The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.

Given,

⇒ QR = 3RS = 6 cm

⇒ 3RS = 6 cm

⇒ RS = $\dfrac{6}{3}$ = 2 cm.

In △PQR,

∠Q = 90° [Angle in semi-circle is a right angle.]

By pythagoras theorem,

⇒ PR² = PQ² + QR²

⇒ PR² = 7² + 6²

⇒ PR² = 49 + 36

⇒ PR² = 85 ……..(1)

In △PSR,

∠S = 90° [Angle in semi-circle is a right angle.]

By pythagoras theorem,

⇒ PR² = PS² + RS²

⇒ 85 = PS² + 2²

⇒ PS² = 85 - 4

⇒ PS² = 81

⇒ PS = $\sqrt{81}$ = 9 cm.

Perimeter of cyclic quadrilateral PQRS = PQ + QR + RS + PS = 7 + 6 + 2 + 9 = 24 cm.

Hence, perimeter of cyclic quadrilateral PQRS = 24 cm.