NCERT Exemplar Solutions Class 9 Mathematics Solutions for Circles - Exercise 10.2 in Chapter 10 - Circles
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.
Facebook
Whatsapp
Copy Link
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
