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(C) Let $r$ be the radius of the incircle.Given sides of the triangle are $a = 11\, cm$,$b = 6\, cm$,and $c = 15\, cm$.First,calculate the semi-perime

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$\frac{5 \sqrt{2}}{4}\, cm$

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(C) Let $r$ be the radius of the incircle.Given sides of the triangle are $a = 11\, cm$,$b = 6\, cm$,and $c = 15\, cm$.First,calculate the semi-perimeter $s$ of the triangle:$s = \frac{a + b + c}{2} = \frac{11 + 6 + 15}{2} = \frac{32}{2} = 16\, cm$.Now,calculate the area of the triangle $(\Delta)$ using Heron's formula:$\Delta = \sqrt{s(s - a)(s - b)(s - c)}$$\Delta = \sqrt{16(16 - 11)(16 - 6)(16 - 15)}$$\Delta = \sqrt{16 	imes 5 	imes 10 	imes 1} = \sqrt{800} = 20 \sqrt{2}\, cm^2$.The radius of the incircle $r$ is given by the formula $r = \frac{\Delta}{s}$.$r = \frac{20 \sqrt{2}}{16} = \frac{5 \sqrt{2}}{4}\, cm$.

The sides of a triangle are $6\, cm$,$11\, cm$,and $15\, cm$. The radius of its incircle is

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