Question

The slant height of the frustum of a cone is $4cm$ and the perimeters of its circular ends are $18cm$ and $6cm$. Find the area of its whole surface and volume.

Answer 1

Given,

Perimeter of top circle having radius (r) is

$2\pi r=6cm$

Perimeter of bottom circle having radius (R) is

$=18cm$

Slant height of frustum of cone $l=4cm$

Hence,

Radius of top circle $r=\frac{6}{2\pi }=\frac{3}{\pi }cm$

Radius of bottom circle $R=\frac{18}{2\pi }=\frac{9}{\pi }cm$

Hight of frustum of cone is

$h=\sqrt{l^2-(R-r{)}^2}$

$h=\sqrt{4^2-(\frac{9}{\pi }-\frac{3}{\pi }{)}^2}cm$

$h=\sqrt{16-(\frac{6}{\pi }{)}^2}cm$

$h=\sqrt{16-\frac{36}{{\pi }^2}}cm$

$h=\sqrt{12.35244}cm$

$h=3.5146cm$

$\text{curved surface area of frustum of cone}$ $=\pi (R+r)l$

$=((R+r)l)\pi$

$=((\frac{9}{\pi }+\frac{3}{\pi })\times 4)\pi c{m}^2$

$=((\frac{12}{\pi })\times 4)\pi c{m}^2$

$=(\frac{48}{\pi })\pi c{m}^2$

$=48c{m}^2$

$\text{Total surface area of the frustum of cone}=48c{m}^2$