Question

A frustum is made by joining edges of following figure. The volume of the frustum thus formed will be $(\pi =22/7)$

• A. (44√2)/3
  
• B. $11\sqrt{2}$
• C. $\frac{11\sqrt{2}}{2}$
• D. $\frac{33\sqrt{2}}{4}$

Answer 1

The correct option is A (44√2)/3

ar $C_1=3\times 120\times \frac{\pi }{180}=2\pi$, ar $C_2=6\times 120\times \frac{\pi }{180}=4\pi$

$2\pi r_1=2\pi \Rightarrow r_1=1,r_2=2$

Volume of frustum $=\frac{1}{3}\pi r_2^2\times 4\sqrt{2}-\frac{1}{3}\pi r_1^2\times 2\sqrt{2}$

$=\frac{1}{3}(\pi \times 2^2\times 4\sqrt{2}-\pi \times 1^2\times 2\sqrt{2})$

$=\frac{1}{3}(16\sqrt{2}\pi -2\sqrt{2}\pi )=\frac{1}{3}\times 14\sqrt{2}\times \frac{22}{7}=\frac{44\sqrt{2}}{3}$