Q12 of 53 Page 153

The sides of a triangle have lengths a2 + b2, 2ab, a2 — b2, where a > b and a, b ϵ R + . Prove that the angle opposite to the side having length a2 + b2 is a right angle.

Let a2 + b2 = p, 2ab = q, a2 — b2 = r


Since a, b ϵ R + , hence


p is the largest side of the triangle


p2 = (a2 + b2)2 = a4 + b4 + 2a2b2


q2 + r2 = 4a2b2 + (a2 – b2)2


⇒ q2 + r2 = a4 + b4 + 2a2b2


So p2 = q2 + r2


Hence the Pythagoras theorem is established


Hence the angle opposite to the side p = a2 + b2 is a right angle


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