Answer :

Given: Radius of circle = 6 cm

QT = 12 cm, TR = 9 cm

Area of the triangle = 189 cm^{2}

To find: Lengths of sides PQ and PR

Let O be the center of the circle. Then join POT, QOU, and ROS.

Since PQ, QR, RP are tangents to the circle at S, T, U

respectively, therefore PQ, QR, and RP are perpendicular to

SO, OT and OU respectively.

Now, tangents drawn from an external point are equal.

Therefore QT = QS = 12 cm

RT = RU = 9 cm

Let PU = PS = x cm

So, we have three triangles POQ, POR and QOR such that:

Base of triangle POQ = (12 + x ) cm

Height of triangle POQ = SO = 6 cm.

Base of triangle POR = (9 + x ) cm

Height of triangle POR = OU = 6 cm.

Base of triangle QOR = (12 + 9) cm

Height of triangle QOR = OT = 6 cm.

Thus, area of the triangle PQR = Area of ∆POQ + Area of ∆POR + Area of ∆QOR

⇒ 189 = 1/2 [(12 + x) × 6] + 1/2 [(9 + x) × 6] + 1/2 [(12 + 9) × 6]

⇒ 189 = 1/2[72 + 6x + 54 + 6x + 72 + 54]

⇒ 189 = 1/2[252 + 12x]

⇒ 189 = 126 + 6x

⇒ 189 - 126 = 6x

⇒ 6x = 63

⇒ x = 63/6 = 10.5

Therefore, PQ = (12 + x) cm = (12 + 10.5) cm = 22.5 cm

PR = (9 + x) cm = (9 + 10.5) cm = 19.5 cm

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