Answer:
1) [tex]h=\frac{8\sqrt{3}}{3}\ units[/tex]
2) [tex]V=\frac{2,048\sqrt{3}}{9}\ units^3[/tex]
Step-by-step explanation:
Part 1) Find the height of the pyramid
we know that
[tex]tan(30^o)=\frac{h}{(b/2)}[/tex]
where
h is the height of the pyramid
b is the length side of the square base
we have
[tex]tan(30^o)=\frac{\sqrt{3}}{3}[/tex]
[tex]b=16\ units[/tex]
substitute
[tex]\frac{\sqrt{3}}{3}=\frac{h}{(16/2)}[/tex]
[tex]\frac{\sqrt{3}}{3}=\frac{h}{8}[/tex]
[tex]h=\frac{8\sqrt{3}}{3}\ units[/tex]
Part 2) Find the volume of the pyramid
we know that
The volume of the pyramid is equal to
[tex]V=\frac{1}{3} Bh[/tex]
where
B is the area of the square base
h is the height of the pyramid
we have
[tex]B=(16^2)=256\ units^2[/tex]
[tex]h=\frac{8\sqrt{3}}{3}\ units[/tex]
substitute
[tex]V=\frac{1}{3} (256)(\frac{8\sqrt{3}}{3})[/tex]
[tex]V=\frac{2,048\sqrt{3}}{9}\ units^3[/tex]
