Question

A right-angled triangle whose sides are $3cm$ and $4cm$ (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose a value of $\pi$ as found appropriate.)

Answer 1

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse, AC = $\sqrt{3^2+4^2}$
$=\sqrt{9+16}=\sqrt{25}=5$ cm

Area of $\Delta ABC=\frac{1}{2}\times AB\times AC$

$\Rightarrow \frac{1}{2}\times AC\times DB=\frac{1}{2}\times 4\times 3$

$\Rightarrow \frac{1}{2}\times 5\times DB=6$

So, DB = $\frac{12}{5}=2.4$ cm

The volume of double cone = Volume of cone 1 + Volume of cone 2

$=\frac{1}{3}\pi r^2{h}_1+\frac{1}{3}\pi r^2{h}_2$

$=\frac{1}{3}\pi r^2[h_1+h_2]=\frac{1}{3}\pi r^2[DA+DC]$

$=\frac{1}{3}\times 3.14\times {2.4}^2\times 5$

$=30.14c{m}^3$

The surface area of double cone = Surface area of cone 1 + Surface area of cone 2

$=\pi r{l}_1+\pi r{l}_2$

$=\pi r[4+3]=3.14\times 2.4\times 7$

$=52.75$ cm${}^2$