# NCERT Exemplar Solutions Class 9 Mathematics Solutions for Circles - Exercise 10.2 in Chapter 10 - Circles

Question 1 Circles - Exercise 10.2

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 :

(A) 17 cm

(B) 15 cm

(C) 4 cm

(D) 8 cm

(D) 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.

Hence, option D is the correct answer.

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