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1. The altitude of an equilateral triangle of side $2 \sqrt{3}$ cm is :

A. $\frac{1}{2}$ cm

B. $\frac{\sqrt{3}}{4}$ cm

C. $\frac{\sqrt{3}}{2}$ cm

D. 3 cm

Solution:

Let ABC be the equilateral triangle and AD be the altitude on base BC
Area mcq solution image

In an equilateral triangle, the altitude and the median coincide.
So, BC = DC =

(\frac{2 \sqrt{3}}{2})

cm =

\sqrt{3}

cm
Let the length of the altitude AD be x cm
Then, in right angled ΔADB,
AB² = AD² + BD²
⇒

{(2 \sqrt{3})}^{2}

= x² +

{(\sqrt{3})}^{2}

⇒ x² = (12 - 3)
⇒ x² = 9
⇒ x = 3 cm

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