Answer:
Volume of the frustum is 76 unit³
Step-by-step explanation:
From the figure attached we have to calculate the volume of the frustum formed by cutting off a square pyramid from the top.
Volume of the frustum = [tex]\frac{1}{3}\text{(area of the base with side 6 units)(height)}-\frac{1}{3}\text{(area of the base with side MN)}(Height)[/tex]
Since ΔAPO' and ΔAOQ are similar so
[tex]\frac{AO}{AO'}= \frac{OQ}{O'P}[/tex]
[tex]\frac{9}{6}= \frac{3}{O'P}[/tex]
O'P = [tex]\frac{3\times6}{9}=2[/tex] units
Therefore, side MN = 2× O'P = 4 units
Now we put these values in the formula
Volume of the frustum = [tex]\frac{1}{3}(6\times6)(9)}-\frac{1}{3}(4\times4)(6)[/tex]
= [tex]\frac{1}{3}(324)-(\frac{1}{3}\times96)[/tex]
= [tex]\frac{324}{3}-\frac{96}{3}[/tex]
= [tex]\frac{324-96}{3}=\frac{228}{3}[/tex]
= 76 unit³
