ABC is a right triangle right …

Let ABC be the right angled triangle such that ∠B = 90° , BC = 6 cm, AB = 8 cm. Let O be the centre and r be the radius of the in circle.

AB, BC and CA are tangents to the circle at P, Nand M.
{tex}\therefore{/tex} OP = ON = OM = r (radius of the circle)
Area of the {tex}\triangle{/tex}ABC = {tex}\frac{1}{2} \times 6 \times 8{/tex}= 24 cm²
By Pythagoras theorem,
CA² = AB² + BC²
{tex}\Rightarrow{/tex} CA² = 8² + 6²
{tex}\Rightarrow{/tex} CA² = 100
{tex}\Rightarrow{/tex} CA = 10 cm
Area of the {tex}\triangle{/tex}ABC = Area {tex}\triangle{/tex}OAB + Area {tex}\triangle{/tex}OBC + Area {tex}\triangle{/tex}OCA
24 = {tex}\frac{1}{2} r \times \mathrm{AB}+\frac{1}{2} r \times \mathrm{B} \mathrm{C}+\frac{1}{2} r \times \mathrm{C} \mathrm{A}{/tex}
24 = {tex}\frac{1}{2} r{/tex}(AB + BC + CA)
{tex}\Rightarrow r=\frac{2 \times 24}{(\mathrm{AB}+\mathrm{BC}+\mathrm{CA})}{/tex}
{tex}\Rightarrow r=\frac{48}{8+6+10}{/tex}
{tex}\Rightarrow r=\frac{48}{24}{/tex}
{tex}\Rightarrow{/tex} r = 2 cm