Class 9 Maths Chapter 5 Exercise 5.6 Solutions – Angles Subtended by an Arc

R K Singh 23/08/2026 No Comments Class 9 Maths Chapter 5 – I’m Up and Down, and Round and Round Class 9 Maths Solutions

Class 9 Maths Chapter 5 Exercise 5.6 Solutions (Ganita Manjari 2026) – Angles Subtended by an Arc

📘 NCERT Ganita Manjari (2026) 📚 CBSE 2026–27 🏆 Step-by-Step Solutions

Class 9 Maths Chapter 5 Exercise 5.6 Solutions

Angles Subtended by an Arc

Prepare with complete Exercise 5.6 solutions from Chapter 5 – I’m Up and Down, and Round and Round based on the latest NCERT Ganita Manjari (2026). This exercise covers arcs, angles subtended by an arc, Theorem 9, angles in the same segment, and the important result for an angle subtended by a diameter. Questions 1, 2 and 3 are explained in a clear, step-by-step CBSE answer-writing style.

📖

Exercise

5.6

❓

Questions

3 Questions

📐

Main Concept

Angles &
Arcs

📏

Key Result

Theorem 9

🧮

Centre Angle
= 2 × Circle Angle

⭐

Difficulty

Moderate

🎯

Exam Importance

★★★★★

🎯 By the End of Exercise 5.6, You Will Be Able To…

✅ Understand Major and Minor Arcs

✅ Understand Angles Subtended by an Arc

✅ Apply Theorem 9 Correctly

✅ Understand Angles in the Same Segment

✅ Use the Diameter–90° Result

✅ Solve Questions 1, 2 and 3 Step by Step

👇 View All Exercise 5.6 Solutions

📑 Table of Contents

📘 Learn

About Exercise Set 5.6
Learn Before We Solve
Reviewed & Prepared By

✍️ Solve

Theorem 9 – Proof (Case1-Case2)
Exercise 5.6 – Question 1
Exercise 5.6 – Question 2
Exercise 5.6 – Question 3

🎯 Revise

Quick Revision Dashboard
Frequently Asked Questions
Continue Learning

📖 About Class 9 Maths Chapter 5 Exercise 5.6

Class 9 Maths Chapter 5 Exercise 5.6 Solutions cover the sixth exercise of Chapter 5 – I’m Up and Down, and Round and Round from the latest NCERT Ganita Manjari (2026). This page is designed for students looking for complete, reliable and easy-to-follow NCERT Class 9 Maths solutions arranged according to the exercise questions.

The page presents Exercise 5.6 in a clear, student-friendly CBSE answer-writing format and is useful for classroom practice, homework, self-study, revision and examination preparation. Each question is organised separately so students can quickly find the solution they need.

🎯 Exercise Snapshot

📘 What You’ll Find

Complete coverage of Exercise Set 5.6.
NCERT-based solutions for all three questions.
Clear, question-wise presentation for Class 9 students.

📚 Page Includes

Step-by-step NCERT solutions.
Complete Exercise 5.6 question coverage.
Simple CBSE answer-writing approach.

🏆 Best For

Students following Ganita Manjari (2026).
Homework and classroom practice.
Revision and CBSE examination preparation.

📚 Learn Before You Solve

✨ Solve करने से पहले कुछ सीखें

Exercise Set 5.6 को आसानी से हल करने से पहले, आइए arcs और angles subtended by an arc के मूल concepts को समझें। यह section NCERT / Ganita Manjari Class 9 Mathematics के अनुसार है, लेकिन concepts को सरल और student-friendly तरीके से समझाया गया है।

यहाँ हम minor और major arcs, angle at the centre, angle at a point on the circle, Theorem 9 और angle in a semicircle को step-by-step समझेंगे।

🎯 इस LBWS में हम क्या सीखेंगे?

🔹 Arc — endpoints, minor arc और major arc

🔹 Angle Subtended at Centre — arc से central angle

🔹 Angle at a Point on Circle — same arc का angle

🔹 Theorem 9 (Case-1) — centre angle और circle angle का संबंध

🔹 Theorem 9 (Case-2) — centre angle और circle angle का संबंध

🔹 Same Arc → Equal Angles — important consequence

🔹 Angle in a Semicircle — 90° का corollary

🔹 Important Condition — point की position क्यों matter करती है

🔹 Concept Connection — पूरे topic को एक साथ समझें

🧠 Quick Revision Dashboard — exam से पहले quick revision

👆 किसी भी concept पर click करके सीधे उस section पर जाएँ।

Arc Basics

Understand arcs, their endpoints, and the difference between minor and major arcs.

