Question

A solid toy s in the form of a hemisphere surrounded by a right circular cone . The height of cone is 4 cm and the diameter of the base is 8 cm . Determine the volume of the toy. If a cube circumscribes the toy , then find the difference of the volumes of cube and the toy .

Answer 1

The height of the cone, h = 4 cm
Diameter of the base, d = 8 cm
Radius of the cone, r = 4 cm
lateral side will be
$$\begin{gathered} l=\sqrt{h^2+r^2} \\ l=\sqrt{{\left(4\right)}^2+{\left(4\right)}^2}=4\sqrt{2} \end{gathered}$$
Volume of the toy = volume of hemisphere + volume of cone
$$\begin{gathered} \Rightarrow V=\frac{2}{3}{\mathrm{\pi r}}^3+\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ =\frac{{\mathrm{\pi r}}^2}{3}\left[2\mathrm{r}+\mathrm{h}\right] \\ =\frac{\mathrm{\pi }{\left(4\right)}^2}{3}\left[2\times 4+4\right] \\ =201.14{\mathrm{cm}}^3 \end{gathered}$$
When the cube circumscribes the toy, then
Volume of the cube = $a^3={\left(8\right)}^3=512c{m}^3$
$\mathrm{Volume}\mathrm{of}\mathrm{cube}-\mathrm{volume}\mathrm{of}\mathrm{toy}=512-201.14=310.86c{m}^3$
Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cone
$$\begin{gathered} =2\mathrm{\pi }{r}^2+\mathrm{\pi }rl \\ =2\mathrm{\pi }{\left(4\right)}^2+\mathrm{\pi }\times 4\times 4\sqrt{2} \\ =\mathrm{\pi }{\left(4\right)}^2\left[2+\sqrt{2}\right] \\ =171.68{\mathrm{cm}}^2 \end{gathered}$$