In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC = 3 cm. Calculate the length of OC.

Let length of OC be x cm.

In right angled triangle AOC,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AC² = AO² + OC²

⇒ 3² = AO² + x²

⇒ AO² = 3² - x²

⇒ AO² = 9 - x²

⇒ AO = $\sqrt{9 - x^2}$ cm.

From figure,

BO = BC + CO = (6 + x) cm.

In right angled triangle AOB,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AO² + BO²

⇒ 8² = $(\sqrt{9 - x^2})^2$ + (6 + x)²

⇒ 64 = 9 - x² + 36 + x² + 12x

⇒ 64 = 45 + 12x

⇒ 12x = 64 - 45

⇒ 12x = 19

⇒ x = $\dfrac{19}{12} = 1\dfrac{7}{12}$.

Hence, OC = $1\dfrac{7}{12}$ cm.