📖 About Class 9 Maths Chapter 5 End-of-Chapter Exercises
Class 9 Maths Chapter 5 End of Chapter Exercises Solutions provide complete, question-wise support for the final exercises 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 who want reliable and easy-to-follow solutions to the End-of-Chapter Exercises in one place.
The complete exercise is presented in the same order as the textbook, with all 26 questions organised clearly for quick access. The page is useful for classroom practice, homework, self-study, revision, doubt-solving, and CBSE examination preparation. Students can use the question navigator to move directly to the question they want to study.
🎯 Exercise Snapshot
📘 What You’ll Find
- Complete coverage of the Chapter 5 End-of-Chapter Exercises.
- Question-wise solutions for all 26 questions.
- Clear presentation following the textbook order.
📚 Page Includes
- Step-by-step NCERT solutions.
- Easy question-wise navigation.
- Student-friendly CBSE answer-writing approach.
🏆 Best For
- Students following Ganita Manjari (2026).
- Homework, classroom practice and self-study.
- Chapter-end revision and CBSE preparation.
⭐ Why Use This Page?
- One place for the complete end exercise.
- Easy navigation between questions.
- Useful for last-minute chapter revision.
📚 Learn Before You Solve
✨ Understand the ideas first — then solve with confidence
Before solving the
End-of-Chapter Exercise — Chapter 5: Circles,
revise the essential circle concepts and theorems you need.
This LBWS focuses only on the
new and important ideas required here.
Concepts already explained in earlier exercises are linked
instead of being repeated.
🎯 What will you learn before solving?
🔹 Concyclic Points
— understand what it means for points to lie
on the same circle
🔹 Theorem 10
— Equal Angles → Concyclic Points
🔹 Cyclic Quadrilateral
— understand the circle-based quadrilateral
🔹 Theorem 11
— Opposite Angles of a Cyclic Quadrilateral
🔹 Theorem 12
— Converse of Theorem 11
🔹 Cyclicity Test
— use the angle condition to identify
a cyclic quadrilateral
🧠 Quick Revision Dashboard
— revise the key theorems, tests and circle properties
👆 Click any concept to jump directly to that section.
Theorem 10
Equal Angles → Concyclic Points
Theorem Statement
If a line segment AB joining two points A and B subtends
equal angles at two other points C and D that lie on the
same side of AB, then the four points lie on a circle.
Given:
Consider segment AB. Points C and D lie on the same side
of AB; points C and D are not on the line AB; and
∠ACB = ∠ADB.
To show:
A, B, C and D lie on the same circle.
Fig. 5.27A — D outside the circle
D is assumed to be outside the circle through A, B and C.
Fig. 5.27B — D inside the circle
D is assumed to be inside the circle through A, B and C.
💡 Why are two figures needed?
If D is not on the circle, it must be either
outside or inside the circle.
NCERT eliminates both possibilities.
✍️ Proof
1
Construct the circle through A, B and C
The points A, B and C are non-collinear
points.
So, using Theorem 1, there is a circle passing through A, B and C.
Let us show that D also lies on this circle.
So, using Theorem 1, there is a circle passing through A, B and C.
Let us show that D also lies on this circle.
2
Assume D is not on the circle
Suppose D is not on the circle.
Then D is either inside or
outside the circle.
Join AD.
Join AD.
Case 1 — D is outside the circle
If D is outside the circle, then AD intersects the
circle again at E, as shown in Fig. 5.27A.
Now C and E are on the same segment of the arc formed by chord AB. Therefore,
Since ∠AEB is an exterior angle of triangle BED,
Now C and E are on the same segment of the arc formed by chord AB. Therefore,
∠ACB = ∠AEB
Angles in the same segment of a circle
Since ∠AEB is an exterior angle of triangle BED,
∠AEB > ∠ADB
Therefore,
∠ACB > ∠ADB
But this contradicts the given condition
∠ACB = ∠ADB
Case 2 — D is inside the circle
If D is inside the circle, extend AD to meet the
circle at E, as shown in Fig. 5.27B.
Again, C and E are on the same segment of the arc formed by chord AB. Therefore,
Since ∠ADB is an exterior angle of triangle BED,
Again, C and E are on the same segment of the arc formed by chord AB. Therefore,
∠ACB = ∠AEB
Angles in the same segment of a circle
Since ∠ADB is an exterior angle of triangle BED,
∠ADB > ∠AEB
Therefore,
∠ADB > ∠ACB
But this contradicts the given condition
∠ACB = ∠ADB
🔍 Both possibilities are impossible
D cannot be outside the circle and D cannot be
inside the circle.
Therefore, D must be on the circle passing through A, B and C.
Therefore, D must be on the circle passing through A, B and C.
✅ Hence Proved
A, B, C and D are concyclic.
🧠 Remember
Equal angles subtended by the same line segment
at two points on the same side
→ the four points are concyclic.
Cyclic Quadrilateral
A four-sided figure whose vertices lie on the same circle.
Once four points are known to lie on the same circle, they form
a special type of quadrilateral called a
cyclic quadrilateral.
📐 Cyclic Quadrilateral ABCD
The four vertices A, B, C and D lie on the same circle.
What is a Cyclic Quadrilateral?
A quadrilateral whose four vertices lie on the
same circle is called a
cyclic quadrilateral.
In the figure, the four vertices A, B, C and D all lie on the circumference of the same circle.
In the figure, the four vertices A, B, C and D all lie on the circumference of the same circle.
🎯 The Key Condition
The important point is not merely that ABCD is a
quadrilateral.
Its four vertices must lie on one common circle.
Its four vertices must lie on one common circle.
🔗 Connection with Concyclic Points
If four points are concyclic, joining them
in order forms a cyclic quadrilateral.
So:
Concyclic vertices → Cyclic quadrilateral
So:
Concyclic vertices → Cyclic quadrilateral
🧠 Remember
Four vertices on one circle → Cyclic quadrilateral
Theorem 11
Opposite Angles of a Cyclic Quadrilateral
Theorem Statement
The sum of two opposite angles of a cyclic quadrilateral
is 180°.
Given:
A, B, C and D are the vertices of a cyclic 4-gon
(Fig. 5.28). The circle has centre O.
To show:
∠BAD + ∠BCD = 180°
Fig. 5.28 — Cyclic quadrilateral ABCD with centre O
💡 Key Idea
Each angle at the circumference is half the angle
subtended by the corresponding arc at the centre.
