A right pyramid has an equilateral triangular base of side 4 cm. If the numerical value of its total surface area is three times the numerical value of its volume, then height is:
8 cm
6 cm
10 cm
12 cm
Solution
Let $a$ be the length of each side of the base, $h$ be the height and $l$ be the slant height of the pyramid.
$\therefore$ slant height $= \sqrt{h^2 + \dfrac{a^2}{12}}$
$\Rightarrow l = \sqrt{h^2 + \dfrac{16}{12}} \Rightarrow l = \sqrt{h^2 + \dfrac{4}{3}}$
It is given that the numerical value of its total surface area is three times the numerical value of its volume.
$\therefore$ Lateral surface area + Area of the base $= 3 \times$ (Volume)
$ \Rightarrow \dfrac{1}{2}(4 + 4 + 4) \times \sqrt{h^2 + \dfrac{4}{3}} + \sqrt{\dfrac{3}{4}} \times 4^2$
$= 3 \times \dfrac{1}{3} \left( \dfrac{\sqrt{3}}{4} \times 4^2 \times h \right)$
$\Rightarrow 6 \sqrt{h^2 + \dfrac{4}{3}} + 4\sqrt{3} = 4\sqrt{3}$h
$\Rightarrow 6 \sqrt{h^2 + \dfrac{4}{3}} = 4\sqrt{3} (h - 1)$
$\Rightarrow 36 \left( h^2 + \dfrac{4}{3}\right) = 48 (h - 1)^2$
$\Rightarrow 3 \left( h^2 + \dfrac{4}{3} \right) = 4 (h - 1)^2$
$\Rightarrow 3h^2 + 4 = 4 (h^2 - 2h + 1)$
$\Rightarrow 8h = h^2 \Rightarrow h = 8$ cm