A circle has a 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 45º.
Solution:
Given, a circle has a radius of √2 cm
The circle is divided into two segments by a chord of length 2 xm
We have to 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 = 2 cm be a chord of a circle
A chord AB is divided by the line OM into two equal segments.
Now, AN = NB = 1 cm
OB = √2 cm
In triangle ONB,
By Pythagorean theorem,
OB² = ON² + NB²
(√2)² = ON² + (1)²
ON² = 2 - 1
ON = 1 cm
Since ON is the perpendicular bisector of the chord AB
∠ONB = 90°
So, ∠NOB = ∠NBO = 45°
Similarly, ∠AON = 45°
Now, ∠AOB = ∠AON + ∠NOB
= 45° + 45°
= 90°
We know that the chord subtends an angle to the circle equal to half the angle subtended by it to the centre.
∠APB = 1/2 ∠AOB
= 1/2 (90°)
= 45°
Therefore, ∠APB = 45°
✦ Try This: In a circle of radius 35cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 10
A circle has a 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 45º.
Summary:
A circle has a radius √2 cm. It is divided into two segments by a chord of length 2 cm. It is proven that the angle subtended by the chord at a point in the major segment is 45º
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