Question
Find the height of an equilateral triangle having side 4 cm. ?

Answer


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Let $\triangle A B C$ be the given equilateral triangle.
$\therefore \angle B=60^{\circ}$ $\quad$ $\ldots$[Angle of an equilateral triangle]
Let $A D \perp B C, B-D-C$.
In $\triangle A B D, \angle B=60^{\circ}, \angle A D B=90^{\circ}$
$\therefore \angle \mathrm{BAD}=30^{\circ}$ $\quad$ $\ldots$[Remaining angle of a triangle]
$\therefore \triangle \mathrm{ABD}$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$\therefore A D=\frac{\sqrt{3}}{2} A B$ $\quad$ $\ldots$ [Side opposite to $60^{\circ}$ ]
$=\frac{\sqrt{3}}{2} \times 4$
$=2\sqrt{3} \text { units }$
$\therefore$ The height of the equilateral triangle is $2 \sqrt{3}$ units.

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