# Question #75d5c

Feb 11, 2018

$V 2 = \frac{3}{4} V 1$ has been proved as shown below

#### Explanation:

A cone circumscribes a sphere and has its slant height equal to the diameter of its base. Show that the volume of the cone is 3/4 the volume of the sphere?

Given:
Slant height equal to diameter of the base.
ie,
the cone has 2 slant heights and one base which are all equal, and forming an equilateral angle
The angle at the vertex is 60 degrees
Semi vertex angle is $\frac{1}{2} \left(60\right) = 30 \mathrm{de} g r e e s$
If the slant height is $l$
radius of the circumscribed sphere is $r$
Angle between the radii is $\theta = 120 \mathrm{de} g r e e s$
Thus ${l}^{2} = {r}^{2} + {r}^{2} - 2 r r \cos \left(120\right)$
$\cos 120 = - \frac{1}{2}$
Now, ${l}^{2} = {r}^{2} + {r}^{2} - 2 r r \left(- \frac{1}{2}\right)$
${l}^{2} = 3 {r}^{2}$
$l = \left(\sqrt{3}\right) r$
Volume of a sphere is V1

$V 1 = \frac{4}{3} \pi {r}^{3}$
radius of the cone is $a = \frac{l}{2}$
${a}^{2} = {\left(\frac{l}{2}\right)}^{2}$
${a}^{2} = {l}^{2} / 4$
${l}^{2} = 3 {r}^{2}$
${a}^{2} = 3 {r}^{2} / 4$
Height of the cone is given by
${l}^{2} = {\left(\frac{l}{2}\right)}^{2} + {h}^{2}$
${h}^{2} = 3 {l}^{2} / 4$
$h = \frac{\sqrt{3}}{2} l$
$h = \frac{\sqrt{3}}{2} \sqrt{3} r$
$h = \frac{3}{2} r$
Volume of the cone is V2
$V 2 = \frac{1}{3} \pi {a}^{2} h$
$V 2 = \frac{1}{3} \pi \frac{3 {r}^{2}}{4} \left(\frac{3}{2} r\right)$
$V 2 = \pi {r}^{3}$

Thus,
Volume of sphere is $V 1 = \frac{4}{3} \pi {r}^{3}$
Volume of the cone is $V 2 = \pi {r}^{3}$
$V 2 = \frac{3}{4} \left(\frac{4}{3}\right) \pi {r}^{3}$
$V 2 = \frac{3}{4} V 1$