Question

From the top of a cone of base radius $24$cm and height $45$cm, a cone of slant height $17$cm is cut off. What is the volume of the remaining frustum of the cone?

Answer 1

$L^2=R^2+H^2\Rightarrow L=\sqrt{{24}^2+{45}^2}=\sqrt{2601}=51cm$

we know that, $△OAB\sim △OCD$, therefore

$\frac{r}{24}=\frac{h}{45}=\frac{17}{51}$

$\Rightarrow \frac{r}{24}=\frac{17}{51}\Rightarrow r=\frac{24}{3}=8cm$

$\Rightarrow \frac{h}{45}=\frac{17}{51}\Rightarrow h=\frac{45}{3}=15cm$

$\therefore$ Height of frustum $=45-15=30cm$

$\therefore$ Volume of frustum = $\frac{1}{3}\pi (R^2+r^2+Rr)h=\frac{1}{3}\times \frac{22}{7}\times 30({24}^2+8^2+24\times 8)$

= $\frac{220}{7}\times 832=26148.57{cm}^3$