A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is $ 21 \mathrm{~cm} $ and its volume is $ 2 / 3 $ of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. (Use $ \pi=22 / 7) $

Given:

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere.

The radius of the base of the cone is \( 21 \mathrm{~cm} \) and its volume is \( 2 / 3 \) of the volume of the hemisphere.

To do:

We have to find the height of the cone and the surface area of the toy.

Solution:

Radius of the base of the conical part $r = 21\ cm$

Volume of the cone $=\frac{2}{3} \times$ Volume of the hemisphere

Let $h$ be the height of the conical part.

Volume of hemispherical part $=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \frac{22}{7} \times (21)^3$

$=19404 \mathrm{~cm}^{3}$

This implies,

Volume of the conical part $=\frac{1}{3} \pi r^{2} h$

Therefore,

$\frac{1}{3} \pi r^{2} h=\frac{2}{3} \times 19404$

$\frac{1}{3} \pi (21)^{2} h=12936 \mathrm{~cm}^{3}$

$h=\frac{12936 \times 3 \times 7}{22 \times 1 \times (21)^2}$

$=28 \mathrm{~cm}$

Surface area of the toy $=$ Curved surface are of the conical part $+$ Surface area of the hemispherical part

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

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

$=\pi r[\sqrt{h^{2}+r^{2}}+2 r]$             [Since $l=\sqrt{h^{2}+r^{2}}$]

$=\frac{22}{7} \times 21[\sqrt{(28)^{2}+(21)^{2}}+2 \times 21]$

$=66[\sqrt{784+441}+42]$

$=66[\sqrt{1225}+42]$

$=66[35+42]$

$=66 \times 77$

$=5082 \mathrm{~cm}^{2}$

The height of the cone is $28\ cm$ and the surface area of the toy is $5082\ cm^2$.

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Updated on: 10-Oct-2022

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