Question

Given: ∆ABC –iso. ∆, m∠BAC = 120°

Answer 1

Answer: $P\triangle ADH = (a+b+\sqrt{a^2+b^2} )$ unit

Step-by-step explanation:

Since Here Δ ABC is a isosceles triangle.

Also, $AH\perp BC$ and $HD\perp AC$

Thus, ∠ ADH = 90°

That is, Δ ADH is the right angle triangle.

In which, AD = a unit and HD = b unit

And, By the definition of Pythagoras theorem,

$AH^2 = HD^2 + AD^2$

⇒ $AH^2 = a^2 +b^2$

⇒ $AH = \sqrt{a^2+b^2}$

Since, the perimeter of a triangle = sum of the all sides of the triangle.

Therefore, perimeter of ΔADH = AD + DH + AH

= $a + b + \sqrt{a^2+b^2}$ unit