Question

If the altitude of an equilateral triangle is $12\sqrt{3} cm$, then its area would be :

• A. 12 sq.cm
• B. $$144\sqrt{3} cm^2$$
• C. 72 sq.cm
• D. $$36\sqrt{3} cm^2$$

Answer 1

Answer:

$$144\sqrt{3} cm^2$$

$AD=12\sqrt{3}cm$

$AB=2xcm$

$BD=xcm$

From$\triangle ABD$

$AD=\sqrt{AB^2-BD^2}$

$=\sqrt{(2x)^2-x^2}$

$=\sqrt{4x^2-x^2}=\sqrt{3x^2}=\sqrt{3}x$

=>\sqrt{3}x=12\sqrt{3}

$x=12cm$

AB = 2x = 2 × 12 = 24 cm.

Area of $\triangle ABC=\frac{\sqrt{3}}{4}\times{side^2}$

$=\frac{\sqrt{3}}{4}\times{24}\times{24}$

$=144\sqrt{3}cm$