In the figure (i) given below, PQ = 24 cm, QR = 7 cm and ∠PQR = 90°. Find the radius of the inscribed circle of △PQR.

Let the sides of triangle PQ, QR and PR meet the circle at L, M and N respectively.

In right-angled triangle PQR

$PR^2 = PQ^2 + QR^2 \\[1em] PR^2 = 24^2 + 7^2 \\[1em] PR^2 = 576 + 49 \\[1em] PR^2 = 625 \\[1em] PR= \sqrt{625} \\[1em] PR = 25 \text{ cm}.$

From figure,

RM = RN = (∵ tangents drawn from a common external point to a circle are equal.)

RM = RQ - QM = (7 - x) cm.

PL = PN = (∵ tangents drawn from a common external point to a circle are equal.)

PL = PQ - QL = (24 - x) cm.

We can see,

PR = PN + RN = PL + RM.

⇒ 25 = 24 - x + 7 - x
⇒ 25 = 31 - 2x
⇒ 2x = 31 - 25
⇒ 2x = 6
⇒ x = 3.

Hence, the radius of the inscribed circle is 3 cm.