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. Calculate the value of x, the radius of the inscribed circle.

Let AB touches the circle at L, AC at N and BC at M.

From figure,

LBMO is a square.

LB = BM = OM = OL = x.

AL = AB - LB = (12 - x) cm.

AL = AN = (12 - x) cm. [∵ Tangents from exterior point are equal in length.]

Since, ABC is a right angled triangle,

∴ AC² = AB² + BC² [By pythagoras theorem]

⇒ 13² = 12² + BC²

⇒ BC² = 13² - 12²

⇒ BC² = 169 - 144

⇒ BC² = 25

⇒ BC = $\sqrt{25}$

⇒ BC = 5 cm.

From figure,

MC = BC - BM = (5 - x) cm.

Also,

CN = CM = (5 - x) cm. [∵ Tangents from exterior point are equal in length.]

Also,

⇒ AC = AN + CN

⇒ 13 = (12 - x) + (5 - x)

⇒ 13 = 17 - 2x

⇒ 2x = 17 - 13

⇒ 2x = 4

⇒ x = $\dfrac{4}{2}$

⇒ x = 2 cm.

Hence, x = 2.