A sector of radius 10.5 cm with the central angle 120° is folded to form a cone by joining the two bounding radii of the sector. What is the volume (in cm3) of the cone so formed?

Answer: Option A

Solution

Answer: Option A

Solution:

$$\begin{array}{rl}& \text{Length of Arc}=L\\ & L=2\times \frac{22}{7}\times \frac{r\times 120}{360}\\ & L=2\times \frac{22}{7}\times 10.5\times \frac{1}{3}\\ & L=22\end{array}$$

$$\begin{array}{rl}& 2\pi R=22\\ & 2\times \frac{22}{7}R=22\\ & R=3.5\\ & h^2+R^2=10.5^2\\ & h^2+3.5^2=10.5^2\\ & h=7\sqrt{2}\\ & V=\frac{1}{3}\pi r^{2}h\\ & V=\frac{1}{3}\pi \times 3.5\times 3.5\times 7\sqrt{2}\\ & V=\frac{343\sqrt{2}}{12}\pi \end{array}$$