}$

The curved surface of the frustum is $48\ cm^2$.

Updated on 10-Oct-2022 13:25:34