Question

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $4cm$ and diameter of the base is $8cm$. Determine the volume of the toy. If a cube circumstances the toy, then find the difference of the volumes of cube and the toy. Also, find the total surface area of the toy.

Answer 1

Let $H=4cm$ and $R=\frac{8}{2}=4cm$ be the height and radius of the cone respectively.

Let $r=4cm$ be the radius of the hemisphere

Therefore, Volume of toy $=$ Volume of cone $+$ Volume of hemisphere

$=\frac{1}{3}\pi R^{2}H+\frac{2}{3}\pi r^3$

$=\frac{1}{3}\pi (4{)}^3+\frac{2}{3}\pi (4{)}^3$

$=\frac{22}{7}\times 64\times (\frac{1}{3}+\frac{2}{3})$

$=\frac{22}{7}\times 64\times 1$

$=201.14c{m}^3$

Since, the cube circumscribes the toy, therefore, side of the cube will be equal height and diameter of the toy ie $8cm$

Volume of cube $=8^3=512c{m}^3$

Thus, difference in volumes $=$ Volume of cube $-$ Volume of toy

$=512-201.14$

$=310.86c{m}^3$

Now, TSA of toy $=$ CSA of cone $+$ CSA of hemisphere

$\therefore$ Slant height $l=\sqrt{R^2+H^2}=5.7cm$

Now, TSA of toy $=\pi Rl+2\pi r^2$

$=\frac{22}{7}[4\times 5.7+2\times 4^2]$

$=\frac{22}{7}\left[54.8\right]$

$=172c{m}^2$