In a $\triangle ABC$ if $\angle B=90^{\circ}$, BC = 8 cm and AB = 15 cm, then radius r of the incircle of $\triangle ABC$ is:

3 cm

Triangle ABC is right angled at B.

AB = 15 cm

BC = 8 cm

Applying Pythagoras theorem, 

AC^2=AB^2+BC^2=(15)^2+(8)^2=(225+64)=289

AC = 17

Since the circle is inscribed, quadrilateral made with the centre of the circle, O, Point B, P and Q make a square

$BP=BQ=xcm$.

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.

The correct answer is Option (3) → 3 cm