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Question

In the given figure, ABC is right-angled at B, in which AB = 15 cm and BC = 8 cm. A circle with center O has been inscribed in ABC. Calculate the value of x, the radius of the inscribed circle. [4 MARKS]

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Solution

Concept: 1 Mark
Application: 3 Marks

Let the inscribed circle touch the sides AB, BC and CA at P, Q and R respectively.

Applying Pythagoras theorem on right ABC, we have

AC2=AB2+BC2=(15)2+(8)2=(225+64)=289

AC=289=17cm.

Clearly, OPBQ is a square.

[OPB=90,PBQ=90,OQB=90 and OP=OQ=xcm]

BP=BQ=xcm.

Since the tangents to a circle from an exterior point are equal in length,

we have AR = AP and CR = CQ.

Now, AR = AP = (AB - BP) = (15 – x) cm

CR = CQ = (BC – BQ) = (8 - x) cm.

AC=AR+CR17=(15x)+(8x)2x=6x=3.

Hence, the radius of the inscribed circle is 3 cm.


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