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Question

In PQR,PQ=24 cm,QR=7 cm and PQR=90. Find the radius (in cm) of the inscribed circle.
582378_7a84d299eb0941f09ec46db40b2b39c1.png

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Solution

By Pythagoras' theorem,
PR2=PQ2+QR2
PR2=242+72=576+49=625
PR=25 cm

Let the inradius of PQR be x cm.
OAQC is a square. Hence QA=x cm and AR=(7x) cm
RA and RB act as tangents to the incircle from point R, hence their lengths are equal. RB=AR=(7x) cm.
Similarly, PB=PC=(24x) cm

PR=PB+RB
25=(24x)+(7x)
25=312x
x=3 cm

630374_582378_ans_07221ff710214a2fbf37ca4128217414.png

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