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Given: ∆ABC –iso. ∆, m∠BAC = 120°
AH ⊥ BC , HD⊥ AC
AD = a cm, HD = b cm
Find: P∆ADH

Solve without Using Pythagorean Theorem

Given: ∆ABC –iso. ∆, m∠BAC = 120° AH ⊥ BC , HD⊥ AC AD = a cm, HD = b cm Find: P∆ADH Solve without Using Pythagorean Theorem class=

Answer :

Answer:

[tex]P = a + b + \sqrt{a^{2} + b^{2}}[/tex]

Step-by-step explanation:

See the attached diagram.

Given that, Δ ADH is a right triangle with ∠ ADH = 90°.

So, applying Pythagoras Theorem  

AH² = AD² + DH² = a² + b²

⇒  [tex]AH = \sqrt{a^{2} + b^{2}}[/tex] {Neglecting the negative root as AH can not be negative}

Therefore, the perimeter of the triangle Δ ADH = AD + AH + DH  

⇒ [tex]P = a + b + \sqrt{a^{2} + b^{2}}[/tex] (Answer)

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