Question

In the given figure, a triangle PQR is drawn to circumscribe a circle of radius 6 cm such that the segments QT and TR into which QR is divided by the point of contant T, are of lengths 12 cm and 9 cm respectively. If the area fo \Delta PQR = 189 cm^2 then the length of side PQ is

(a) 17.5 cm (b) 20 cm (c) 22.5 cm (d) 25 cm

Answer 1

(c) 22.5 cm


Let PQ and PR touch the circle at points S and U respectively. Join O with P, Q, R, S and U.

We have, OS = OT = OU = 6 cm (Radii of the circle)

QT = 12 cm and TR = 9 cm

QR = QT + TR = 12 cm + 9 cm = 21 cm

Now, QT = QS = 12 cm (Tangents from the same point)

TR = RU = 9 cm

Let PS= PU = x cm

Then, PQ = PS + SQ = (12 + x) cm and PR = PU + RU = (9 + x) cm

It is clear that

ar (∆OQR) + ar (∆OPR) + ar (∆OPQ) = ar (∆PQR)

Thus, PQ = (12 + 10.5) cm = 22.5 cm