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)
