The perimeters of the two circular ends of a frustum of a cone are 48 cm

Solution:

We have,

Perimeter of upper end, $C=48 \mathrm{~cm}$,

Perimeter of lower end, $c=36 \mathrm{~cm}$ and

Height, $h=11 \mathrm{~cm}$

Let the radius of upper end be $R$ and the radius of lower end be $r$.

As, $C=48 \mathrm{~cm}$

$\Rightarrow 2 \pi R=48$

$\Rightarrow R=\frac{48}{2 \pi}$

$\Rightarrow R=\frac{24}{\pi} \mathrm{cm}$

Similarly, $\mathrm{c}=36 \mathrm{~cm}$

$\Rightarrow r=\frac{36}{2 \pi}$

$\Rightarrow r=\frac{18}{\pi} \mathrm{cm}$

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

$=\sqrt{\left(\frac{24}{\pi}-\frac{18}{\pi}\right)^{2}+11^{2}}$

$=\sqrt{\left(\frac{6}{\pi}\right)^{2}+11^{2}}$

$=\sqrt{\left(\frac{6 \times 7}{22}\right)^{2}+11^{2}}$

$=\sqrt{\left(\frac{21}{11}\right)^{2}+11^{2}}$

$=\sqrt{\frac{441+121 \times 121}{121}}$

$=\sqrt{\frac{441+14641}{121}}$

$=\frac{\sqrt{15082}}{11} \mathrm{~cm}$

Now,

Volume of the frustum $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \pi \times 11 \times\left[\left(\frac{24}{\pi}\right)^{2}+\left(\frac{18}{\pi}\right)^{2}+\left(\frac{24}{\pi}\right) \times\left(\frac{18}{\pi}\right)\right]$

$=\frac{11 \pi}{3} \times\left[\frac{576}{\pi^{2}}+\frac{324}{\pi^{2}}+\frac{432}{\pi^{2}}\right]$

$=\frac{11 \pi}{3} \times \frac{1332}{\pi^{2}}$

$=\frac{11}{3} \times \frac{1332}{\pi}$

$=\frac{11}{3} \times \frac{1332 \times 7}{22}$

$=1554 \mathrm{~cm}^{3}$

Also,

Curved surface area of the frustum $=\pi(R+r) l$

$=\frac{22}{7} \times\left(\frac{24}{\pi}+\frac{18}{\pi}\right) \times \frac{\sqrt{15082}}{11}$

$=\frac{22}{7} \times \frac{42}{\pi} \times \frac{\sqrt{15082}}{11}$

$=\frac{22}{7} \times \frac{42 \times 7}{22} \times \frac{\sqrt{15082}}{11}$

$\approx 42 \times 11.164436$

$\approx 468.91 \mathrm{~cm}^{2}$