A conical tent has an angle of 60° at the vertex. If the curved surface area is 100 m^2, then what is the volume of the tent?

  1. A. \frac{250\sqrt{2}}{\sqrt{3}\pi} m^3 ✓
  2. B. \frac{500\sqrt{2}}{\sqrt{3}\pi} m^3
  3. C. \frac{1000\sqrt{2}}{\sqrt{3}\pi} m^3
  4. D. \frac{1000\sqrt{3}}{\sqrt{2}\pi} m^3

Correct Answer: A. \frac{250\sqrt{2}}{\sqrt{3}\pi} m^3

Explanation

Since the vertical angle is 60^\circ, the semi-vertical angle is 30^\circ. We have \frac{r}{l} = \sin(30^\circ) = \frac{1}{2} \Rightarrow l = 2r. The curved surface area \pi r l = 100 \Rightarrow 2\pi r^2 = 100 \Rightarrow r^2 = \frac{50}{\pi}. The height h = \sqrt{l^2 - r^2} = \sqrt{3}r. Volume = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{50}{\pi}\right) \sqrt{3} \sqrt{\frac{50}{\pi}} = \frac{250\sqrt{2}}{\sqrt{3}\pi} m^3.

Related questions on Mensuration

Practice more CDS Elementary Mathematics questions