In the given diagram, AD is a diameter of the circle and AB is a chord. If AO= 17 cm, AB = 30 cm, the distance of AB from the centre of the circle is
A
17 cm
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B
15 cm
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C
8 cm
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D
4 cm
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Solution
The correct option is C 8 cm Given, AD = 34 cm and AB = 30 cm
Consider the figure given below,
Draw OL ⊥ AB
Since the perpendicular from the centre of a circle to a chord bisects the chord, ∴AL=LB=12AB=15cm
In right angled Δ OLA, OA2=OL2+AL2 ∴(17)2=OL2+(15)2 ⇒289=OL2+225 ⇒289=OL2+225 ⇒OL2=289−225=64 ∴OL=8cm
[taking positive square root, because length is always positive]
Hence, the distance of the chord from the centre is 8cm.
Therefore, option c is correct.