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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
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Solution
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.
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