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

Answer: (A)

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, $Areaof\Delta ABC=\frac{1}{2}\times 20\times OB=96⟹OB=9.6cm$