A triangle PQR is drawn to circumscribe a cirle with radius 6cm. such that the segments QT and TR into which QR is divided at the point of contact T, are the lenghts of 12cm and 9cm respectively. if the area of PQR is 189cm2, then find the lenghts of side PQ and PR
Answers
Answered by
0
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!
Was this answer helpful
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!
Was this answer helpful
Similar questions