Question

Find the volume and lateral surface area of the frustum of a regular square pyramid whose altitude is 38 cm and whose base edges are 10 cm and 20 cm respectively.

FINAL ANSWERS: _____________________ ______________________

Answer 1

$V = \frac {1}{3}h(S_1+\sqrt{S_1S_2}+S_2)$

$\text{where h is the height of the pyramid;} S_1,S_2\text{the area of the bases}$

$\text{since a regular square pyramid:}$

$S_1 = 10*10 =100$

$S_2 = 20*20=400$

$V = \frac {1}{3}*38*(100+\sqrt{100*400}+400) \approx8867\ cm^3$

$S_{sur}= \frac{1}{2}(p_1+p_2)*f$

$\text{where } p_1,p_2 \text{ the perimeters of the bases } f \text{ apothem}$

$p_1 = 10 *4 =40$

$p_2 = 20*4=80$

$f = \sqrt{h^2+{[\frac{b-a}{2}]}^2}$

$\text{where }h \text{ is the height of the pyramid}$

$a,b \text{ base side length}$

$f = \sqrt{38^2+{[\frac{20-10}{2}]}^2}\approx38.33$

$S_{sur}= \frac{1}{2}(40+80)*38.33\approx2299.8\ cm^2$

Aswer: $V \approx8867\ cm^3$ $S_{sur}\approx2299.8\ cm^2$