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Let AB be a chord of a circle with centre O and radius 13 cm such that AB = 10 cm.
From O, draw OL ⊥ AB. Join OA.
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = LB = ½ AB = 5 cm
Now, in right triangle OLA, we have
OA2 = OL2 + AL2
⇒ 132 = OL2 = 52
⇒ 132 – 52 = OL2
⇒ OL2 = 144
⇒ OL = 12 cm.
Hence, the distance of the chord from the centre is 12 cm.
From O, draw OL ⊥ AB. Join OA.
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = LB = ½ AB = 5 cm
Now, in right triangle OLA, we have
OA2 = OL2 + AL2
⇒ 132 = OL2 = 52
⇒ 132 – 52 = OL2
⇒ OL2 = 144
⇒ OL = 12 cm.
Hence, the distance of the chord from the centre is 12 cm.
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Answer:
Ans = O is the centre of circle.
Length of chord is 5+ 5 = 10
Radius of circle = 13
/_ONA = 90°
by hypotenuse theorem,
OA²= ON²+ AN²
(13)²= ON²+ (5)²
169= ON² + 25
169 - 25= ON²
144 = ON²
Taking square root,
12 = ON
....The distance of the chord from the centre is 12 cm.
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