Before learning how an arc subtends an angle, we first need to understand what an arc is and how two points on a circle can determine two different arcs.

📐 Minor Arc and Major Arc

Two arcs are formed between the same endpoints A and B.

What is an Arc?

An arc is a connected portion of a circle.

It is identified by its two points on the circle, called the endpoints, together with the curved part of the circle joining them.

📍 Endpoints of an Arc

The two points on the circle that mark the ends of an arc are called its endpoints.
In the figure, A and B are the endpoints.

Minor Arc

The smaller of the two arcs between the same endpoints is called the minor arc.
Here: AXB

Major Arc

The larger of the two arcs between the same endpoints is called the major arc.
Here: AYB

💡 Key Observation

The same two endpoints A and B can determine two arcs:
one minor arc and one major arc.

So, whenever a question refers to an arc, first identify which route between the two endpoints is being considered.

🧠 Remember

Same endpoints → Two arcs
Smaller arc → Minor arc | Larger arc → Major arc

Angle Subtended by an Arc at the Centre

Understand how a chosen arc of a circle determines an angle at its centre.

Once we understand an arc, the next step is to see how that arc subtends an angle at the centre of the circle. The angle depends on the particular arc chosen between the two endpoints.

📐 Arc and Central Angle

The minor arc AXB subtends ∠AOB at the centre O.

What Does “Angle Subtended by an Arc” Mean?

Consider an arc AXB of a circle with centre O.

The endpoints of the arc are A and B. Join the centre O to A and B.

The angle formed at the centre is:

Angle Subtended at the Centre

∠AOB

The arc AXB subtends ∠AOB at the centre O.

🔄 Think of It as an Angle Being Swept

Imagine moving from OA to OB along the chosen arc. The angle swept during this movement is the angle subtended by that arc at the centre.

Minor Arc

The smaller arc gives the smaller central angle.

Major Arc

The larger arc gives the larger central angle.

📌 How to Identify the Arc

Central angle < 180° Minor arc

Central angle > 180° Major arc

🧠 Remember

An arc subtends an angle at the centre.
Minor arc → smaller central angle | Major arc → larger central angle

Angle Subtended at a Point on the Circle

See how the same fixed arc can subtend the same angle at suitable points on the circle.

We have seen how an arc subtends an angle at the centre. Now let us move the vertex from the centre to a point on the circumference. If A and B are the endpoints of a fixed arc and P, Q or R is a suitable point on the circle, the arc subtends angles such as ∠APB, ∠AQB and ∠ARB.

📐 Fixed Arc and Angles at P, Q and R

P, Q and R are suitable points on the circle outside the fixed arc AKB.

What Happens When the Point Moves?

Keep the arc AKB fixed. Now choose different points P, Q and R on the circle, outside the chosen arc.

Join each point to the endpoints A and B. We then obtain:

📐 Angles Formed

∠APB ∠AQB ∠ARB

Each of these angles is subtended by the same fixed arc AKB.

🔬 Observe the Activity

Follow the idea of the NCERT activity:

Fix the arc AKB.
Choose suitable points P, Q and R on the circle outside the chosen arc.
Join each point to A and B.
Measure ∠APB, ∠AQB and ∠ARB.
Compare the three measured angles.

🔎 Main Observation

Although the point of observation changes from P to Q to R, the fixed arc remains the same.

The angles subtended by this same arc at these suitable points are equal.

∠APB = ∠AQB = ∠ARB

⚠️ Important Condition — Do Not Miss This

It is not enough to remember the statement as:

❌ “Same chord → always equal angles.”

The point where the angle is formed must be a point on the circle and must lie in the required position outside the chosen arc.

✅ Same fixed arc + suitable points on the circle → equal angles subtended by that arc

🧠 Remember

Same fixed arc → Same angle at suitable points on the circle

🔵 Theorem 9

Angle subtended by an arc at the centre and at a point on the circle

Theorem 9

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.