✍️ Proof
Consider arc BCD. Point A
is on the circle and lies outside the arc BCD.
So, ∠BAD subtends at the centre O the reflex angle BOD.
Similarly, ∠BCD is half the angle subtended by the arc BAD at the centre O.
So, ∠BAD subtends at the centre O the reflex angle BOD.
∠BAD = ½ × reflex angle BOD
To move from OB to OD
along C gives the reflex angle BOD.
Similarly, ∠BCD is half the angle subtended by the arc BAD at the centre O.
∠BCD = ½ × ∠BOD
The reflex angle BOD and the remaining angle BOD together
make one complete angle at the centre O.
reflex angle BOD + ∠BOD = 360°
Therefore,
∠BAD + ∠BCD
= ½ × 360°
Hence,
∠BAD + ∠BCD = 180°
✅ Hence Proved
∠BAD + ∠BCD = 180°
🧠 Remember
In a cyclic quadrilateral,
opposite angles are supplementary.
Opposite angles → 180°
Opposite angles → 180°
Theorem 12
Converse of Theorem 11 — Cyclic Quadrilateral
Theorem Statement
If two opposite angles of a quadrilateral add up to
180°, then the vertices of the quadrilateral lie on a
circle, i.e., they are concyclic.
Given:
ABCD is a 4-gon.
∠BAD + ∠BCD = 180°.
∠ABC + ∠CDA = 180°.
∠BAD + ∠BCD = 180°.
∠ABC + ∠CDA = 180°.
To show:
ABCD is cyclic.
📐 Figure 5.29 — C outside the circle
Fig. 5.29 — Circle through A, B and D; CD meets the circle again at E
📐 Figure 5.29B — C inside the circle
C lies inside the circle through A, B and D; CD meets the circle again at E.
💡 Why two cases?
If C does not lie on the circle through
A, B and D, then C must be
either outside or inside
the circle.
We eliminate both possibilities.
We eliminate both possibilities.
✍️ Proof
Suppose ABCD is not cyclic.
Since A, D and B are non-collinear, there is a circle passing through A, D and B.
If this circle does not pass through C, there are two cases:
C is outside the circle or C is inside the circle.
Since A, D and B are non-collinear, there is a circle passing through A, D and B.
If this circle does not pass through C, there are two cases:
C is outside the circle or C is inside the circle.
Case 1 — C is outside the circle
Let E be the point where
CD meets the circle again.
Then ABED is a cyclic 4-gon.
So, from Theorem 11,
Therefore, by the exterior angle property,
Then ABED is a cyclic 4-gon.
So, from Theorem 11,
∠BAD + ∠BED = 180°
We are also given
∠BAD + ∠BCD = 180°
Therefore,
∠BCD = ∠BED
But ∠BED is the exterior angle
of triangle BEC.
Therefore, by the exterior angle property,
∠BED > ∠BCE
Since B, C, D are arranged as shown
in the figure,
∠BCE = ∠BCD
Hence,
∠BED > ∠BCD
But we already obtained
∠BED = ∠BCD
This is impossible.
Case 2 — C is inside the circle
Now suppose C lies inside the circle.
Extend CD to meet the circle again at E.
Then ABED is a cyclic 4-gon.
Therefore, by Theorem 11,
Extend CD to meet the circle again at E.
Then ABED is a cyclic 4-gon.
Therefore, by Theorem 11,
∠BAD + ∠BED = 180°
We are given
∠BAD + ∠BCD = 180°
Hence,
∠BED = ∠BCD
Since D, C and E
are collinear, the rays ED and
EC are in the same direction.
Therefore,
∠BED = ∠BEC
Hence,
∠BEC = ∠BCD
Now, since CD and CE
are opposite rays,
∠BCD + ∠BCE = 180°
Therefore,
∠BEC + ∠BCE = 180°
But in triangle BCE,
∠BEC + ∠BCE + ∠CBE = 180°
Comparing the two equations,
∠CBE = 0°
This is impossible because B, C
and E are not collinear.
Therefore, C cannot be inside
the circle.
🔍 Both possibilities are impossible
C cannot be outside the circle.
C cannot be inside the circle.
Therefore, our assumption that ABCD is not cyclic is impossible.
Hence, C must lie on the circle passing through A, B and D.
C cannot be inside the circle.
Therefore, our assumption that ABCD is not cyclic is impossible.
Hence, C must lie on the circle passing through A, B and D.
✅ Hence Proved
A, B, C and D are concyclic.
🧠 Remember
If the opposite angles of a quadrilateral add up to
180°,
→ the quadrilateral is cyclic.
This is the converse of Theorem 11.
→ the quadrilateral is cyclic.
This is the converse of Theorem 11.
Cyclicity Test
Use the opposite-angle condition to recognise a cyclic quadrilateral.
Cyclic Quadrilateral
If A, B, C and D lie on
the same circle,
opposite angles are supplementary.
Theorem 11
opposite angles are supplementary.
Theorem 11
⇄
Opposite Angles
If
∠A + ∠C = 180°
then the four vertices are concyclic.
Theorem 12
∠A + ∠C = 180°
then the four vertices are concyclic.
Theorem 12
🔍 Quick Cyclicity Test
To test whether four points form a cyclic quadrilateral,
check whether a pair of opposite angles adds up to
180°.
🧠 Remember
Cyclic → Opposite angles = 180°
⇄
Opposite angles = 180° → Cyclic
⚡ Quick Revision Dashboard
Chapter 5 — Circles | End-of-Chapter Exercise
🎯 CORE IDEA
Equal Angles ↔ Concyclic Points
|
Opposite Angles ↔ 180°
🔵 1. Concyclic Points
Points are called concyclic when they
all lie on the same circle.
Three non-collinear points determine one unique circle. A fourth point is concyclic with them when it also lies on that circle.
Three non-collinear points determine one unique circle. A fourth point is concyclic with them when it also lies on that circle.
Same circle → Concyclic points
⭐ Theorem 10 — Equal Angles → Concyclic
If a line segment AB subtends equal
angles at two points C and
D on the same side of AB, then
A, B, C and D lie on a circle.
∠ACB = ∠ADB
⇒ A, B, C, D are concyclic
⇒ A, B, C, D are concyclic
🔷 3. Cyclic Quadrilateral
A quadrilateral whose four vertices lie on the same
circle is called a cyclic quadrilateral.
