Question

A regular square pyramid has base edges of length 16 and its lateral faces are inclined 30­° to the base of the pyramid. What is the: 1) height of the pyramid? 2) volume of the pyramid?

Answer 1

Answer:

1) $h=\frac{8\sqrt{3}}{3}\ units$

2) $V=\frac{2,048\sqrt{3}}{9}\ units^3$

Step-by-step explanation:

Part 1) Find the height of the pyramid

we know that

$tan(30^o)=\frac{h}{(b/2)}$

where

h is the height of the pyramid

b is the length side of the square base

we have

$tan(30^o)=\frac{\sqrt{3}}{3}$

$b=16\ units$

substitute

$\frac{\sqrt{3}}{3}=\frac{h}{(16/2)}$

$\frac{\sqrt{3}}{3}=\frac{h}{8}$

$h=\frac{8\sqrt{3}}{3}\ units$

Part 2) Find the volume of the pyramid

we know that

The volume of the pyramid is equal to

$V=\frac{1}{3} Bh$

where

B is the area of the square base

h is the height of the pyramid

we have

$B=(16^2)=256\ units^2$

$h=\frac{8\sqrt{3}}{3}\ units$

substitute

$V=\frac{1}{3} (256)(\frac{8\sqrt{3}}{3})$

$V=\frac{2,048\sqrt{3}}{9}\ units^3$