⭐ Core Relationship

Angle at centre = 2 × Angle at circle
Angle at circle = ½ × Angle at centre

Case 1 — Complete Proof

Given,

AFB is an arc of a circle.
∠ACB is the angle subtended by arc AFB at the centre C.
D is a point on the circle outside arc AFB.
The extension of DC meets the circle at E on arc AFB.

To show,

∠BCA = 2∠BDA

📐 Figure — Case 1

Case 1: D lies outside the chosen arc AFB and D, C, E are collinear.

✍️ Proof

Since CB and CD are radii of the same circle,

CB = CD

[Radii of the same circle]

Therefore, △DCB is an isosceles triangle.

∠CBD = ∠CDB

[Angles opposite equal sides of an isosceles triangle]

Since CE is an extension of CD, ∠BCE is an exterior angle of △BCD.

∠BCE = ∠CBD + ∠CDB

[Exterior angle theorem]

∠BCE = 2∠BDC

[Since ∠CBD = ∠CDB]

Similarly, CA and CD are radii of the same circle.

CA = CD

[Radii of the same circle]

Therefore, △ADC is an isosceles triangle.

∠CAD = ∠CDA

[Angles opposite equal sides of an isosceles triangle]

Since CE is an extension of CD, ∠ACE is an exterior angle of △ACD.

∠ACE = ∠CAD + ∠CDA

[Exterior angle theorem]

∠ACE = 2∠CDA

[Since ∠CAD = ∠CDA]

Now, CE lies inside ∠BCA. Therefore,

∠BCA = ∠BCE + ∠ECA

Substituting the two results obtained above,

∠BCE = 2∠BDC

∠ECA = 2∠CDA

Hence,

∠BCA = 2∠BDC + 2∠CDA

∠BCA = 2(∠BDC + ∠CDA)

But,

∠BDA = ∠BDC + ∠CDA

[DC lies inside ∠BDA]

Therefore,

∠BCA = 2∠BDA

✅ Hence Proved

∠BCA = 2∠BDA

🧠 Theorem 9 — Key Idea

Angle subtended by an arc at the centre = 2 × angle subtended by the same arc at a suitable point on the circle.
∠BCA = 2∠BDA

Case 2 — Extension of DC Meets the Circle at E

Given,

AFB is an arc. ∠ACB is the angle subtended by arc AFB at centre C. D is a point on the circle outside arc AFB such that when we extend DC it cuts the circle at some point E outside the arc AFB, as shown in Fig. 5.22.

To show,

∠ACB = 2∠ADB

📐 Figure

Fig. 5.22: Angle subtended by arc AFB

✍️ Proof

Since CA = CD, triangle △ADC is an isosceles triangle.

∠ADC = ∠CAD

[Angles opposite equal sides of an isosceles triangle]

∠ACE is the exterior angle of △ADC.

∠ACE = ∠ADC + ∠CAD

[Exterior angle theorem]

∠ACE = 2∠ADC

[Since ∠ADC = ∠CAD]

Similarly, △BCD is an isosceles triangle because

CB = CD

[Radii of the same circle]

∠BDC = ∠CBD

[Angles opposite equal sides of an isosceles triangle]

∠BCE is the exterior angle of △BCD.

∠BCE = ∠BDC + ∠CBD

[Exterior angle theorem]

∠BCE = 2∠BDC

[Since ∠BDC = ∠CBD]

🔎 Now combine the two angle relations

Since E lies outside arc AFB, the central angle is obtained by subtracting the two exterior angles.

Also,

∠ACB = ∠ACE − ∠BCE

= 2∠ADC − 2∠BDC

= 2(∠ADC − ∠BDC)

Also,

∠ADB = ∠ADC − ∠BDC

Therefore,

∠ACB = 2(∠ADB)

Hence,

∠ACB = 2∠ADB

✅ Hence Proved

∠ACB = 2∠ADB

Same Arc → Equal Angles

A useful consequence of Theorem 9

Once we know Theorem 9, we can understand an important pattern in a circle: the same arc subtends equal angles at suitable points on the circle.
In the figure, the fixed arc is AXB. Points D, E and F lie on the circle outside this arc.

📐 Same Arc Subtends Equal Angles

Same arc AXB subtends equal angles at D, E and F.

🔎 Why are the three angles equal?

The arc AXB is fixed. Therefore, the central angle subtended by this arc, ∠ACB, is the same for every point chosen on the circle outside the arc.

