Question

Problem1 a right square frustum is formed by cutting a right square pyramid with a plane parallel to its base. Suppose the original pyramid has base length 6 and height 9. and that plane cutting the pyramid to form the frustum is 3 units from the base of pyramid. What is the volume, in cubic units, of the right square frustum that is cut off by this plane?

Answer 1

Volume of the frustum is 76 unit³

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 =
`(1)/(3)\text{(area of the base with side 6 units)(height)}-(1)/(3)\text{(area of the base with side MN)}(Height)`

Since ΔAPO' and ΔAOQ are similar so

`(AO)/(AO')= (OQ)/(O'P)`

`(9)/(6)= (3)/(O'P)`

O'P =
`(3*6)/(9)=2` units

Therefore, side MN = 2× O'P = 4 units

Now we put these values in the formula

Volume of the frustum =
`(1)/(3)(6*6)(9)}-(1)/(3)(4*4)(6)`

=
`(1)/(3)(324)-((1)/(3)*96)`

=
`(324)/(3)-(96)/(3)`

=
`(324-96)/(3)=(228)/(3)`

= 76 unit³