In the figure (i) given below, PR is a diameter of the circle, PQ = 7 cm, QR = 6 cm and RS = 2 cm. Calculate the perimeter of the cyclic quadrilateral PQRS.

We know that ∠PQR = 90° as angle in semicircle is equal to 90°.

So, △PQR is a right angled triangle.

By pythagoras theorem,

$PR^2 = PQ^2 + QR^2 \\[1em] PR^2 = 7^2 + 6^2 \\[1em] PR^2 = 49 + 36 \\[1em] PR^2 = 85.$

In △PRS,

∠PSR = 90° as angle in semicircle is equal to 90°.

So, △PRS is a right angled triangle.

By pythagoras theorem,

$PR^2 = PS^2 + RS^2 \\[1em] 85 = PS^2 + 2^2 \\[1em] PS^2 = 85 - 4 \\[1em] PS^2 = 81 \\[1em] PS = 9.$

Perimeter of PQRS = PQ + QR + RS + SP = 7 + 6 + 2 + 9 = 24 cm.

Hence, the perimeter of cyclic quadrilateral is 24 cm.