Therefore, whenever four points are shown on one circle, the quadrilateral formed by them is cyclic.
Therefore, whenever four points are shown on one circle, the quadrilateral formed by them is cyclic.
4 vertices
→
Same circle
→
Cyclic quadrilateral
⭐ Theorem 11 — Opposite Angles
In a cyclic quadrilateral, the sum of two opposite
angles is 180°.
∠BAD + ∠BCD = 180°
∠ABC + ∠ADC = 180°
∠ABC + ∠ADC = 180°
⭐ Theorem 12 — Converse of Theorem 11
If a pair of opposite angles of a quadrilateral
has sum 180°, then the quadrilateral
is cyclic.
Opposite angles = 180°
⇒ Quadrilateral is cyclic
⇒ Quadrilateral is cyclic
🧠 6. Cyclicity Test
To test whether four points form a cyclic quadrilateral,
look for either of the two key conditions:
Equal subtended angles
OR
Opposite angles sum to 180°
OR
Opposite angles sum to 180°
Either valid condition → Concyclicity
🔗 Theorem Connection
Equal Angles
→
Concyclic Points
→
Cyclic Quadrilateral
→
Opposite Angles = 180°
🔄 Forward + Converse
Cyclic
→
Opposite angles = 180°
Opposite angles = 180°
→
Cyclic
Theorem 11 → forward direction
Theorem 12 → converse direction
Theorem 12 → converse direction
⚠️ Don’t Mix the Conditions
Check before applying a theorem
- Equal angles must subtend the same line segment AB.
- For Theorem 10, the relevant points lie on the same side of AB.
- For Theorem 11, the quadrilateral is already known to be cyclic.
- For Theorem 12, the opposite angles summing to 180° establish cyclicity.
🚦 Quick Decision Test
Ask this first:
Do I know the four points are on one circle?
→ Use cyclic quadrilateral properties.
Do I know equal angles subtending AB?
→ Think Theorem 10.
Do opposite angles add to 180°?
→ Think Theorem 12.
→ Use cyclic quadrilateral properties.
Do I know equal angles subtending AB?
→ Think Theorem 10.
Do opposite angles add to 180°?
→ Think Theorem 12.
✅ Last-Minute Exam Checklist
✓ Identify the four relevant points.
✓ Check whether they are concyclic.
✓ Look for equal subtended angles.
✓ Check the same-side condition for Theorem 10.
✓ Remember opposite angles = 180°.
✓ Use Theorem 12 for the converse.
⭐ Equal Angles → Concyclic
|
Cyclic → Opposite Angles = 180°
|
Opposite Angles = 180° → Cyclic
Class 9 Maths Chapter 5 End Exercise Question 1 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 1: Finding the Length of a Chord
In a circle, a chord is 5 cm away from the centre.
If the radius of the circle is 13 cm, what is the
length of the chord?
✍️ Solution
Given,
Perpendicular distance of the chord from the centre = 5 cm
Radius of the circle = 13 cm
To find,
Length of the chord
📐 Figure
OM ⟂ AB, OM = 5 cm, OA = 13 cm
✍️ Solution
Let AB be the chord of the circle and O be its centre.
Draw OM ⟂ AB, where M lies on the chord AB.
Since the perpendicular from the centre of a circle
to a chord bisects the chord,
AM = MB
Therefore, M is the midpoint of AB and AM is half
the length of the chord.
In right-angled △OMA,
OA = 13 cm, OM = 5 cm
By the Baudhāyana–Pythagoras theorem,
OA² = OM² + AM²
13² = 5² + AM²
169 = 25 + AM²
AM² = 144
AM = 12 cm
Hence, the length of the complete chord AB is
AB = AM + MB
AB = 12 + 12 = 24 cm
✅ Final Answer
Length of the chord = 24 cm
Class 9 Maths Chapter 5 End Exercise Question 2 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 2: Angle Subtended by an Arc
An arc of a circle subtends an angle of 70° at the centre.
What is the measure of the angle subtended by the arc
at a point on the circle?
✍️ Solution
Given,
Angle subtended by the arc at the centre = 70°
To find,
Angle subtended by the same arc at a point on the circle
📐 Figure
Central angle = 70°; angle subtended at C = 35°
✍️ Solution
Let the arc be AB and let C be a point on the
remaining part of the circle.
By
Theorem 9
,
the angle subtended by an arc at the centre of a circle
is twice the angle subtended by the same arc at a point
on the remaining part of the circle.
∠AOB = 2 × ∠ACB
Given,
∠AOB = 70°
70° = 2 × ∠ACB
∠ACB = 70° ÷ 2
∠ACB = 35°
✅ Final Answer
Angle subtended by the arc at the point on the
circle = 35°
Class 9 Maths Chapter 5 End Exercise Question 3 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 3: Distance of a Chord from the Centre
The diameter of a circle is 26 cm. A chord of length
24 cm is drawn in the circle. Find the distance from
the centre of the circle to the chord.
✍️ Solution
Given,
Diameter of the circle = 26 cm
Length of chord AB = 24 cm
To find,
Distance of the chord from the centre
📐 Figure
OA = 13 cm, AM = MB = 12 cm, OM = ?
✍️ Solution
Let AB be the chord and O be the centre of the circle.
Draw OM ⟂ AB, where M lies on the chord AB.
Since the perpendicular from the centre of a circle
to a chord bisects the chord,
AM = MB
Given that the length of chord AB is 24 cm,
AM = MB = 24 ÷ 2 = 12 cm
The diameter of the circle is 26 cm, so its radius is
OA = 26 ÷ 2 = 13 cm
In right-angled △OMA, by the Baudhāyana–Pythagoras theorem,
OA² = OM² + AM²
13² = OM² + 12²
169 = OM² + 144
OM² = 25
OM = 5 cm
✅ Final Answer
Distance from the centre to the chord = 5 cm
Class 9 Maths Chapter 5 End Exercise Question 4 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 4: Length of a Chord
A circle has a radius of 15 cm. A chord is drawn.
The distance from the centre of the circle to the chord
is 9 cm. What is the length of the chord?
