Question

PROBLEM 3. Find the radius of a cone of revolution of slant height 6 inches in which the total area is twice the lateral area.
FINAL ANSWER: _______________________________________

Answer 1

Consider a cone of radius $r$ , height $h$ and slant height $l$ .

The total area of the cone is given by $A_{T}=\pi r^2 +\pi r l$

The lateral area of the cone is $A_{L}=\pi r l$

It is given that, $A_{T}=2A_{L}$

Therefore,

$\pi r^2+\pi r l=2\pi r l$

$\pi r^2=\pi r l$

$r^2=rl$

$r=l=6$

So, the radius of the cone is $r=6$ inches.

Here, the radius of the cone is equal to the slant height of the cone, which is practically impossible as the slant height must be larger than the radius or vertical height of the cone.

So, the possible question may be "lateral area is twice the base area of the cone".

The lateral area of the cone is $A_L=\pi r l$

Area of circular base of radius $r$ is $A_B=\pi r^2$

Given,

$A_L=2A_B$

$\pi r l=2 \pi r^2$

$l=2r$

$r=\frac{l}{2}$

$r=\frac{6}{2}=3$

Therefore, the radius of the cone is $r=3$ inches.