ΔABC is a right-angled triangle in which ∠A = 90°, AC = 12 cm and BC = 13 cm. A circle with centre 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,

ANLO is a square

AL = LO = ON = AN = x

NC = AC - AN = (12 - x) cm

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

Since, ABC is a right angled triangle,

∴ BC² = AC² + AB² [By pythagoras theorem]

⇒ 13² = 12² + AB²

⇒ AB² = 13² - 12²

⇒ AB² = 169 - 144

⇒ AB² = 25

⇒ AB = $\sqrt{25}$

⇒ AB = 5 cm.

From figure,

LB = AB - AL = (5 - x) cm.

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

Then,

⇒ BC = BM + CM

⇒ 13 = (5 - x) + (12 - x)

⇒ 13 = 17 - 2x

⇒ 2x = 17 - 13

⇒ 2x = 4

⇒ x = $\dfrac{4}{2}$

⇒ x = 2 cm.

Hence, x = 2 cm.