# Question #06a16

Oct 29, 2017

see explanation.

#### Explanation:

see Fig 1,
given that height $h$ of cone $= 20$ cm and slant height ${h}_{s} = 25$ cm,
let ${r}_{c}$ be the radius of the base of the cone,
${r}_{c} = \sqrt{A {C}^{2} - A {D}^{2}} = \sqrt{{25}^{2} - {20}^{2}} = 15$ cm
See Fig 2,
the largest possible hemisphere inside the cone touches the cone at $E$, so $A B$ is tangent to the hemisphere at $E$,
$\implies \Delta A E D \mathmr{and} \Delta A D B$ are similar,
$\implies \frac{E D}{A D} = \frac{D B}{A B}$
$\implies$ radius of hemisphere ${r}_{h} = E D = \frac{20 \times 15}{25} = 12$ cm

Volume of the cone ${V}_{c} = \frac{1}{3} \cdot \pi \cdot {r}_{c}^{2} \cdot h$
$= \frac{1}{3} \cdot \pi \cdot {15}^{2} \cdot 20 = 1500 \pi {\text{ cm}}^{3}$
Volume of hemisphere ${V}_{h} = \frac{2}{3} \cdot \pi \cdot {r}_{h}^{3}$
$= \frac{2}{3} \cdot \pi \cdot {12}^{3} = 1152 \pi {\text{ cm}}^{3}$
Volume of the remaining portion $= {V}_{c} - {V}_{h} = 1500 \pi - 1152 \pi = 348 \pi {\text{ cm}}^{3}$