Let ABC be a right triangle right angled at B with AB = 16 cm and BC = 8 cm. Then, the largest square BRSP which can be inscribed in this triangle will be as shown in Fig.
Let PB = x cm. So., AP = (16–x) cm. In ∆APS and ∆ABC, ∠A = ∠A and ∠APS = ∠ABC (Each 90°)
So, ∆APS ~ ∆ABC (AA similarity)
Thus, the side of the required square is of length 16/3 cm.