
- Пирамида.png (71.49 KiB) Прегледано 295 пъти
$\frac{H}{a}=tg60^\circ=\sqrt{3};\ a=\frac{H}{\sqrt{3}}$
$\frac{H}{b}=tg30^\circ=\frac{\sqrt{3}}{3};\ b=H\sqrt{3}$
$B=a.b=9cm^2\Rightarrow\frac{H}{\sqrt{3}}\cdot H\sqrt{3}=9\Rightarrow H=\sqrt{9}=3cm$
$V=\frac{1}{3}B.H=\frac{1}{3}\cdot9.3=9cm^3$