Question

Given a right triangle PQR which is right-angled at Q. QR = 12 cm, PQ = 5 cm. The
radius of the circle which is inscribed in triangle PQR will be?

Answer 1

In ΔPQR

(PR)2=(PQ)2+(QR)2

(PR)2=(12)2+(5)2

(PR)2=169

PR=13

Now  AOBQ is a square

So, QB=x

Then, BR=5−x

Similarly AQ=x

Then AP=12−x

Also, CR=BR

CR=5−x

[∵ Length of tangents drawn from external point are equal]

And, CP=AP=12−x

[∵ Length of tangents drawn from external point are equal]

PR=PC+CR

13=5−x+12−x

2x=4⇒x=2 cm.