✍️ Solution
Given,
Radius of the circle = 15 cm
Perpendicular distance of the chord from the centre = 9 cm
To find,
Length of the chord
📐 Figure
OM ⟂ AB, OM = 9 cm, OA = 15 cm
✍️ Solution
Let AB be the chord of the circle and O be its centre.
Draw OM ⟂ AB, where M lies on the chord AB.
Since the perpendicular from the centre of a circle
to a chord bisects the chord,
AM = MB
Therefore, M is the midpoint of AB.
Since AB = 24 cm is not yet known, let
AM = MB = x cm.
In right-angled △OMA,
OA = 15 cm, OM = 9 cm
By the Baudhāyana–Pythagoras theorem,
OA² = OM² + AM²
15² = 9² + AM²
225 = 81 + AM²
AM² = 144
AM = 12 cm
Since M is the midpoint of AB,
AB = AM + MB
AB = 12 + 12 = 24 cm
✅ Final Answer
Length of the chord = 24 cm
Class 9 Maths Chapter 5 End Exercise Question 5 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 5: Perpendicular Bisector of a Chord
Prove that the perpendicular bisector of a chord
passes through the centre of the circle.
✍️ Solution
Given,
AB is a chord of a circle with centre O.
DE is the perpendicular bisector of AB,
meeting AB at D.
Therefore, AD = DB and DE ⟂ AB.
To prove,
The perpendicular bisector DE passes through
the centre O of the circle.
📐 Figure
Assume DE does not pass through O; join OD.
✍️ Proof
Let AB be a chord of the circle with centre O.
Let DE be the perpendicular bisector of AB,
meeting AB at D.
Since DE is the perpendicular bisector of AB,
AD = DB
DE ⟂ AB
Assume, for contradiction:
Suppose that the perpendicular bisector DE
does not pass through the centre O.
Join OD.
Since D is the midpoint of chord AB, the line
joining the centre O to the midpoint D of the chord
is perpendicular to the chord.
OD ⟂ AB
Therefore,
∠ADO = 90°
∠ADE = 90°
Thus, through the same point D, both OD and DE
are perpendicular to the same line AB.
But through a given point, only one perpendicular
can be drawn to a given line.
Contradiction
Hence, OD and DE must be the same line.
Therefore O must lie on DE.
This contradicts our assumption that DE does not
pass through O.
Therefore, our assumption is false.
Hence, the perpendicular bisector DE of chord AB
passes through the centre O of the circle.
✅ Hence Proved
The perpendicular bisector of a chord passes through
the centre of the circle.
Class 9 Maths Chapter 5 End Exercise Question 6 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 6: Angle Subtended by a Diameter
The diameter of a circle is AB. Point C is on the
circumference. What is the measure of ∠ACB?
Explain your reasoning.
✍️ Solution
Given,
AB is the diameter of the circle.
O is the centre of the circle.
C is a point on the circumference.
To find,
Measure of ∠ACB
📐 Figure
AB is a diameter, O is the centre and C lies on the circumference.
✍️ Solution
Since AB is the diameter of the circle and O is
its centre, A, O and B lie on the same straight line.
Therefore, the angle subtended by the diameter AB
at the centre O is a straight angle.
∠AOB = 180°
The angles ∠AOB and ∠ACB are subtended by the
same chord AB at the centre and at the circumference,
respectively.
By the
Theorem 9
,
the angle subtended by a chord at the circumference is half
the angle subtended by the same chord at the centre.
∠ACB =
1
2
× ∠AOB
∠ACB =
1
2
× 180°
∠ACB = 90°
Thus, the angle made by the diameter AB at the
point C on the circumference is a right angle.
✅ Final Answer
∠ACB = 90°
Class 9 Maths Chapter 5 End Exercise Question 7 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 7: Cyclic Quadrilateral Opposite Angles
ABCD is a cyclic quadrilateral inscribed in a circle.
If ∠A measures 75°, what is the measure of ∠C?
If ∠B measures 110°, what is the measure of ∠D?
✍️ Solution
Given,
ABCD is a cyclic quadrilateral.
∠A = 75°
∠B = 110°
To find,
∠C and ∠D
📐 Figure
ABCD is a cyclic quadrilateral; opposite angles are supplementary.
✍️ Solution
Since ABCD is a cyclic quadrilateral, its opposite
angles are supplementary.
Therefore, ∠A and ∠C are opposite angles.
∠A + ∠C = 180°
75° + ∠C = 180°
∠C = 180° − 75°
∠C = 105°
Similarly, ∠B and ∠D are opposite angles.
∠B + ∠D = 180°
110° + ∠D = 180°
∠D = 180° − 110°
∠D = 70°
Thus, in a cyclic quadrilateral, each pair of
opposite angles adds up to 180°.
✅ Final Answer
∠C = 105° and ∠D = 70°
Question 8: Cyclic Quadrilateral Opposite Angles
Quadrilateral PQRS is inscribed in a circle.
If ∠P = (2x + 10)° and ∠R = (3x − 20)°,
find the value of x and the measures of ∠P and ∠R.
✍️ Solution
Given,
PQRS is a cyclic quadrilateral.
∠P = (2x + 10)°
∠R = (3x − 20)°
To find,
The value of x and the measures of ∠P and ∠R
📐 Figure
P and R are opposite angles of the cyclic quadrilateral.
✍️ Solution
Since PQRS is a cyclic quadrilateral, its opposite
angles are supplementary.
∠P and ∠R are opposite angles. Therefore,
∠P + ∠R = 180°
Substituting the given values,
(2x + 10)° + (3x − 20)° = 180°
2x + 10 + 3x − 20 = 180
5x − 10 = 180
5x = 180 + 10
5x = 190
x = 38
Now, substituting x = 38 in ∠P = (2x + 10)°,
∠P = (2 × 38 + 10)°
∠P = 76° + 10°
∠P = 86°
Similarly, substituting x = 38 in
∠R = (3x − 20)°,
∠R = (3 × 38 − 20)°
∠R = 114° − 20°
∠R = 94°
Check:
∠P + ∠R = 86° + 94° = 180°
Hence, the result satisfies the property of
opposite angles of a cyclic quadrilateral.
✅ Final Answer
x = 38 ;
∠P = 86° ;
∠R = 94°
Question 9: Radius of a Circle from a Chord
The distance of a chord of length 16 cm from the centre
of a circle is 6 cm. Find the radius of the circle.
