Question

A right angled triangle, whose sides are $3$ cm and $4$ cm (other than hypotenuse) is made to revolve about its hypotenuse. The volume of the double cone so formed is _________.

• A. $9\pi {\text{cm}}^3$
• B. $\frac{48}{5}\pi {\text{cm}}^3$
• C. $10\pi {\text{cm}}^3$
• D. $\frac{46}{5}\pi {\text{cm}}^3$

Answer 1

The correct option is B $\frac{48}{5}\pi {\text{cm}}^3$

Refer figure,

In right angle $△ABC$ using Pythagoras theorem

$AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{4^2+3^2}=\sqrt{25}=5cm$

$area\left(△ABC\right)=\frac{1}{2}\times AB\times AC$

$\Rightarrow \frac{1}{2}\times AC\times OB=\frac{1}{2}\times 4\times 3\Rightarrow \frac{1}{2}\times 5\times OB=6$

$\Rightarrow OB=\frac{12}{5}cm$

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

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

$=\frac{1}{3}\pi r^2(OA+OC)=\frac{1}{3}\times \pi \times {\left(\frac{12}{5}\right)}^2\times AC$

$=\frac{1}{3}\times \pi \times \frac{144}{25}\times 5=\frac{48\pi }{5}c{m}^3$