Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.6 Solutions

Angle subtended by an arc means the angle made by joining the arc’s endpoints to a point on the circle or to the centre. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.6 connects this idea with chord length, angles in the same segment and cyclic figures.

Exercise 5.6 of Chapter 5, I’m Up and Down and Round and Round Class 9, moves from chord length to angle relationships inside a circle. Students use the result that an arc subtends double the angle at the centre compared to a point on the circle. The exercise also uses the corollary that a diameter subtends a right angle at any point on the circle. Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.6 Solutions cover a 60° central angle chord problem, equal angles standing on the same chord, and a find x in circle Class 9 question from Fig. 5.26. Exercise Set 5.6 has three questions based on angles subtended by arcs and points on a circle.

Key Takeaways

• Central Angle: A 60° central angle with radius 12 cm gives chord length 12 cm.
• Same Segment: Points on the same side of chord AB make equal angles with A and B.
• Diameter Angle: A diameter subtends 90° at any point on the circle.
• Cyclic Angle: Opposite angles in Fig. 5.26 add up to 180°.

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.6 Solutions Structure 2026

Class 9 Maths Ganita Manjari Chapter 5 Exercise 5.6 Solutions for Angles and Chords

Exercise 5.6 uses the connection between a chord and the angle it subtends at the centre or on the circle. A chord becomes easier to measure when the central angle and radius are known.

Q1. 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?

The length of chord AB is 12 cm.

Given:

Central angle AOB = 60°

Radius = 12 cm

So:

OA = 12 cm

OB = 12 cm

To find:

AB

Solution:

1. OA and OB are radii of the same circle.

Equation:

OA = OB = 12 cm

1. The central angle is given.

Equation:

∠AOB = 60°

1. In ΔAOB, two sides are equal.

Equation:

OA = OB

1. Since the angle between the equal sides is 60°, the triangle is equilateral.

Therefore:

OA = OB = AB

1. Substitute the radius.

Equation:

AB = 12 cm

Answer:

The length of chord AB is 12 cm.

Ganita Manjari Class 9 Chapter 5 Exercise 5.6: Central Angle and Chord Length

A 60° angle at the centre creates an equilateral triangle when the two joining sides are radii. This is why the chord becomes equal to the radius in Question 1.

Central Angle and Chord Length

If O is the centre and A, B are points on the circle, then OA and OB are radii.

Copy-friendly result:

OA = OB

If ∠AOB = 60°, then:

ΔAOB is equilateral

So:

AB = OA = OB

For Question 1:

AB = 12 cm

This is the shortest solution for central angle and chord length questions.

Class 9 Maths Chapter 5 Exercise 5.6 Solutions for Angles in Same Segment

Angles in the same segment are equal because they stand on the same chord or the same arc. In this exercise, the chord is AB and the points X and Y are checked using ∠AXB and ∠AYB.

Q2. Let A and B be two points on a circle with centre O.

Q2(i). Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?

No, ∠AXB and ∠AYB cannot be different if X and Y lie on the same side of AB on the circle.

Reason:

1. A and B are fixed points on the circle.
2. X and Y lie on the same side of AB.
3. Both angles stand on the same chord AB.
4. Angles in the same segment of a circle are equal.

Equation:

∠AXB = ∠AYB

Answer:

No, such points X and Y do not exist on the same side of AB.

Q2(ii). Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of AB?

No, this is not always true.

Reason:

1. Equal angles do not always force X and Y to lie on the same side of AB.
2. If AB is a diameter, any point on the circle makes a right angle at the circumference.
3. X and Y may lie on opposite sides of AB and still make equal angles.

Copy-friendly example:

If AB is a diameter, then:

∠AXB = 90°

∠AYB = 90°

So:

∠AXB = ∠AYB

But:

X and Y can be on opposite sides of AB

Answer:

No, equal angles do not always mean that X and Y lie on the same side of AB.

Q2(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?

Not always. The circle through A, B and X passes through Y only when Y lies on the same circle determined by A, B and X.

Reason:

1. Three non-collinear points A, B and X determine one unique circle.
2. Equal angles alone are not enough in every position.
3. A same-side condition is needed for the standard concyclicity result.
4. If X and Y are on the same side of AB and ∠AXB = ∠AYB, then A, B, X and Y are concyclic.

Copy-friendly conclusion:

If X and Y are on the same side of AB:

∠AXB = ∠AYB

Therefore:

A, B, X and Y lie on one circle

Without that condition, the result is not guaranteed.

Answer:

No, the circle through A, B and X does not always pass through Y.

Class 9 Maths Ganita Manjari Chapter 5 Solutions: Angle Subtended by Diameter Is 90

The chapter states that the angle subtended by a diameter at any point on the circle is 90°. This follows because the angle at the centre is 180°, and the angle at the circle is half of it.

Angle Subtended by Diameter Is 90

If AB is a diameter and X lies on the circle, then ∠AXB is a right angle.

Copy-friendly result:

∠AXB = 90°

Reason:

Angle at centre = 180°

Angle at circle = 180° ÷ 2

Angle at circle = 90°

This result is useful in Question 2(ii).

Class 9 Maths Chapter 5 Exercise 5.6 Solutions: Find x in Fig. 5.26

Fig. 5.26 shows a circle with an angle of 100° and an opposite angle marked x. Since the points lie on the circle, the opposite angles form a cyclic angle relation.

Q3. Find x in Fig. 5.26.

The value of x is 80°.

Given:

One angle = 100°

Opposite angle = x

To find:

x

Solution:

1. The figure shows four points on a circle.
2. Opposite angles of a cyclic quadrilateral add up to 180°.
3. The given angle and x are opposite angles.

Equation:

x + 100° = 180°

x = 180° - 100°

x = 80°

Answer:

x = 80°

I’m Up and Down and Round and Round Class 9: Concepts Used in Exercise 5.6

Exercise 5.6 uses circle angle facts rather than length formulas. The main idea is that arcs, chords and points on the circle create fixed angle relationships.

Angle Subtended by Arc Class 9

An arc subtends an angle when its endpoints are joined to a point.

At the centre:

Angle subtended by arc AB = ∠AOB

At the circle:

Angle subtended by arc AB = ∠AXB

Main result:

Angle at centre = 2 × angle at circle

Copy-friendly form:

∠AOB = 2∠AXB

Angles in Same Segment Class 9

Angles in the same segment stand on the same chord and lie on the same side of it.

Copy-friendly result:

∠AXB = ∠AYB

Condition:

X and Y lie on the same side of AB.

This result is used in Question 2(i).

Cyclic Quadrilateral Opposite Angles

When four points lie on one circle, they form a cyclic quadrilateral.

Copy-friendly result:

Opposite angles add up to 180°.

For Fig. 5.26:

x + 100° = 180°

x = 80°

Class 9 Maths Circles Solutions: Formula Pattern for Exercise 5.6

Exercise 5.6 becomes easier when students identify whether the question is about a centre angle, a same-segment angle or a cyclic angle pair.

Useful Results for Exercise 5.6

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5