✍️ Solution
Given,
Length of chord AB = 16 cm
Distance of chord AB from the centre = 6 cm
To find,
Radius of the circle
📐 Figure
OM ⟂ AB, OM = 6 cm, AB = 16 cm
✍️ Solution
Let O be the centre of the circle and AB be the
chord. Draw OM ⟂ AB, where M lies on AB.
Since the perpendicular from the centre of a circle
to a chord bisects the chord,
AM = MB
Therefore, M is the midpoint of AB.
AM = MB =
16
2
= 8 cm
In right-angled △OMA,
OM = 6 cm, AM = 8 cm
Also, OA is the radius of the circle.
By the Baudhāyana–Pythagoras theorem,
OA² = OM² + AM²
OA² = 6² + 8²
OA² = 36 + 64
OA² = 100
Since OA is a length,
OA = 10 cm
Therefore, the radius of the circle is 10 cm.
✅ Final Answer
Radius of the circle = 10 cm
Question 10: Area of a Cyclic Quadrilateral
A cyclic quadrilateral has sides 5, 5, 12, 12 units.
Find its area.
✍️ Solution
Given,
ABCD is a cyclic quadrilateral.
AB = BC = 5 units
CD = DA = 12 units
To find,
Area of quadrilateral ABCD
📐 Figure
BD divides the cyclic quadrilateral into two congruent right triangles.
✍️ Solution
Join diagonal BD.
In △ABD and △CBD,
AB = BC = 5 units
AD = CD = 12 units
BD = BD
Therefore,
△ABD ≅ △CBD
by SSS congruence.
By CPCT,
∠BAD = ∠BCD
Since ABCD is a cyclic quadrilateral, its opposite
angles are supplementary.
∠BAD + ∠BCD = 180°
But ∠BAD = ∠BCD. Therefore,
2∠BAD = 180°
∠BAD = 90°
∠BCD = 90°
Hence, △ABD and △CBD are right-angled triangles,
right-angled at A and C respectively.
Area of △ABD
= ½ × AB × AD
= ½ × 5 × 12
= 30 square units
Similarly,
Area of △CBD = 30 square units
Therefore,
Area of ABCD
= Area of △ABD + Area of △CBD
= 30 + 30
= 60 square units
✅ Final Answer
Area of the cyclic quadrilateral = 60 square units
Question 11: Position of the Circumcentre of a Cyclic Quadrilateral
Consider a cyclic quadrilateral. Without drawing its
circumcircle, how can we find out whether the centre of
the circumcircle lies inside the quadrilateral or outside?
What is the best way of finding out?
✍️ Solution
Given,
ABCD is a cyclic quadrilateral.
To find,
Whether the centre O of the circumcircle lies
inside, on, or outside the quadrilateral.
Case 1: ∠ACB = ∠ADB < 90°
Case 2: ∠ACB = ∠ADB = 90°
Case 3: ∠ACB = ∠ADB > 90°
✍️ Explanation
Let ABCD be a cyclic quadrilateral and take AB as
a chord.
The chord AB subtends ∠ACB at C and ∠ADB at D.
Since both angles stand on the same chord AB,
∠ACB = ∠ADB
We now compare this common angle with 90°.
Case 1: ∠ACB = ∠ADB < 90°
The angle subtended by chord AB at the
circumference is less than 90°.
Therefore, the corresponding central angle
is less than 180°. Hence the centre O lies on
the same side of AB as C and D.
O lies INSIDE ABCD
Case 2: ∠ACB = ∠ADB = 90°
A chord that subtends a right angle at the
circumference is a diameter.
AB is a diameter
O lies ON AB
Case 3: ∠ACB = ∠ADB > 90°
The angle subtended by chord AB at the
circumference is greater than 90°.
Therefore, the centre O lies on the opposite
side of AB from C and D.
O lies OUTSIDE ABCD
Thus, by observing the angles subtended by the
same chord AB, we can determine the position of
the centre O without drawing the circumcircle.
✅ Final Answer
∠ACB = ∠ADB < 90° → O lies inside ABCD
∠ACB = ∠ADB = 90° → O lies on AB
∠ACB = ∠ADB > 90° → O lies outside ABCD
∠ACB = ∠ADB = 90° → O lies on AB
∠ACB = ∠ADB > 90° → O lies outside ABCD
Question 12: Equal Intersecting Chords
When two chords intersect, each of them is divided into
two line segments. Show that if the intersecting chords
are of equal length, then the line segments of one chord
are equal to the corresponding line segments of the
other chord.
✍️ Solution
Given,
Chords AB and CD of a
circle intersect at P.
AB = CD
To prove that,
AP = CP and
PB = PD
Construction,
Draw OM ⟂ AB and
ON ⟂ CD, where O
is the centre of the circle.
📐 Construction
Proof
In △OMP and △ONP,
OM = ON
Equal chords of a circle are equidistant from the
centre.
OP = OP
(Common)
∠OMP = ∠ONP = 90°
(By construction)
∴ △OMP ≅ △ONP
(By RHS congruence rule)
Therefore, by CPCT,
MP = PN
— (i)
Also,
AB = CD
Since the perpendicular from the centre of a circle
bisects a chord,
MB = ½AB
DN = ½CD
∴ MB = DN
— (ii)
From (i) and (ii),
MB − MP = DN − PN
Therefore,
PB = PD
— (iii)
Again, since
AB = CD
and
MB = DN
and
MP = PN,
the corresponding remaining parts are equal:
AP = CP
Hence, the line segments of one chord are equal to the
corresponding line segments of the other chord.
✅ Hence Proved
AP = CP and PB = PD
Question 13: Construction of a Circle with a Chord
Draw a circle in which a chord of 6 cm length stands at a
distance of 3 cm from the centre.
Hint: Is it a circumcircle of a suitable triangle?
Hint: Is it a circumcircle of a suitable triangle?
✍️ Solution
Given,
Chord AB = 6 cm.
Distance of the chord from the centre
= 3 cm.
To construct,
A circle having a chord of length 6 cm at a
perpendicular distance of 3 cm from its centre.
Construction
1. Draw a line segment
AB = 6 cm.
2. Construct the perpendicular bisector
of AB using equal arcs with A and B as centres.
Let it meet AB at M.
3. On the perpendicular bisector, take a
point O such that
OM = 3 cm.
4. With O as centre and
OA as radius, draw the circle.
