Question

A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone. Round your answers to the nearest hundredth.

Answer 1

Answer:

Surface area of the cone $\approx 52 in^{2}$

Volume of the cone $\approx 25 in^{2}$

Step-by-step explanation:

Inscribing a sphere into a cone in simple terms, means placing the sphere into the circular base of the right cone, in such a way that the external circumference of the sphere just touches the internal circumference of the base of the cone.

The fact that inscription can occur perfectly, shows that the radius of the sphere and the base circle of the cone are equal.

Hence, we can obtain the dimensions of the cone as:

radius, r = 2 inches

height, h = 6 inches.

Surface area of the cone is got by the following formula:

$A = \pi r (r+ \sqrt{h^{2}+r^{2}})$

substituting in the values, we will have

$A = \pi \times 2 (2+ \sqrt{6^{2}+2^{2}})=52.304 in^{2} \approx 52in^{2}$

Volume of the cone =

$V= \pi r^{2}\frac{h}{3}$

substituting in the values, we will have

$V= \pi 2^{2} \times \frac{6}{3} =25.13 \approx 25in^{2}$