Question

A triangle Pqr is drawn to circumscribe a circle to radius 6 cm such that the segments QT and TR into which QR is divided by the point of contact T, are of lenghts 12cm and 9 cm respectively. If the area of Triangle PQR = 189cm, then find the lenghts of sides PQ and PR

Answer 1

Here is the answer to your question,

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 and PR = (9 + 10.5) cm = 19.5 cm.

Hope! This will help you.
Cheers!

Answer 2

Answer: PQ=22.5cm

PR= 19.5cm

Step-by-step explanation:

{Go for the attachment!!}