Question

If ABC is an equilateral triangle of side a, prove that its altitude = $\frac{\sqrt{3}}{2}$a

Answer 1

ΔABC is an equilateral triangle.

We are given that AB = BC = CA = a. AD is the altitude, i.e., AD $\perp$ BC.

Now, in right angled triangles ABD and ACD, we have

AB = AC [Given]

and AD = AD [Common side]

$\Rightarrow$ ΔABD ~ ΔACD [ By RHS congruence]

$\Rightarrow$BD = CD ⇒ BD = DC = $\frac{1}{2}$ BC = $\frac{a}{2}$

From right triangle ABD,

${AB}^2$ = ${AD}^2$ + ${BD}^2$ ⇒ $a^2$ = ${AD}^2$ + $({\frac{a}{2})}^2$

$\Rightarrow$${AD}^2$ = $a^2$ - $\frac{a^2}{4}$ = $\frac{3}{4}$ $a^2$

$\Rightarrow$ AD =$\frac{\sqrt{3}a}{2}$