$$\triangle ABC$$ is a right angled triangle with $$AB=6 cm$$, $$BC = 8 cm$$. O is the in-centre of the triangle. The radius of the in-circle is: 

Let the inradius of the triangle be $$r$$ cm

In right $$\triangle$$ ABC,

=> $$(AC)^=(AB)^2+(BC)^2$$

=> $$(AC)^=(6)^2+(8)^2$$

=> $$(AC)^2=36+64=100$$

=> $$AC=\sqrt{100}=10$$ cm

Area of triangle = $$\triangle=r\times s$$, where $$r$$ is inradius and $$s$$ is semi-perimeter.

=> Area = $$\triangle=\frac{1}{2}\times8\times6=24$$ $$cm^2$$

Semi-perimeter = $$s=\frac{(10+8+6)}{2}=\frac{24}{2}=12$$ cm

$$\therefore$$ Inradius of triangle = $$r=\frac{\triangle}{s}=\frac{24}{12}=2$$ cm

=> Ans - (C)