Question

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is ______.

Options

• 17 cm

• 15 cm

• 4 cm

• 8 cm

Answer 1

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is 8 cm.

Explanation:

Given: Diameter of the circle = d = AD = 34 cm

∴ Radius of the circle = r = `d/2` = AO = 17 cm

Length of chord AB = 30 cm

Since the line drawn through the center of a circle to bisect a chord is perpendicular to the chord, therefore AOL is a right angled triangle with L as the bisector of AB.

∴ AL = `1/2`(AB) = 15 cm

In right angled triangle AOB, by Pythagoras theorem, we have:

(AO)² = (OL)² + (AL)²

⇒ (17)² = (OL)² + (15)²

⇒ (OL)² = (17)² – (15)²

⇒ (OL)² = 289 – 225

⇒ (OL)² = 64

Take square root on both sides:

⇒ (OL) = 8

∴ The distance of AB from the center of the circle is 8 cm.