NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 5 I’m Up and Down, and Round and Round

A circle is the set of all points in a plane that are at a fixed distance from a fixed point called the centre. In Class 9 Maths Chapter 5, students learn about radius, chord, diameter, circumcircle, equal chords, arcs, angles in circles and cyclic quadrilaterals.

NCERT Solutions for Class 9 Maths Ganit Manjari Chapter 5 help students solve textbook questions from I’m Up and Down, and Round and Round in the new Ganita Manjari Class 9 Maths textbook. This chapter explains the properties of circles through diagrams, constructions, chord theorems, perpendicular bisectors, arcs, angle subtended by an arc and cyclic quadrilaterals.

Through constructions, symmetry, chord relationships and angle reasoning, students develop the main circle theorems needed for CBSE 2026-27 exams. The exercise-wise solutions help students solve proof-based and numerical questions on equal chords, perpendiculars from the centre, angle in a semicircle and opposite angles of cyclic quadrilaterals.

NCERT Solutions for Class 9 Maths Chapter 5 PDF Download

The NCERT Solutions for Class 9 Maths Chapter 5 PDF helps students revise circle geometry with clear steps and theorem-based explanations for CBSE 2026-27. It is useful for practising construction questions, chord-length problems, cyclic quadrilateral questions and diagram-based proofs.

Students can use the PDF to revise:

1. Definition of circle
2. Centre, radius, chord and diameter
3. Symmetries of a circle
4. Circles through two points
5. Circumcircle and circumcentre
6. Equal chords and angles at the centre
7. Perpendicular from centre to chord
8. Distance of chords from centre
9. Angles subtended by arcs
10. Angle in a semicircle
11. Concyclic points
12. Cyclic quadrilaterals

Access Exercise-Wise NCERT Solutions for Class 9 Maths Chapter 5

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.1

Exercise Set 5.1 focuses on drawing circumcircles of triangles and understanding where the circumcentre lies.

Question 1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Answer:
Steps of construction:

1. Draw AB = 5 cm.
2. At A, construct ∠A = 70°.
3. At B, construct ∠B = 60°.
4. Let the two rays meet at C. This gives ΔABC.
5. Draw the perpendicular bisectors of AB and AC.
6. Let them meet at O.
7. With O as centre and OA as radius, draw a circle.

Since:

∠C = 180° - 70° - 60°
= 50°

All angles of the triangle are less than 90°, so the triangle is acute-angled. The circumcentre of an acute-angled triangle lies inside the triangle.

Final answer:
The circumcentre lies inside the triangle.

Question 2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Answer:
Steps of construction:

1. Draw AB = 5 cm.
2. At A, construct ∠A = 100°.
3. On this ray, mark AC = 4 cm.
4. Join B to C to form ΔABC.
5. Draw perpendicular bisectors of AB and AC.
6. Let them meet at O.
7. With O as centre and OA as radius, draw the circumcircle.

Since ∠A = 100°, the triangle is obtuse-angled. The circumcentre of an obtuse-angled triangle lies outside the triangle.

Final answer:
The circumcentre lies outside the triangle.

Question 3. Draw ΔABC with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

Answer:
Steps of construction:

1. Draw AB = 6 cm.
2. With A as centre and radius 7 cm, draw an arc.
3. With B as centre and radius 7 cm, draw another arc cutting the first arc at C.
4. Join AC and BC.
5. Draw perpendicular bisectors of AB and AC.
6. Let them meet at O.
7. Draw a circle with O as centre and OA as radius.

Since O is the circumcentre, it is equidistant from all three vertices.

Final answer:
OA = OB = OC.
The measured value depends on the construction, but all three distances will be equal.

Question 4. What is the least possible radius of a circle through two points A and B?

Answer:
For a circle to pass through A and B, its centre must be equidistant from A and B. The smallest such circle is obtained when AB is the diameter.

If AB is the diameter, then radius = AB/2.

Final answer:
The least possible radius is AB/2.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.2

Exercise Set 5.2 focuses on chords and the angles they subtend at the centre.

Question 1. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Answer:
Let AB be a chord of a circle with centre O.

Join OA and OB.

Since A and B lie on the circle:

OA = radius
OB = radius

So:

OA = OB

Therefore, triangle OAB has two equal sides.

