Question 10
A circle has radius √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 the major segment is 45∘.
Draw a circle having centre O. Let AB = 2cm be the chord of a circle. The chord AB is divided by the line OM into two equal segments.
To prove that ∠APB=45∘
Here, AN=NB=1 cm
And OB=√2 cm
In ΔONB, OB2=ON2+NB2 [use Pythagoras theorem]
⇒ (√2)2=ON2+(1)2
⇒ ON2=2−1=1
ON=1 cm
[taking positive square root, because distance is always positive]
Also, ∠ONB=90∘
[ON is the perpendicular bisector of the chord AB]
∴ ∠NOB=∠NBO=45∘
[ON =NB =1cm,angles opposite to equal sides are also equal]
Similarly,
∠AON=45∘
Now,
∠AOB=∠AON+∠NOB
=45∘+45∘=90∘
We know that, chord subtends an angle on the circumference is half the angle subtended by it on the centre.
∴ ∠APB=12 ∠AOB
∠APB=90∘2=45∘
Hence proved.