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A right triangle, whose sides are $ 3 \mathrm{~cm} $ and $ 4 \mathrm{~cm} $ (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of $ \pi $ as found appropriate.)
Given:
A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse.
To do:
We have to find the volume and surface area of the double cone so formed.
Solution:
Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of the right triangle $ABC$.
This implies, using Pythagoras theorem,
$AC^2=AB^2+BC^2$
$AC^2=3^2+4^2$
$=9+16$
$=25$
$\Rightarrow AC=\sqrt{25}$
$=5\ cm$
When the triangle $ABC$ is revolved about the hypotenuse $AC$, the following double cone is formed.
In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BDC}$,
$\angle \mathrm{ABC}=\angle \mathrm{CDB}=90^{\circ}$ ($\mathrm{BD} \perp \mathrm{AC})$)
$\angle \mathrm{BCA}=\angle \mathrm{BCD}$ (Common)
Therefore, by AA similarity,
$\triangle \mathrm{ABC} \sim \triangle \mathrm{BDC}$
This implies,
$\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$ (Corresponding sides of similar triangles are in
proportion)
$\mathrm{BD}=\frac{\mathrm{AB} \times \mathrm{BC}}{\mathrm{AC}}$
$=\frac{3 \mathrm{~cm} \times 4 \mathrm{~cm}}{5 \mathrm{~cm}}$
$=\frac{12}{5} \mathrm{~cm}$
$=2.4 \mathrm{~cm}$
We know that,
Volume of a cone of height $h$ and radius $r$ is $\frac{1}{3} \pi r^{2} h$
The volume of the double cone in the figure $=$ Volume of cone ABB' $+$ Volume of cone BCB'
$=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AD}+\frac{1}{3} \pi(\mathrm{BD})^{2} \times \mathrm{DC}$
$=\frac{1}{3}\times \pi(\mathrm{BD})^{2}(\mathrm{AD}+\mathrm{DC})$
$=\frac{1}{3} \times \pi(\mathrm{BD})^{2} \times \mathrm{AC})$
$=\frac{1}{3}\times 3.14 \times (2.4)^2 \mathrm{~cm} \times 5 \mathrm{~cm}$
$=30.144 \mathrm{~cm}^{3}$
The curved surface area of a cone $=\pi rl$
The curved surface area of the double cone $=$ curved surface area of the cone ABB' $+$ curved surface area of the cone BCB'
$=\pi \times \mathrm{BD} \times \mathrm{AB}+\pi \times \mathrm{BD} \times \mathrm{BC}$
$=\pi \times \mathrm{BD}(\mathrm{AB}+\mathrm{BC})$
$=3.14 \times 2.4 \mathrm{~cm} \times(3 \mathrm{~cm}+4 \mathrm{~cm})$
$=3.14 \times 2.4 \mathrm{~cm} \times 7 \mathrm{~cm}$
$=52.752 \mathrm{~cm}{ }^{2}$
$=52.75 \mathrm{~cm}{ }^{2}$
The volume and surface area of the double cone so formed are $30.144 \mathrm{~cm}^{3}$ and $52.75 \mathrm{~cm}{ }^{2}$ respectively.
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