Final answer:
The triangle formed by a chord and the centre is isosceles.

Question 2. Show that if two such isosceles triangles have equal base length, they are congruent to each other.

Answer:
Let AB and CD be two equal chords of the same circle with centre O.

Consider triangles OAB and OCD.

OA = OC because both are radii.
OB = OD because both are radii.
AB = CD because the chord lengths are equal.

Therefore, by SSS congruence:

ΔOAB ≅ ΔOCD

Final answer:
The two isosceles triangles are congruent by SSS congruence.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.3

Exercise Set 5.3 focuses on the perpendicular from the centre to a chord and the line joining the centre to the midpoint of a chord.

Question 1. Explain why the perpendicular from the centre of a circle to a chord bisects the chord.

Answer:
Let AB be a chord of a circle with centre C. Let CM be perpendicular to AB.

Join CA and CB.

Since A and B lie on the circle:

CA = CB

Also:

CM = CM

And:

∠CMA = ∠CMB = 90°

Therefore, by RHS congruence:

ΔCMA ≅ ΔCMB

So:

AM = BM

Final answer:
The perpendicular from the centre to a chord bisects the chord.

Question 2. An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.

Answer:
Since AB = AC, triangle ABC is isosceles.

The altitude from A to BC also bisects BC.

Let the altitude meet BC at M. Then:

BM = CM
AM ⟂ BC

Since M is the midpoint of chord BC, the line joining the centre of the circle to M is perpendicular to BC.

But only one perpendicular can be drawn to BC at M. Therefore, the centre of the circle lies on AM.

Final answer:
The altitude from A to BC passes through the centre of the circle.

Question 3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius is 5 cm, find the distance between the midpoints of the chords.

Answer:
For a chord of length 6 cm:

Half chord = 3 cm
Radius = 5 cm

Distance from centre:

d₁ = √(5² - 3²)
= √(25 - 9)
= √16
= 4 cm

For a chord of length 8 cm:

Half chord = 4 cm

Distance from centre:

d₂ = √(5² - 4²)
= √(25 - 16)
= √9
= 3 cm

Since the chords are on opposite sides of the centre, the distance between their midpoints is:

4 + 3 = 7 cm

Final answer:
The distance between the midpoints is 7 cm.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.4

Exercise Set 5.4 focuses on equal chords and chords equidistant from the centre.

Question 1. Use the Baudhāyana-Pythagoras theorem to show why chords of equal length are equidistant from the centre.

Answer:
Let AB and CD be equal chords of a circle with centre O.

Drop perpendiculars OM and ON to AB and CD respectively.

Since perpendicular from centre to chord bisects the chord:

AM = AB/2
CN = CD/2

Given:

AB = CD

So:

AM = CN

Also:

OA = OC because both are radii.

In right triangles OMA and ONC:

OA² = OM² + AM²
OC² = ON² + CN²

Since OA = OC and AM = CN:

OM² = ON²

Therefore:

OM = ON

Final answer:
Equal chords are equidistant from the centre.

Question 2. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF.

Answer:
Since CE ⟂ AB, CE bisects chord AB.

So:

AE = EB

Since CH ⟂ GF, CH bisects chord GF.

So:

GH = HF

Now consider right triangles CEA and CHG.

CA = CG because both are radii.
CE = CH given.
∠CEA = ∠CHG = 90°.

By RHS congruence:

ΔCEA ≅ ΔCHG

So:

AE = GH

Therefore:

AB = 2AE
GF = 2GH

Since AE = GH:

AB = GF

Final answer:
AB = GF.

Question 3. Solve the previous question using the Baudhāyana-Pythagoras theorem.

Answer:
Let the radius be r.

For chord AB:

AE² = r² - CE²

For chord GF:

GH² = r² - CH²

Given:

CE = CH

Therefore:

r² - CE² = r² - CH²

So:

AE² = GH²

Hence:

AE = GH

Therefore:

AB = 2AE and GF = 2GH

So:

AB = GF

Final answer:
The two chords are equal.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.5

Exercise Set 5.5 focuses on the relation between radius, perpendicular distance from centre and chord length.

Question 1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Answer:
Radius = 7 cm
Distance from centre to chord = 6 cm

Half chord = √(r² - d²)

= √(7² - 6²)
= √(49 - 36)
= √13

Chord length = 2√13 cm

Final answer:
2√13 cm

Question 2. Explain why, if the perpendicular distance of a chord from the centre is d and the radius is r, then chord length is 2√(r² - d²).

