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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.
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 $48\ cm^2$.