A solid toy is in the form of a hemisphere surmounted by a right circular cone. Th height of cone is $ 4 \mathrm{~cm} $ and the diameter of the base is $ 8 \mathrm{~cm} $. Determine the volume the toy. If a cube circumscribes the toy, then find the difference of the volume cube and the toy. Also, find the total surface area of the toy.

Given:

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Th height of cone is \( 4 \mathrm{~cm} \) and the diameter of the base is \( 8 \mathrm{~cm} \). 

A cube circumscribes the toy.

To do:

We have to find the volume the toy, the difference of the volumes of cube and the toy and the total surface area of the toy.

Solution:

Let $r$ be the radius of the hemisphere and the cone and $h$ be the height of the cone.

Volume of the toy $=$ Volume of the hemisphere $+$ Volume of the cone

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

$=(\frac{2}{3} \times \frac{22}{7} \times 4^{3}+\frac{1}{3} \times \frac{22}{7} \times 4^{2} \times 4)$

$=\frac{1408}{7} \mathrm{cm}^{3}$

The edge of the cube $=8\ cm$

Volume of the cube $= 8^3\ cm3$

$= 512\ cm^3$

Difference in the volume of the cube and the toy $=(512-\frac{1408}{7}) \mathrm{cm}^{3}$

$=310.86 \mathrm{~cm}^{3}$

Total surface area of the toy $=$ Curved surface area of the cone $+$ Curved surface area of the hemisphere

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

$=\pi r(l+2 r)$

$=\frac{22}{7} \times 4 \sqrt{4^{2}+4^{2}}+2 \times 4$                   [Since $l=\sqrt{r^2+h^2}$]

$=\frac{22}{7} \times 4 \times 4\sqrt{2}+8$

$=\frac{88 \times 4}{7} \sqrt{2}+2$

$=171.68 \mathrm{~cm}^{2}$.

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Updated on: 10-Oct-2022

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