Question

A right circular triangles with sides $12$ cm and $16$cm is revolved around its hypotenuse. Find the volume of the double cone so formed.

• A. $1930.97c{m}^3$
• B. $1234.97c{m}^3$
• C. $1710.95c{m}^3$
• D. $1536.2c{m}^3$

Answer 1

The correct option is A $1930.97c{m}^3$

Given $AB$ is $12cm$ and $BC$ is $16cm$. So, by Pythagoras theorem, $AC$ is $20cm.$
Having $AB$ as height and $BC$ as base, the area of $\Delta ABC=\frac{1}{2}\times 12\times 16=96sq.cm$
Having OB as height and AC as base, area of ABC is same as 96 sq cm.
Hence, $Areaof\Delta ABC=\frac{1}{2}\times 20\times OB=96⟹OB=9.6cm$
In triangle AOB, ${AO}^2+{OB}^2={AB}^2⟹{AO}^2+{9.6}^2={12}^2⟹{AO}^2={12}^2-{9.6}^2=51.84orAO=7.2cm$
Hence, OC $=20-7.2=12.8cm$.
Volume of double cone $=volumeofconeI+volumeofconeII=\frac{1}{3}\pi \times {9.6}^2\times 7.2+\frac{1}{3}\pi \times {9.6}^2\times 12.8=1930.97c{m}^3$