Hint:
The hint asks us to think whether the required circle can be
regarded as the circumcircle of a suitable triangle. It is a
guiding idea connecting the construction with the circumcircle
concept; we do not need to calculate the radius separately.
📐 Construction Figure
✓ Verification
Since M is the midpoint of
AB,
AM = MB = 3 cm
Therefore,
AB = AM + MB = 3 + 3 = 6 cm
Also, by construction,
OM = 3 cm
and
OM ⟂ AB
Hence, the perpendicular distance of chord AB from
the centre O is exactly 3 cm.
Therefore, the constructed circle satisfies both the
given conditions.
✅ Required Circle Constructed
Chord AB = 6 cm
and
distance of AB from centre O = 3 cm
Question 14: Cyclic Parallelogram and Rectangle
Show that rectangle is the only parallelogram that can be
inscribed in a circle.
✍️ Solution
Given,
ABCD is a parallelogram inscribed in a circle.
To Prove,
ABCD is a rectangle.
📐 Cyclic Parallelogram ABCD
Proof
Since ABCD is a parallelogram,
∠A = ∠C
because opposite angles of a parallelogram are equal.
Also, ABCD is a cyclic quadrilateral. Therefore,
∠A + ∠C = 180°
because opposite angles of a cyclic quadrilateral are
supplementary.
Since
∠A = ∠C
we get
∠A + ∠A = 180°
2∠A = 180°
∠A = 90°
Therefore,
∠C = 90°
Also, adjacent angles of a parallelogram are supplementary.
Therefore,
∠A + ∠B = 180°
Since
∠A = 90°
we get
∠B = 90°
Similarly,
∠D = 90°
Hence all four angles of ABCD are right angles.
Therefore, ABCD is a rectangle.
✅ Hence Proved
A parallelogram that can be inscribed in a circle
is necessarily a rectangle.
Question 15: Rectangle Inscribed in a Circle
Show that if a rectangle is inscribed in a circle, then
the point of intersection of its diagonals is the centre
of the circle.
✍️ Solution
Given,
ABCD is a rectangle inscribed in a circle.
Its diagonals AC and BD
intersect at O.
To Prove,
O is the centre of the circle.
📐 Rectangle ABCD Inscribed in a Circle
Proof
Since ABCD is a rectangle,
∠ADC = 90°
The angle ∠ADC is an angle at the circumference standing
on chord AC.
An angle in a semicircle is 90°.
Therefore, the chord AC is a diameter of
the circle.
Similarly, since ABCD is a rectangle,
∠DAB = 90°
The angle ∠DAB stands on chord DB.
Since an angle in a semicircle is 90°,
DB is also a diameter of the circle.
Thus, AC and BD are two
diameters of the same circle.
The two diameters intersect at O.
The point of intersection of two diameters of a circle is
its centre.
Therefore, O is the centre of the circle.
✅ Hence Proved
The point of intersection of the diagonals of a rectangle
inscribed in a circle is the centre of the circle.
Question 16: Midpoints of Equal Chords
Consider all chords of a circle of a fixed length.
What is the shape formed by the midpoints of all these chords?
✍️ Solution
Given,
A circle with centre O is given.
Consider all chords of the circle having the same
fixed length.
To Find,
The shape formed by the midpoints of all these
equal chords.
📐 Midpoints of Equal Chords
Proof
Let AB and CD be any two
equal chords of the circle, and let their midpoints be
M and N respectively.
By
Theorem 6
,
equal chords of a circle are equidistant from the centre.
Therefore,
OM = ON
Since all the chords have the same fixed length, the midpoint
of every such chord is at the same fixed distance from
O.
OM = ON = OP = OQ = …
Thus, all the midpoints are equidistant from the fixed point
O.
The locus of all points at a fixed distance from a fixed point
is a circle.
Hence, the midpoints of all these chords form a circle with
centre O. It is therefore concentric
with the given circle.
✅ Final Answer
The midpoints of all chords of a fixed length form a
circle concentric with the given circle.
Class 9 Maths Chapter 5 End Exercise Question 17 Solution
Ganita Manjari | Chapter 5 End-of-Chapter Exercise | Circle
Question 17: Congruent Chords and Angle Bisector
In a circle with centre O, chords AB and AC are congruent.
Explain why this statement is true:
“The centre of the circle lies on the angle bisector
of ∠BAC”.
✍️ Solution
Given,
In a circle with centre O,
AB and AC are
congruent chords.
To Prove,
O lies on the angle bisector of
∠BAC.
Construction
Draw OM ⟂ AB and
ON ⟂ AC. Join OA.
📐 Congruent Chords AB and AC
Proof
The two chords are congruent.
Therefore,
AB = AC
½ AB = ½ AC
AM = AN
[Perpendicular from the centre bisects the chord]
In triangles △AOM and
△AON,
OA = OA
[Common side]
OM = ON
[Equal chords are equidistant from the centre]
AM = AN
[Already proved]
Hence,
△AOM ≅ △AON
[SSS]
Therefore, corresponding angles are equal:
∠MAO = ∠OAN
Since M lies on AB and N lies on AC,
∠BAO = ∠OAC
Thus, AO divides ∠BAC into two equal
angles. Hence, the centre O lies on
the angle bisector of ∠BAC.
✅ Hence Proved
The centre O of the circle lies on the angle bisector
of ∠BAC.
Question 18:Ditance between chords
Two parallel chords of lengths 10 cm and 24 cm are on the
same side of the centre of a circle. The distance between
the chords is 7 cm. Find the radius of the circle.
✍️ Solution
Given,
AB = 10 cm, CD = 24 cm
Distance between chords = 7 cm
Distance between chords = 7 cm
To Find,
Radius of the circle = R
📐 Parallel Chords
Solution
AM = ½ AB = ½ × 10 = 5 cm
[Perpendicular from centre bisects the chord]
CN = ½ CD = ½ × 24 = 12 cm
[Perpendicular from centre bisects the chord]
Since the two chords are on the same side
of the centre,
OM − ON = 7
ON = OM − 7
In right triangle △OMA,
OA² = OM² + AM²
R² = OM² + 5²
In right triangle △ONC,
OC² = ON² + CN²
R² = ON² + 12²
Since OA = OC = R,
OM² + 5² = ON² + 12²
Substituting ON = OM − 7,
OM² + 25 = (OM − 7)² + 144
OM² + 25 = OM² − 14OM + 49 + 144
14OM = 168
OM = 12 cm
Using right triangle △OMA,
R² = 12² + 5²
R² = 144 + 25 = 169
R = 13 cm
✅ Final Answer
Radius of the circle = 13 cm
Chapter 5 End Exercise: Question 19
A regular hexagon is inscribed in a circle of radius
r. Find the length of the sides of the
hexagon and the distance of each side from the centre
of the circle.
