Question

25. Find the midsection of a frustum of a square pyramid if its lower base is 20 cm on a side, its upper base is 12 cm on a side, and the altitude of the frustum is 14 cm.

Answer 1

Given $a=20cm, b=12cm,h=14cm$

$$\dfrac{x}{b/2}=\dfrac{x+h/2}{m/2}=\dfrac{x+h}{a/2}$$

$$mx=bx+bh/2$$

$$ax=bx+bh$$

$$x=\dfrac{bh}{a-b}$$

$$m=b+\dfrac{bh}{2bh}(a-b)$$

$$m=\dfrac{20+12}{2}=16(cm)$$

$$A_m=m^2=(16)^2=256(cm^2)$$

$256\ cm^2$