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  • A circle has radius sqrt{2} cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45º.

A circle has radius \sqrt{2} cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45º.

Answers (1)

A circle with radius \sqrt{2} cm is divided into two segments by a chord of length 2 cm.

We have drawn a circle having centre O

Let AB = 2 cm be the required chord which divides the circle into two segment.

To Prove: the angle subtended by the chord at a point in major segment is 45º.

Which means we have to prove that \angleAPB = 45°

We have, radius = \sqrt{2} cm

OA = OB = \sqrt{2} cm                (given)

OA2 = OB2 = 2 cm

Also, chord AB = 2 cm           (given)

AB2 = 4cm

Now, we can see that in DAOB,

OA2 + OB2 = AB2

As, 2 + 2 = 4

So this follows Pythagoras theorem.

Hence \angleAOB = 90°

Now we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

\\\because \angle APB = \angle AOB\\ \Rightarrow \angle APB = 45^{\circ}

Hence proved.

Posted by

infoexpert24

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