Math, asked by itsrudrak, 3 months ago

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Answered by vvd4
1
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.
Answered by somnathsalunkhe1496
0

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.

Step-by-step explanation:

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