✍️ Solution
Given,
Regular hexagon ABCDEF is inscribed in a circle
of radius r.
To Find,
1. Length of each side.
2. Distance of each side from the centre.
2. Distance of each side from the centre.
📐 Regular Hexagon Inscribed in a Circle
Solution
A regular hexagon divides the circle into
6 equal central angles.
∠AOB =
360°
6
= 60°
In triangle △AOB,
OA = OB = r
[Radii of the same circle]
∠AOB = 60°
Therefore, the other two angles of
△AOB are also 60°.
Hence, △AOB is an equilateral triangle.
AB = OA = r
Therefore, the length of each side of the regular hexagon is
r
Now draw OM ⟂ AB, where M is the midpoint
of AB.
AM =
1
2
AB =
r
2
[Perpendicular from centre bisects the chord]
In right triangle △OMA,
OA² = OM² + AM²
r² = OM² +
(
r
2
)²
OM² = r² −
r²
4
OM² =
3r²
4
OM =
√3r
2
By symmetry, all six sides of the regular hexagon are
equidistant from the centre.
Therefore, the distance of each side from the centre is
√3r
2
✅ Final Answer
Length of each side =
r
|
Distance of each side from centre =
√3r
2
Question 20: A quadrilateral MNOP
A quadrilateral MNOP is inscribed in a circle.
If MN is a diameter, what can you say about
∠MOP and ∠MNP?
Explain your reasoning.
✍️ Solution
Given,
Quadrilateral MNOP is inscribed in a circle
and MN is a diameter.
To Find,
Relation between ∠MOP and
∠MNP.
📐 Cyclic Quadrilateral MNOP
Solution
The angles ∠MOP and ∠MNP
stand on the same chord MP.
Also, the points O and N
lie on the same segment of the circle with respect to
chord MP.
Therefore, by the theorem
“Angles in the same segment of a circle are equal”,
∠MOP = ∠MNP
[Angles in the same segment of a circle]
Since MN is a diameter, the angle subtended
by it at P is a right angle.
∠MPN = 90°
[Angle in a semicircle is a right angle]
This additional fact is not required for the required
relationship between ∠MOP and
∠MNP.
✅ Final Answer
∠MOP = ∠MNP
They are equal because they stand on the same chord MP and lie in the same segment of the circle.
They are equal because they stand on the same chord MP and lie in the same segment of the circle.
Question 21: Exterior Angle of a Cyclic Quadrilateral
Let ABCD be a cyclic quadrilateral.
Explain why the exterior angle at any vertex is equal
to the interior opposite angle.
Example: ∠ADE = ∠ABC, where E is a point on the extension of side AD.
Example: ∠ADE = ∠ABC, where E is a point on the extension of side AD.
✍️ Solution
Given,
ABCD is a cyclic quadrilateral and
AD is extended to E.
To Prove,
∠ADE = ∠ABC
Solution
Since AD is extended to E,
∠ADE + ∠ADC = 180°
[Linear pair]
Also, ABCD is a cyclic quadrilateral.
Therefore,
∠ADC + ∠ABC = 180°
[Opposite angles of a cyclic quadrilateral]
Comparing the two equations,
∠ADE + ∠ADC = ∠ADC + ∠ABC
∠ADE = ∠ABC
[Subtracting ∠ADC from both sides]
✅ Hence Proved
∠ADE = ∠ABC
Thus, the exterior angle at a vertex of a cyclic quadrilateral is equal to its interior opposite angle.
Thus, the exterior angle at a vertex of a cyclic quadrilateral is equal to its interior opposite angle.
Question 22: Longest Chord of a Circle
“There is no chord of a circle that is longer than its
diameter.” How do you justify this statement?
✍️ Solution
Given,
Let AB be any chord of a circle with
centre O and radius r.
To Prove,
No chord of the circle is longer than its diameter.
Solution
Join OA and OB.
In triangle △AOB,
OA + OB > AB
[Triangle inequality]
Since OA = OB = r,
2r > AB
But,
Diameter = 2r
Therefore,
AB < Diameter
If the chord passes through the centre O,
then it becomes the diameter itself.
AB = Diameter
[A diameter is also a chord]
✅ Hence Proved
The diameter is the longest chord of a circle.
Therefore, no chord can be longer than the diameter.
Therefore, no chord can be longer than the diameter.
Question 23: Shortest Chord Through an Interior Point
Let A be any point within a given circle
with centre O. Show that the shortest chord
of the circle that passes through point A
is the one that is perpendicular to OA.
✍️ Solution
Given,
A is any point inside a circle with
centre O.
To Prove,
The shortest chord passing through A
is perpendicular to OA.
Solution
Let BC be any chord passing through
A, and let M be the midpoint
of BC.
Since M is the midpoint of chord
BC,
OM ⟂ BC
[Perpendicular from the centre bisects a chord]
In right triangle △OMA,
OA² = OM² + AM²
Therefore,
OM² = OA² − AM²
If the radius of the circle is r, then
BC = 2√(r² − OM²)
Substituting OM² = OA² − AM²,
BC = 2√(r² − OA² + AM²)
Here, r² − OA² is fixed.
Therefore, BC is shortest when
AM = 0
Hence,
M = A
Since OM ⟂ BC,
OA ⟂ BC
[As M = A]
✅ Hence Proved
The shortest chord passing through A
is the chord perpendicular to OA.
Question 24: Angle in a Semicircle
How would you use the following figure to justify the
statement that the angle in a semicircle is 90°?
✍️ Solution
Given,
BC is a diameter of the circle with
centre O, and A is a
point on the circle.