∠ACB = 100°
∠ADB = ∠AEB = ∠AFB = ½ × 100° = 50°

This follows directly from Theorem 9:

Angle at the centre = 2 × angle at the circle

So, for the same arc, every suitable point on the circle gives the same angle.

Same arc AXB
↓
Same central angle ∠ACB
↓
∠ADB = ∠AEB = ∠AFB = 50°

⚠️ Important Condition

The points D, E and F must be on the circle and in the appropriate position outside the chosen arc AXB.

So do not remember this as simply: “Same chord → always equal angles.”

The correct idea is: Same arc + suitable points on the circle → equal angles.

🧠 Student Memory

Same arc + suitable points on the circle → Equal angles
∠ADB = ∠AEB = ∠AFB

🔵 Corollary — Angle in a Semicircle

A 90° angle follows directly from Theorem 9

The angle subtended by a diameter at any point on the circle is 90°.
This result is called the angle in a semicircle.

📐 Angle in a Semicircle

AB is a diameter, so the angle ADB is a right angle.

🔎 Why is the angle 90°?

Let AB be a diameter of the circle and C be its centre.

Since AB is a diameter, the radii CA and CB form a straight angle at the centre.
∠ACB = 180°

By Theorem 9, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at a point on the circle.
Therefore,
∠ADB = ½ × ∠ACB

∠ADB = ½ × 180°
= 90°

Therefore, ∠ADB = 90° .

Hence, the angle subtended by a diameter at any point on the circle is a right angle.

💡 What is a Corollary?

A corollary is a fact that follows immediately from an already proved result.

Here, the 90° result follows directly from Theorem 9.

🧠 Quick Memory

Diameter → Central angle = 180°
Theorem 9 → Angle at circle = ½ × 180°
Therefore → Angle in a semicircle = 90°

🟠 Same Arc Segment: What Changes and What Does Not?

Same arc does not mean “same chord → always equal angles”

The important question is not only which chord is used. We must also look at which arc is being considered and where the vertex of the angle lies.
Let us compare the correct situation with other positions.

✅ Suitable Points — Same Arc, Equal Angles

D, E and F lie on the circle in the required position relative to the selected arc AXB.

The arc AXB is fixed.

When suitable points D, E and F are taken on the circle outside this selected arc, the same arc subtends equal angles:

∠ADB = ∠AEB = ∠AFB

This is the situation described by Theorem 9: the angle subtended by the same arc at suitable points on the circle remains the same.

⚠️ Other Positions — Do Not Apply Automatically

The position of the vertex matters; do not apply the equal-angle rule without checking the conditions.

The statement “same chord → always equal angles” is too broad.

The theorem concerns the angle subtended by a specified arc at a point on the circle in the appropriate position relative to that arc.

Therefore, simply seeing the same endpoints A and B is not enough to apply the result.

🔍 Before You Use “Same Arc → Equal Angles”

Which arc is being considered?
Where is the vertex of the angle?
Is the vertex on the circle?
Is it in the required position outside the selected arc?

🚫 Important: Do Not Memorise the Wrong Rule

❌ Wrong shortcut: “Same chord always gives equal angles.”

✅ Correct understanding: “Angles subtended by the same arc at appropriate points on the circle are equal.”

🧠 Student Memory Rule

Same arc + Suitable points on the circle → Equal angles
Always check the arc and the position of the vertex first.

🔗 Concept Connection

अब तक सीखे गए सभी concepts एक-दूसरे से जुड़े हुए हैं। इस connection को समझ लेने पर Exercise 5.6 के questions को solve करना बहुत आसान हो जाता है।

Step 1 Arc

Step 2 Minor Arc / Major Arc

Step 3 Central Angle

Step 4 Angle at a Point on the Circle

⭐ Theorem 9 Centre angle = 2 × angle at the circle

Consequence Same arc → Equal angles

🔵 Corollary Diameter → 90°

Remember: Arc → Central Angle → Angle on Circle → Theorem 9 → Equal Angles → Diameter gives 90°.

⚡ Quick Revision Dashboard

Exercise 5.6 — Angles Subtended by an Arc

🎯 CORE IDEA

Central Angle = 2 × Angle at the Circle

🔵 1. Arc

An arc is a connected portion of a circle between two points on the circle.

The smaller arc is the minor arc and the larger arc is the major arc.

