Question

If the altitude of an equilateral triangle is $\sqrt{6}$ $cm$, find its area.

• A. $2\sqrt{3}c{m}^2$
• B. $2\sqrt{2}c{m}^2$
• C. $3\sqrt{3}c{m}^2$
• D. $6\sqrt{2}c{m}^2$

Answer 1

The correct option is A $2\sqrt{3}c{m}^2$

In right angle $△ADB$, Using Pythagones theorem,

$\Rightarrow {\left(\frac{a}{2}\right)}^2+{\left(\sqrt{6}\right)}^2=a^2$

$\Rightarrow \frac{a^2}{4}+6=a^2\Rightarrow \frac{{3a}^2}{4}=6\Rightarrow a^2=8\Rightarrow a=2\sqrt{2}cm$

$\therefore$ $△ABC$ (equilateral $△$) = $\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times 2\sqrt{2}\times 2\sqrt{2}=2\sqrt{3}{cm}^2$