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Ex.10.5 Q2 Circles Solution - NCERT Maths Class 9

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Question

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

 Video Solution
Circles
Ex 10.5 | Question 2

Text Solution

What is known?

Chord’s length is equal to the radius.

What is unknown?

Angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Reasoning:

  • The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
  • A quadrilateral \(ABCD\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^{\circ}.\end {align}\)

Steps:

Draw a circle with any radius and centre \(O\). Let \(AO\) and \(BO\) be the \(2\) radii of the circle and let \(AB\) be the chord equal to the length of radius. Join them to form a triangle.

\(OA = OB = AB\)

Hence \(\begin{align} \Delta {ABO} \end {align}\) becomes an equilateral triangle.

Draw \(2\) points \(C\) and \(D\) on the circle such that they lie on major arc and minor arc respectively.

Since \(\begin{align} \Delta {ABO} \end {align}\) is an equilateral triangle, we get \(\begin{align} \angle {ABO}=60^{\circ} \end {align}\)

For the arc \(AB,\) \(\begin {align} \angle {AOB}=2 \angle {ACB} \end {align}\) as we know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

\[\begin{align}\angle ACB &= \frac{1}{2}\angle AOB\\&= \frac{1}{2} \times 60 \\&= 30^\circ \end{align}\]

As you can notice the points \(A, B, C\) and \(D\) lie on the circle. Hence \(A B C D\) is a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^\circ.\end {align}\)

Therefore,

\[\begin{align}\angle ACB + \angle ADB &= 180^\circ \\30 + \angle ADB &= 180\\\angle ADB &= 150^\circ \end{align}\]

So when the chord of a circle is equal to the radius of the circle, the angle subtended by the chord at a point on the minor arc is \(\begin{align}150^\circ.\end {align}\) and also at a point on the major arc is \(\begin{align}30^\circ.\end {align}\)

 Video Solution
Circles
Ex 10.5 | Question 2
  
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