#### Question

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

#### Solution 1

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.

#### Solution 2

OM ⊥ QR

ON ⊥ PQ ....( Tangents and radius perpendicular to each other.)

OM = ON = r

QM = QN ....( Tangents from an external point)

⇒ QMON is a square.

⇒ QM = OM = ON = QN = x cm

So, mR = (7 - x) cm

PN = (24 - x) cm

PT = PN = 24 - x

and, mR = RT = 7 - x ....(Tangents from an external point)

⇒ PR = PT + RT

PR = 24 - x + 7 - x

PR = 31 - 2x

Now, In ΔPQR,

PR^{2} = PQ^{2} + QR^{2}

PR^{2} = 24^{2} + 7^{2}

PR^{2} = 576 + 49 = 625

PR = 25 cm

⇒ 31 - 2x = 25

⇒ 2x = 31 - 25

⇒ 2x = 6

⇒ x = 3.