Question

The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, find its volume and curved surface area.

Answer 1

We have,

Perimeter of upper end, C = 48 cm,

Perimeter of lower end, c = 36 cm and

Height, h = 11 cm

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

As, C = 48 cm

$\Rightarrow 2\pi r=48\Rightarrow R=\frac{48}{2\pi }\Rightarrow R=\frac{24}{\pi }$ cm

Similarly, c = 36 cm

$\Rightarrow r=\frac{36}{2\pi }\Rightarrow r=\frac{18}{\pi }cm$

And, $l=\sqrt{(R-r{)}^2+h^2}=\sqrt{(\frac{24}{\pi }-\frac{18}{\pi }{)}^2+{11}^2}=\sqrt{(\frac{6}{\pi }{)}^2+{11}^2}=\sqrt{(\frac{6\times 7}{22}{)}^2+{11}^2}=\sqrt{(\frac{21}{11}{)}^2+{11}^2}=\sqrt{\frac{441+121\times 121}{121}}=\sqrt{\frac{441+14641}{121}}=\frac{\sqrt{15082}}{11}cm$
Now,

Volume of the frustum $=\frac{1}{3}\pi h(R^2+r^2+Rr)=\frac{1}{3}\times \pi \times 11\times \left[\right(\frac{24}{\pi }{)}^2+(\frac{18}{\pi }{)}^2+(\frac{24}{\pi })\times (\frac{18}{\pi }\left)\right]=\frac{11\pi }{3}\times [\frac{576}{{\pi }^2}+\frac{324}{{\pi }^2}+\frac{432}{{\pi }^2}]=\frac{11\pi }{3}\times \frac{1332}{\pi 2}=\frac{11}{3}\times \frac{1332}{\pi }=\frac{11}{3}\times \frac{1332\times 7}{22}=1554c{m}^3$

Also,

Curved surface area of the frustum $=\pi (R+r)l=\frac{22}{7}\times (\frac{22}{\pi }+\frac{18}{\pi })\times \frac{\sqrt{15082}}{11}=\frac{22}{7}\times \frac{42}{\pi }\times \frac{\sqrt{15082}}{11}=\frac{22}{7}\times \frac{42\times 7}{22}\times \frac{\sqrt{15082}}{11}\approx 42\times 11.164436\approx 468.91c{m}^2$