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

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

Since ΔAPO' and ΔAOQ are similar so

$\frac{AO}{AO'}= \frac{OQ}{O'P}$

$\frac{9}{6}= \frac{3}{O'P}$

O'P = $\frac{3\times6}{9}=2$ units

Therefore, side MN = 2× O'P = 4 units

Now we put these values in the formula

Volume of the frustum = $\frac{1}{3}(6\times6)(9)}-\frac{1}{3}(4\times4)(6)$

= $\frac{1}{3}(324)-(\frac{1}{3}\times96)$

= $\frac{324}{3}-\frac{96}{3}$

= $\frac{324-96}{3}=\frac{228}{3}$

= 76 unit³