Smaller → Minor Arc

Larger → Major Arc

📐 2. Angle Subtended at Centre

If an arc has endpoints A and B, the angle subtended by the arc at centre O is the central angle ∠AOB.

The angle depends on the arc being swept from OA to OB.

Minor arc → smaller central angle
Major arc → larger central angle

🟠 3. Angle at a Point on the Circle

If A and B are the endpoints of an arc and D is a suitable point on the circle outside the selected arc, then the arc subtends ∠ADB at D.

The angle does not depend on which suitable point on the circle is selected.

Same arc + suitable point on circle → same subtended angle

⭐ Theorem 9

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.

∠Centre = 2 × ∠Circle
∠Circle = ½ × ∠Centre

🧠 Theorem 9 — Quick Use

Whenever a question gives the angle at the centre and asks for the angle at a suitable point on the circle:

Angle at circle = ½ × central angle

If the angle at the circle is given, then:

Central angle = 2 × angle at circle

🔗 Same Arc → Equal Angles

For a fixed arc AB, if D, E, F are suitable points on the circle outside that arc, then:

∠ADB = ∠AEB = ∠AFB

Each angle is half of the same central angle.

⚠️ Important Condition

Do not memorise the wrong rule!

❌ Same chord always gives equal angles.
✅ Check which arc is being considered.
✅ The vertex must be on the circle.
✅ The point must be in the required position outside the selected arc.

🔵 Corollary — Angle in a Semicircle

A diameter subtends a straight angle of 180° at the centre.

By Theorem 9:

∠ at circle = ½ × 180° = 90°

Diameter → 90°
The angle in a semicircle is a right angle.

🔗 Concept Connection

Arc

→

Minor / Major Arc

→

Central Angle

→

Angle on Circle

→

Theorem 9

→

Same Arc → Equal Angles

→

Diameter → 90°

📝 Exercise Set 5.6 — Questions at a Glance

A circle has centre O, central angle ∠AOB = 60° and radius 12 cm. Find the length of chord AB.

For points A and B on a circle, analyse whether points X and Y can produce different/equal subtended angles and whether equal angles imply the required cyclic position.

In Fig. 5.26, the angle at D is 100°. Find x.

🧩 Q3 — Key Idea

In the cyclic quadrilateral of Fig. 5.26, the opposite angles are supplementary.

x + 100° = 180°

x = 80°

Final Answer: x = 80°

✅ Exam Checklist

✓ Identify the arc first.

✓ Identify the central angle.

✓ Check the position of the point on the circle.

✓ Use Theorem 9 carefully.

✓ Same arc → equal angles only under the required condition.

✓ Diameter → 90°.

⭐ Arc → Central Angle → Angle on Circle | Centre Angle = 2 × Circle Angle | Diameter → 90°

📝 Class 9 Maths Chapter 5 Exercise 5.6 Solutions

Solve all three questions of Class 9 Maths Chapter 5 Exercise 5.6 with clear, step-by-step NCERT Ganita Manjari (2026) solutions. This exercise focuses on angles subtended by an arc, Theorem 9, angles in the same arc segment and the important result for an angle subtended by a diameter.

📝 Step-by-Step Solutions 🎯 NCERT & CBSE Aligned ⭐ Complete Exercise 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

Given,

Radius of the circle = 12 cm

∠AOB = 60°

To find,

Length of chord AB

📐 Figure

OA = OB = 12 cm and ∠AOB = 60°.

✍️ Solution

Since OA and OB are radii of the same circle,

OA = OB

Therefore, △AOB is an isosceles triangle.

Given,

∠AOB = 60°

Since OA = OB, the angles opposite to these equal sides are equal.

∠OAB = ∠OBA

In △AOB, the sum of the three angles is 180°.

∠OAB + ∠OBA + ∠AOB = 180°

∠OAB + ∠OBA + 60° = 180°

∠OAB + ∠OBA = 120°

But,

∠OAB = ∠OBA

∠OAB = ∠OBA = 60°

Thus all three angles of △AOB are 60°.

∠OAB = ∠OBA = ∠AOB = 60°

Therefore, △AOB is an equilateral triangle.

Key idea: In an equilateral triangle, all three sides are equal.

Hence,

AB = OA = OB

Since the radius is 12 cm,

AB = 12 cm

✅ Final Answer

Length of chord AB = 12 cm

Question 2

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

Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of AB?
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?