To Prove,
∠BAC = 90°
Solution
Since OA = OB (radii of the circle),
∠OAB = ∠ABO = a
[Angles opposite equal sides of an isosceles triangle]
Also, OA = OC (radii of the circle),
∠OAC = ∠ACO = b
[Angles opposite equal sides of an isosceles triangle]
Therefore,
∠BAC = ∠BAO + ∠OAC
∠BAC = a + b
In triangle △ABC,
a + (a + b) + b = 180°
[Angle sum property of a triangle]
2(a + b) = 180°
a + b = 90°
But,
∠BAC = a + b
∠BAC = 90°
✅ Hence Proved
The angle in a semicircle is
90°.
Question 25: Perpendicular Chords and Their Midpoints
In a circle, two chords CC′ and
DD′ are drawn perpendicular to a diameter
AB. Prove that the segment MM′
joining the midpoints of the chords CD and
C′D′ is perpendicular to AB.
✍️ Solution
Given,
AB is a diameter of the circle.
CC′ and DD′ are chords
perpendicular to AB.
M and M′ are the midpoints of CD and C′D′, respectively.
M and M′ are the midpoints of CD and C′D′, respectively.
To Prove,
MM′ ⟂ AB
Proof
We have,
CC′ ⟂ AB
[Given]
DD′ ⟂ AB
[Given]
Since CC′ and DD′
are both perpendicular to AB,
CC′ ∥ DD′
[Perpendicular to the same line]
Therefore, CC′D′D is a trapezium,
with CC′ and DD′
as its parallel sides.
M is the midpoint of CD,
and M′ is the midpoint of
C′D′.
Thus, CD and C′D′ are the two non-parallel sides of the trapezium.
Thus, CD and C′D′ are the two non-parallel sides of the trapezium.
The segment joining the midpoints of the two
non-parallel sides of a trapezium is parallel to
its parallel sides.
MM′ ∥ CC′
[Midpoint theorem for a trapezium]
But,
CC′ ⟂ AB
[Given]
Since MM′ ∥ CC′,
MM′ ⟂ AB
[Since MM′ ∥ CC′ and CC′ ⟂ AB]
✅ Hence Proved
MM′ ⟂ AB
Question 26: Opposite Angles of a Cyclic Quadrilateral
How would you use the following figure to justify the
statement that the sum of the opposite angles of a cyclic
quadrilateral is 180°?
✍️ Solution
Given,
ABCD is a cyclic quadrilateral.
AC and BD pass through
the centre O of the circle.
To Prove,
∠DAB + ∠BCD = 180°
Proof
Since OA = OB = OC = OD, the four triangles
AOB, BOC, COD and DOA are
isosceles triangles.
Therefore, their equal base angles are respectively
p, p; q, q;
u, u; and v, v.
The four angles of quadrilateral ABCD are:
∠DAB = p + v
[From the figure]
∠ABC = p + q
[From the figure]
∠BCD = q + u
[From the figure]
∠CDA = u + v
[From the figure]
The sum of the four angles of a quadrilateral is
360°.
(p + v) + (p + q) + (q + u) + (u + v) = 360°
2(p + q + u + v) = 360°
p + q + u + v = 180°
Rearranging the above result,
(p + v) + (q + u) = 180°
But,
∠DAB = p + v
[From the figure]
∠BCD = q + u
[From the figure]
∠DAB + ∠BCD = 180°
✅ Hence Proved
The sum of a pair of opposite angles of a cyclic
quadrilateral is 180°.
📚 Continue Learning
Congratulations! You have completed the Class 9 Maths Chapter 5 End-of-Chapter Exercises. You have now revised the complete Chapter 5 through its 26 end-of-chapter questions. Continue your preparation by revisiting any exercise for practice or moving to another chapter.
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❓ Frequently Asked Questions
Find quick answers to common questions about Class 9 Maths Chapter 5 End of Chapter Exercises Solutions. These FAQs help students understand the exercise, its coverage, and how to use the complete solutions effectively for homework, revision, self-study, and CBSE examination preparation.
What are the End-of-Chapter Exercises in Class 9 Maths Chapter 5?
The End-of-Chapter Exercises are the final set of practice questions given at the end of Chapter 5 – I’m Up and Down, and Round and Round in the NCERT Ganita Manjari (2026) textbook. They bring together the major concepts and applications covered throughout the chapter.
How many questions are there in Chapter 5 End-of-Chapter Exercises?
There are 26 questions in the End-of-Chapter Exercises. The questions are presented in the same order as the textbook and include different types of practice, proof, application, construction, and higher-order questions.
What topics are covered in the Chapter 5 End Exercises?
The End-of-Chapter Exercises revise a wide range of Chapter 5 topics, including chords, distance from the centre, angles subtended by arcs, diameter-related results, concyclicity, cyclic quadrilaterals, intersecting chords, constructions, and applications involving circles. The complete set also includes proof-based and higher-order questions.
Are the Chapter 5 End Exercises useful for revision?
Yes. The End-of-Chapter Exercises are especially useful for complete Chapter 5 revision because they bring together different types of questions from the chapter. Students can use the question-wise solutions to identify weak areas and practise before tests and examinations.
Are there higher-order questions in the Chapter 5 End Exercises?
Yes. The exercise includes several questions marked with an asterisk (*) in the textbook. These questions require deeper reasoning, proof, construction, or application of the concepts developed in Chapter 5.
Are these solutions suitable for CBSE Class 9 students?
Yes. The solutions are presented in a clear, student-friendly, step-by-step CBSE answer-writing style. They can be used for classroom practice, homework, self-study, revision, and examination preparation.
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Yes. A dedicated Jump to Any Question navigator is provided on this page. Students can select any question from Q1 to Q26 and move directly to its solution without scrolling through the complete exercise.
Are these Class 9 Maths Chapter 5 End-of-Chapter Solutions based on Ganita Manjari 2026?
Yes. These Class 9 Maths Chapter 5 End of Chapter Exercises Solutions are prepared according to the latest NCERT Ganita Manjari (2026) textbook. The questions are followed in textbook order and the solutions are presented in a student-friendly format for learning, revision, and CBSE preparation.
👨🏫 Reviewed & Prepared By
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Founder, Maths Gurukulam &
Newton Study Point
Teaching Mathematics Since 2006
These Class 9 Maths Chapter 5 End-of-Chapter Exercises Solutions are carefully prepared according to the latest NCERT Ganita Manjari (2026) and the CBSE curriculum. The complete set of 26 questions is presented in a clear, step-by-step format to help students practise, revise, and prepare confidently for Class 9 Mathematics examinations.
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