Question

ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with center O, has been inscribed inside the triangle.Find it's radius.

Answer 1

Using Pythagoras theorem we have;
$(BC{)}^2=(AC{)}^2-(AB{)}^2\Rightarrow B{C}^2={13}^2-{12}^2$

$BC=\sqrt{25}=5$
Now, AB, BC and CA are tangents to the circle at P, N and M respectively.
∴ OP = ON = OM = r (radius of the circle)
Area of triangle ABC = $\frac{1}{2}\times BC\times AB=\frac{1}{2}\times 5\times 12=30c{m}^2$
Area of ∆ABC = Area ∆OAB + Area ∆OBC + Area ∆OCA
30 = 1/2 r×AB+1/2 r×BC+1/2 r×CA
⇒30 = 1/2 r(AB+BC+CA)
⇒r = 2×30/(AB+BC+CA)
⇒r = 60/12+5+13 = 2
Therefore the radius of inscribed circle is 2 cm