NCERT Solutions for Class 9 Maths Chapter 5 Exercise 5.6 (2026-27)

Exercise Set 5.6

Question 1.
In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?

Solution:

OA and OB are radii of the same circle.

Therefore:

OA = OB = 12 cm

So, ∆AOB is an isosceles triangle.

Given:

∠AOB = 60°

Let:

∠OAB = ∠OBA = x

Using the angle-sum property of a triangle:

x + x + 60° = 180°

2x = 120°

x = 60°

Thus, all three angles of ∆AOB are 60°. Therefore, ∆AOB is equilateral.

Hence:

AB = OA = OB = 12 cm

Therefore, the length of chord AB is 12 cm.

Question 2.
Let A and B be two points on a circle with centre O.

(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?

Solution:
No. Points X and Y lie on the same segment of the circle formed by chord AB.

Angles subtended by the same chord at points lying in the same segment are equal.

Therefore:

∠AXB = ∠AYB

Hence, these two angles cannot be different.

(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?

Solution:
No, this is not always true.

For example, if AB is a diameter, then the angle subtended by AB at any point on the circle is 90°. Therefore, points X and Y may lie on opposite sides of AB and still satisfy:

∠AXB = ∠AYB = 90°

Hence, equal angles do not always mean that X and Y lie on the same side of AB.

(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B, and X also pass through Y?

Solution:
If X and Y lie on the same side of AB and

∠AXB = ∠AYB, then A, B, X, and Y are concyclic.

Therefore, the circle passing through A, B, and X will also pass through Y.

However, the condition that X and Y lie on the same side of AB is important.

Question 3.
Find x in Fig. 5.26.

Solution:
Given:

∠ADC = 100°

The angle subtended by arc ABC at the centre is twice the angle subtended by it at the circumference.

Therefore:

Reflex central angle corresponding to arc ABC
= 2 × 100°
= 200°

The complete angle around the centre is 360°.

Therefore, the central angle corresponding to the remaining arc ADC is:

360° - 200° = 160°

The angle x subtended by this arc at the circumference is half the corresponding central angle.

x = 160°/2

x = 80°

Therefore, x = 80°.

Make the Most of Vedantu’s Class 9 Maths Chapter 5 Exercise 5.6 Solutions

Exercise 5.6 is especially useful when you want to connect circle properties with actual angle problems. The Class 9 Maths Chapter 5 Exercise 5.6 solutions can help you:

• Understand how a 60° central angle leads to an equilateral triangle.

• Work out questions involving the same chord and different points on a circle.

• Recognise how concyclic points are identified through angle relationships.

• Use the given figure to decide which circle property applies.

• Refer to the Class 9 Maths Chapter 5 Exercise 5.6 solutions CBSE material when practising theorem-based questions.

• Keep the Class 9 Maths Chapter 5 PDF or Class 9 Maths Chapter 5 solutions PDF handy to practise the full chapter.

• Use the Class 9 Maths Chapter 5 Solutions PDF Ganita Manjari material when you want to go back to the chapter and check related exercises.

For more explanations, solutions, and study material, students can visit Vedantu and explore NCERT Solutions Class 9 Maths.

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Chapter-Specific NCERT Solutions for Class 9 Maths

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