Answer 1

Let ABC be the Δ with

AB = (a – 1) cm BC = 2`sqrta` cm, CA = (a + 1) cm

Hence, AB² = (a − 1)2 = a2 + 1 − 2𝑎

BC² = (2`sqrta`)² = 4a

CA² = (a + 1)² = a² + 1 + 2a

Hence AB² + BC² = AC²

So Δ ABC is right angled Δ at B.