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]

Answer 1

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

${AC}^2={AB}^2+{BC}^2={\left(15\right)}^2+{\left(8\right)}^2=(225+64)=289$

$\Rightarrow AC=\sqrt{289}=17cm$.

Clearly, OPBQ is a square.

$[\angle OPB={90}^{\circ },\angle PBQ={90}^{\circ },\angle OQB={90}^{\circ }andOP=OQ=xcm]$

$BP=BQ=xcm$.

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.

$AC=AR+CR\Rightarrow 17=(15–x)+(8–x)\Rightarrow 2x=6\Rightarrow x=3$.

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