#### Question

In triangle PQR, PQ = 24 cm, QR = –7 cm and ∠PQR = 90°. Find the radius of the inscribed circle.

#### Solution

Since ΔPQR is a right-angled angle,

PR = `sqrt(7^2 + 24^2) = sqrt(49 + 576) = sqrt625 = 25 cm`

Let the given inscribed circle touches the sides of the given triangle at points A, B and C respectively.

Then, clearly, OAQB is a square.

=> AQ = BQ = x cm

PA = PQ – AQ = (24 – x) cm

RB = QR – BQ = (7 – x) cm

Since tangents from an exterior point to a circle are equal,

PC = PA = (24 – x) cm

And, RC = RB = (7 – x) cm

PR = PC + CR

=> 25 = (24 – x) + (7 – x)

=> 25 = 31 – 2x

=> 2x = 6

=> x = 3 cm

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

Is there an error in this question or solution?

#### APPEARS IN

Solution In Triangle Pqr, Pq = 24 Cm, Qr = –7 Cm and ∠Pqr = 90°. Find the Radius of the Inscribed Circle. Concept: Circumscribing and Inscribing a Circle on a Triangle.