Let ∆PQR be the given right angled isosceles triangle.
PQ = QR = x.
In ∆PQR, ∠Q = 90° [Pythagoras theorem]
∴ PR2 = PQ2 + QR2
= x2 + x2
= 2x2
∴ PR = \(\sqrt{2x^2}\) [Taking square root of both sides]
= x√2 units
∴ The hypotenuse of the right angled isosceles triangle is x√2 units.
∴ The hypotenuse of the right angled isosceles triangle is x√2 units.