A three-quarter sector of a circle of radius $4$ inches together with its interior can be rolled up to form the lateral surface of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches? $\textbf{(A)}\ 3\pi \sqrt5 \qquad$ $\textbf{(B)}\ 4\pi \sqrt3 \qquad$ $\textbf{(C)}\ 3 \pi \sqrt7 \qquad$ $\textbf{(D)}\ 6\pi \sqrt3 \qquad$ $\textbf{(E)}\ 6\pi \sqrt7$
Solution
Arc length of the $\frac34$ sector of radius $4$ is $\frac34\cdot2\pi(4)=6\pi$, equal to the base circumference $2\pi r$, so $r=3$.
Slant height is $4$, so height is $\sqrt{16-9}=\sqrt7$.
Volume $=\frac13\pi(9)(\sqrt7)=3\pi\sqrt7$. The answer is (\textbf{C}).
A circle of radius $3\sqrt3$ inches has a $120^\circ$ sector cut out. The remaining $240^\circ$ sector, together with its interior, is rolled up to form the lateral surface of a right circular cone by taping its two straight edges together. What is the volume of the cone in cubic inches?
$\textbf{(A)}\ 3\pi \sqrt5 \qquad$ $\textbf{(B)}\ 4\pi \sqrt3 \qquad$ $\textbf{(C)}\ 3 \pi \sqrt7 \qquad$ $\textbf{(D)}\ 6\pi \sqrt3 \qquad$ $\textbf{(E)}\ 6\pi \sqrt7$ 