Question 2 (i)

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

✍️ Solution

Given,

A and B are two points on a circle.
X and Y are points on the circle on the same side of chord AB.

📐 Figure

X and Y lie on the same segment of the circle.

Solution

Since X and Y lie on the same side of chord AB, the angles ∠AXB and ∠AYB are subtended by the same chord AB in the same segment of the circle.

Angles subtended by the same chord in the same segment of a circle are equal.

∠AXB = ∠AYB

Therefore, ∠AXB cannot be different from ∠AYB.

∴ No, there are no such points X and Y on the same side of AB for which ∠AXB is different from ∠AYB.

Question 2 (ii)

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

✍️ Solution

Given,

A and B are two fixed points on a circle.
∠AXB = ∠AYB.
We have to determine whether X and Y must lie on the same side of AB.

📐 Figure

X and Y lie on opposite sides of diameter AB, yet both subtend a right angle at AB.

Solution

No, it is not necessarily true.

Consider the special case in which AB is a diameter of the circle.

By the theorem on the angle in a semicircle, the angle subtended by diameter AB at any point on the circle is a right angle.

∠AXB = 90°

∠AYB = 90°

Therefore,
∠AXB = ∠AYB

But X and Y are on opposite sides of AB. Hence, equality of ∠AXB and ∠AYB does not imply that X and Y must lie on the same side of AB.

∴ No, X and Y need not lie on the same side of AB.

Question 2 (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

Given,

A and B are two fixed points.
X and Y are points not lying on the circle.
∠AXB = ∠AYB.

📐 Figure

The circle through A, B and X also passes through Y.

Solution

Let the circle passing through A, B and X be drawn. We are given that

∠AXB = ∠AYB

The angle ∠AXB is the angle subtended by chord AB at X. Since ∠AYB is equal to ∠AXB, the point Y lies on the same circle through A, B and X.

Therefore, the four points A, B, X and Y lie on one circle.

Key idea: Equal angles standing on the same chord AB identify points lying on the same circle.

∴ Yes, the circle through A, B and X also passes through Y.

Question 3

Find x in Fig. 5.26.

✍️ Solution

Given,

A, B, C and D are points on the same circle.
∠ADC = 100°.
∠ABC = x.

📐 Figure 5.26

Fig. 5.26 — A, B, C and D lie on the same circle.

Solution

Since A, B, C and D lie on the same circle, ABCD is a cyclic quadrilateral.

Property used: Opposite angles of a cyclic quadrilateral are supplementary.

Therefore,

∠ADC + ∠ABC = 180°

100° + x = 180°

x = 180° − 100°

x = 80°

∴ x = 80°

📚 Continue Learning

Congratulations! You have completed the Class 9 Maths Chapter 5 Exercise 5.6 Solutions. You have now worked through Exercise 5.6 of Chapter 5 – I’m Up and Down, and Round and Round. Continue your learning journey by moving to the previous or next exercise, or revisiting Exercise 5.6 whenever you need a quick revision.

← Exercise 5.5 ✅ Exercise 5.6 End of Chapter Exercise →

📖 Explore More Class 9 Maths Chapters

Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5

❓ Frequently Asked Questions

Find quick answers to common questions about Class 9 Maths Chapter 5 Exercise 5.6 Solutions. These FAQs help students understand what is covered in the exercise, how the questions are approached, and how to use the solutions effectively for learning, revision, and examination preparation.

What is the main focus of Class 9 Maths Exercise 5.6?

Exercise 5.6 of Class 9 Maths Chapter 5 focuses on angles subtended by an arc. The exercise applies the ideas developed in this part of the chapter, including Theorem 9, angles in the same arc segment, and the special result for an angle subtended by a diameter.

How many questions are there in Exercise 5.6?

There are three questions in Exercise 5.6. Question 2 also contains three parts, so the page presents Question 1, Question 2(i), Question 2(ii), Question 2(iii), and Question 3 separately for easier learning and revision.

What is Theorem 9 in Class 9 Maths Chapter 5?

Theorem 9 states that the angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the circle outside the arc. Thus, if the angle at the centre is known, the corresponding angle at the circle is half of it.

What should I know before attempting Exercise 5.6?

Students should be familiar with the basic meaning of an arc, major arc, minor arc, and an angle subtended by an arc at the centre or at a point on the circle. The preceding pages of Chapter 5 also introduce the ideas needed to understand Theorem 9 and its applications.

