Question

The base of a right pyramid is a regular hexagon whose area is $243$ square cm. If the area of a side face the pyramid is $46$ square cm, what should be its volume (in cc.) in the nearest interger value?

Answer 1

Volume of hexagon = $\frac{1}{3}$ $\times$ area of base $\times$ height

Area of base = 243 ${cm}^2$

$\Rightarrow 6\times \frac{\sqrt{3}}{4}\times a^2=243\Rightarrow a^2=\frac{243\times 4}{6\sqrt{3}}=93.53$

$\Rightarrow a=9.67cm$

area of side force = $\frac{1}{2}\times a\times h=46{cm}^2$

$\Rightarrow \frac{1}{2}\times 9.67\times h=46\Rightarrow h=\frac{46\times 2}{9.67}=9.81cm$

Height of prism = $\sqrt{{\left(9.81\right)}^2-{\left(9.67\right)}^2}=1.65cm$

Volume = $\frac{1}{3}\times 243\times 1.65=133.65{cm}^3$