A circular sheet of paper of radius 6 inches is cut into 4 equal sectors, and each sector is formed into a cone with no overlap. What is the height in inches of the cone?
+14903 A circular sheet of paper of radius 6 inches is cut into 4 equal sectors, and each sector is formed into a cone with no overlap. What is the height in inches of the cone?
Hello Guest!
\(C_{cone}=\frac{2\cdot \pi \cdot 6in}{4}=\pi \cdot 3in\)
\(r_{cone}=\frac{C_{cone}}{2\cdot \pi}=\frac{\pi \cdot 3in}{2\cdot \pi}\\ r_{cone}=\frac{3}{2}in\)
\(h_{cone}=\sqrt{(6in)^2-(r_{cone})^2}=\sqrt{36in^2-(\frac{3}{2})^2in^2}\)
\(h_{cone}=5.809\ inches\)
The height of the cone in inches is 5.809 inches.
!
