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In a triangle PQR, PQ=24cm, QR=7cm and angle PQR=90°. A circle is inscribed in the triangle. Find the radius of the circle.
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Answer:
In triangle PQR,
By Pythagoras' theorem,
PR² = PQ² + QR²
PR² = 24² + 7²
= 576 + 49
= 625
PR = 25 cm
Let the radius of circle be x cm
OAQC is a square. Hence
QA = x cm and AR = (7-x) cm
RA and RB act as tangents to the in circle from point R,
hence their lengths are equal.
RB = AR = (7-x) cm.
Similarly, PB = PC = (24 - x) cm.
PR =PB+RB
25 = (24-x) + (7-x)
25 = 31 -2x
x = 3cm
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