Question

if the altitude of an equilateral triangle is root 6 its area​

Answer 1

Given:

∆ABC is an equilateral triangle.

Altitude CD is $\sqrt{6}$cm.

To find:

Area of ∆ABC

Proof:

In ∆CDB,

$\implies sin60°=\frac{CD}{BC}\\\implies \frac{ \sqrt{3} }{2} = \frac{ \sqrt{6} }{BC} \\\implies \sqrt{3} BC = 2 \sqrt{6} \\\implies BC = \frac{2 \sqrt{6} }{ \sqrt{3} } \\\implies BC = 2 \sqrt{2} cm$

Since,∆ABC is equilateral.

$\implies AB=BC=AC=2 \sqrt{2} cm$

$\boxed{\sf{Area\:of\: triangle=\frac{1}{2} \times base \times height}}$

$\implies Area\:of\: \triangle ABC = \frac{1}{2} \times AB \times CD \\ \implies Area\:of\: \triangle ABC = \frac{1}{2} \times 2 \sqrt{2} \times \sqrt{6} \\ \implies Area\:of\: \triangle ABC = \sqrt{12} \\ \implies Area\:of\: \triangle ABC = 2 \sqrt{3} c {m}^{2}$

Hence, area of ∆ABC is 2√3 sq.cms.