Question

Please help me!!! Geometry question.

Answer 1

Answer:

The volume of the figure is $(\frac{l^{3}}{3})[\frac{\pi }{2}-1]\ units^{3}$

Step-by-step explanation:

we know that

The volume of the figure is equal to the volume of the cone minus the volume of the square pyramid

step 1

Find the volume of the cone

The volume of the cone is equal to

$V=\frac{1}{3}\pi r^{2}h$

we have

$r=\frac{l\sqrt{2}}{2}\ units$ ----> the diagonal of the square base of pyramid is equal to the diameter of the cone

$h=l\ units$

substitute

$V=\frac{1}{3}\pi (\frac{l\sqrt{2}}{2})^{2}(l)$

$V=\frac{1}{6}\pi (l)^{3}\ units^{3}$  

step 2

Find the volume of the square pyramid

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$

where

B is the area of the base

h is the height of the pyramid

we have

$h=l\ units$

$B=l^{2}\ units^{2}$

substitute

$V=\frac{1}{3}l^{2}(l)$

$V=\frac{1}{3}l^{3}\ units^{3}$

step 3

Find the volume of the figure

$\frac{1}{6}\pi (l)^{3}\ units^{3}-\frac{1}{3}l^{3}\ units^{3}=(\frac{l^{3}}{3})[\frac{\pi }{2}-1]\ units^{3}$