Question

In the figure,O is the centre of a circle chord AB=10cm and diameter AC=26cm. The distance of the chord AB of the circle from the centre is:

Answer 1

Answer:

hey here's Ur ans

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ans-12 cm

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Answer 2

The distance of center from the chord=12cm

Step-by-step explanation:

Diameter AC=d=26cm

Radius of circle=$\frac{d}{2}=\frac{26}{2}=13cm$

Chord=AB=10cm

We know that the perpendicular drawn from center to the chord bisect the chord.

Therefore, AP=PB

AP=$\frac{1}{2}(10)=5cm$

In triangle OAP

$OA^2=OP^2+AP^2$

By using Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(perpendicular\;side)^2$

Substitute the values then we get

$(13)^2=5^2+OP^2$

$169=25+OP^2$

$169-25=OP^2$

$144=OP^2$

$OP=\sqrt{144}=12cm$

The distance of center from the chord=12cm

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https://brainly.in/question/6653415:Answered by Brainly user