A circle is inscribed in a right triangle ABC, right angled at B. If the lengths of the two sides containing the right…

AC = $\sqrt{(15)^2 + (8)^2} = 17$ cm ($\frac{1}{2}$ Mark)
Let '$r$' be the radius of the circle.
Since, radius is perpendicular to the tangent through the point of contact.
$\therefore$ OP is perpendicular to AB and OQ is perpendicular to BC.
Thus, OPBQ is a square.
$\Rightarrow$ OP = PB = BQ = OQ = $r$ ($\frac{1}{2}$ Mark)
Thus, AR = AP = $8 - r$
and CR = CQ = $15 - r$ } (1 Mark)
Now, AC = AR + CR ($\frac{1}{2}$ Mark)
$r = 3$ cm ($\frac{1}{2}$ Mark)