Answer :

Given: Radius of circle = 6 cm


QT = 12 cm, TR = 9 cm


Area of the triangle = 189 cm2


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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