Question

$\text{The area of an equilateral triangle is}81\sqrt{3}c{m}^2.\text{Its height is}\left(a\right)9\sqrt{3}cm\left(b\right)6\sqrt{3}cm\left(c\right)18\sqrt{3}cm\left(d\right)9cm$

Answer 1

Area of an equilateral triangle = 81$\sqrt{3}c{m}^2$
Let the side of the equilateral triangle be 'a' cm
⇒ $\sqrt{3/4}{a}^2=\left(\surd 3/4\right)a2=81\sqrt{3}$
⇒ $\frac{1}{4}{a}^2$ = 81
⇒ $a^2$= 81 x 4
⇒ a = $\sqrt{(}81\times 4)$
⇒ a = 9 x 2
⇒ a = 18.

If a side of the a triangle is considered as a base:
Area of a triangle = $\frac{1}{2}$ x base x height
$81\sqrt{3}=\frac{1}{2}x18xh$
$81\sqrt{3}=9xh$
$h=9\sqrt{3}$

Therefore, height of the equilateral triangle is $9\sqrt{3}$cm.
option a is correct