Question

In a ∆ABC, AB = BC = CA = 2a and AD ⊥ BC. Prove that

(i) $\mathrm{AD}=a\sqrt{3}$
(ii) $Area\left(∆ABC\right)=\sqrt{3}{a}^2$

Answer 1

In and

$\angle B=\angle C$ (60º each)

$\angle ADB=\angle ADC$ (90º each)

(AAS congruence theorem)

But therefore, we get,

………(1)

Now we will divide both sides of the equation (1) by 2, we get,

Now we will use Pythagoras theorem in right angled triangle ADB.

Now we will substitute the values of AB and BD we get,

Therefore, .

We have given an equilateral triangle and we know that the area of the equilateral triangle is.

Here, side is 2a

Therefore, .