What is the relationship between the angle at the centre and the angle at the circle?

According to Theorem 9, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at a point on the circle outside the arc. Therefore, angle at the circle = half the angle at the centre.

Why is the angle subtended by a diameter 90°?

A diameter subtends a straight angle of 180° at the centre. By Theorem 9, the angle subtended by the same diameter at any point on the circle is half of this angle: 180° ÷ 2 = 90°. Thus, the angle subtended by a diameter at any point on the circle is 90°.

What does it mean that angles in the same arc segment are equal?

For the same arc, the angles subtended at different suitable points on the circle in the same arc segment are equal. The textbook illustrates this using points on the circle and shows that the corresponding subtended angles have the same measure.

Are these Class 9 Maths Exercise 5.6 Solutions based on Ganita Manjari 2026?

Yes. These Class 9 Maths Chapter 5 Exercise 5.6 Solutions are prepared for the latest NCERT Ganita Manjari (2026) textbook and follow the concepts, terminology, theorem and questions covered in Exercise Set 5.6. The solutions are presented in a student-friendly, step-by-step format for classroom practice, self-study, revision and CBSE examination preparation.

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Continue your preparation with more Class 9 Maths resources from Maths Gurukulam, or visit the official NCERT and CBSE websites for the latest textbooks, syllabus, and academic updates.

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Find Class 9 Maths Chapter 5 Exercise 5.6 Solutions helpful? For students who need personal support beyond self-study, join Newton Study Point for concept-based Maths coaching by an experienced teacher. Small batches, individual attention, regular tests, and complete CBSE exam preparation are available for Classes 8, 9, 10, 11 & 12.

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These Class 9 Maths Chapter 5 Exercise 5.6 Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026) and the CBSE curriculum. The solutions follow a clear, step-by-step approach to help students understand the questions in Exercise 5.6 and build confidence in solving circle and angle-based problems.

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Class 9 Maths Chapter 5 Exercise 5.6 Solutions (Ganita Manjari 2026) – Angles Subtended by an Arc

Class 9 Maths Chapter 5 Exercise 5.6 Solutions

Angles Subtended by an Arc

📑 Table of Contents

📖 About Class 9 Maths Chapter 5 Exercise 5.6

🎯 Exercise Snapshot

📚 Learn Before You Solve

Arc Basics

Angle Subtended by an Arc at the Centre

Angle Subtended at a Point on the Circle

🔵 Theorem 9

Case 1 — Complete Proof

Case 2 — Extension of DC Meets the Circle at E

Same Arc → Equal Angles

🔎 Why are the three angles equal?

🔵 Corollary — Angle in a Semicircle

🔎 Why is the angle 90°?

🟠 Same Arc Segment: What Changes and What Does Not?

🔗 Concept Connection

⚡ Quick Revision Dashboard

📝 Class 9 Maths Chapter 5 Exercise 5.6 Solutions

📚 Continue Learning

❓ Frequently Asked Questions

📚 Useful Learning Resources

Need Personal Guidance in Maths?

Rakesh Kumar Singh

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Class 9 Maths Chapter 5 Exercise 5.6 Solutions (Ganita Manjari 2026) – Angles Subtended by an Arc

Class 9 Maths Chapter 5 Exercise 5.6 Solutions

Angles Subtended by an Arc

📑 Table of Contents

📖 About Class 9 Maths Chapter 5 Exercise 5.6

🎯 Exercise Snapshot

📚 Learn Before You Solve

Arc Basics

Angle Subtended by an Arc at the Centre

Angle Subtended at a Point on the Circle

🔵 Theorem 9

Case 1 — Complete Proof

Case 2 — Extension of DC Meets the Circle at E

Same Arc → Equal Angles

🔎 Why are the three angles equal?

🔵 Corollary — Angle in a Semicircle

🔎 Why is the angle 90°?

🟠 Same Arc Segment: What Changes and What Does Not?

🔗 Concept Connection

⚡ Quick Revision Dashboard

📝 Class 9 Maths Chapter 5 Exercise 5.6 Solutions

📚 Continue Learning

❓ Frequently Asked Questions

📚 Useful Learning Resources

Need Personal Guidance in Maths?

Rakesh Kumar Singh

Leave A Reply Cancel reply

You May Also Like

Categories

Lists

Maths gurukulam

Feature

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