ABC is a right angled triangle, right angled at A. A circle is inscribed in it. The lengths of two sides containing the right angle are 48 cm and 14 cm. The radius of the inscribed circle is:

Using the pythagoras theorem,

BC$$^2$$ = AB$$^2$$ + AC$$^2$$

$$\Rightarrow$$  BC$$^2$$ = 48$$^2$$ + 14$$^2$$

$$\Rightarrow$$  BC$$^2$$ = 2304 + 196

$$\Rightarrow$$  BC$$^2$$ = 2500

$$\Rightarrow$$  BC = 50 cm

Let the radius of the circle = r

$$\Rightarrow$$  OD = OI = OJ = r

AB, BC, AC are tangents to the circle

Area of $$\triangle$$ABC = Area of $$\triangle$$OAC + Area of $$\triangle$$OBC + Area of $$\triangle$$OAB

$$\Rightarrow$$  $$\frac{1}{2}$$ x 48 x 14 = $$\frac{1}{2}$$ x AC x OD + $$\frac{1}{2}$$ x BC x OI + $$\frac{1}{2}$$ x AB x OJ

$$\Rightarrow$$  336 = $$\frac{1}{2}$$ x 14 x r + $$\frac{1}{2}$$ x 50 x r + $$\frac{1}{2}$$ x 48 x r

$$\Rightarrow$$  336 = 7r + 25r + 24r

$$\Rightarrow$$  56r = 336

$$\Rightarrow$$  r = 6 cm

$$\therefore\ $$Radius of the inscribed circle = 6 cm

Hence, the correct answer is Option C