Answer:
Let AB be the chord and O be the centre.

Drop perpendicular OM to AB.

Then:

OM = d
OA = r

Since perpendicular from centre to chord bisects the chord:

AM = AB/2

In right triangle OMA:

OA² = OM² + AM²

r² = d² + AM²

AM² = r² - d²

AM = √(r² - d²)

Therefore:

AB = 2AM
= 2√(r² - d²)

Final answer:
Chord length = 2√(r² - d²).

Question 3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, can we conclude that CD = 2AB?

Answer:
No.

Chord length does not vary directly with distance from the centre. The formula is:

Chord length = 2√(r² - d²)

If one chord is at distance d and another is at distance 2d, their lengths are:

2√(r² - d²) and 2√(r² - 4d²)

These are not in the ratio 1:2.

Final answer:
No, CD = 2AB cannot be concluded. Chord length depends on √(r² - d²), not directly on d.

NCERT Solutions for Class 9 Maths Chapter 5 Exercise Set 5.6

Exercise Set 5.6 focuses on angles subtended by arcs at the centre and at points on the circle.

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 chord AB?

Answer:
OA = OB = 12 cm

∠AOB = 60°

Triangle AOB has two equal sides OA and OB. Since the included angle is 60°, the remaining angles are also 60°.

So, ΔAOB is an equilateral triangle.

Therefore:

AB = OA = OB = 12 cm

Final answer:
Chord AB = 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?

Answer:
No.

Angles subtended by the same chord AB at points on the same segment of the circle are equal.

Final answer:
No, ∠AXB and ∠AYB will be equal.

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

Answer:
If X and Y are on the same segment of the circle with respect to chord AB, then the angles are equal. However, equal angles can also occur in special geometric situations, so the statement needs the condition that X and Y lie on the same circle with A and B.

Final answer:
For points on the same circle, equal angles subtended by AB indicate that the points lie on the same segment.

(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?

Answer:
If X and Y lie on the same side of AB and ∠AXB = ∠AYB, then A, B, X and Y are concyclic.

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

Final answer:
Yes, if X and Y are on the same side of AB.

Question 3. Find x in the given figure.

Answer:
From the figure, ABCD is a cyclic quadrilateral and one angle is 100°.

Opposite angles of a cyclic quadrilateral add up to 180°.

So:

x + 100° = 180°
x = 80°

Final answer:
x = 80°

NCERT Solutions for Class 9 Maths Chapter 5 End-of-Chapter Exercises

The end-of-chapter exercises include chord-length questions, arc-angle questions, cyclic quadrilateral problems, proof-based questions and construction problems.

Question 1. In a circle, a chord is 5 cm away from the centre. If the radius is 13 cm, what is the length of the chord?

Answer:
Radius = 13 cm
Distance from centre = 5 cm

Half chord = √(13² - 5²)
= √(169 - 25)
= √144
= 12 cm

Chord length = 2 × 12
= 24 cm

Final answer:
24 cm

Question 2. 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?

Answer:
Angle subtended by an arc at the centre is twice the angle subtended at a point on the circle.

So:

Angle at point on circle = 70°/2
= 35°

Final answer:
35°

Question 3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn. Find the distance from the centre of the circle to the chord.

Answer:
Diameter = 26 cm
Radius = 13 cm
Chord = 24 cm

Half chord = 12 cm

Distance from centre:

d = √(13² - 12²)
= √(169 - 144)
= √25
= 5 cm

Final answer:
5 cm

Question 4. A circle has radius 15 cm. A chord is drawn. The distance from the centre to the chord is 9 cm. What is the length of the chord?

Answer:
Radius = 15 cm
Distance from centre = 9 cm

Half chord = √(15² - 9²)
= √(225 - 81)
= √144
= 12 cm

Chord length = 2 × 12
= 24 cm

Final answer:
24 cm

Question 5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Answer:
Let AB be a chord and let its perpendicular bisector meet AB at M.

Every point on the perpendicular bisector of AB is equidistant from A and B.

The centre O of the circle is also equidistant from A and B because:

OA = OB = radius

Therefore, O lies on the perpendicular bisector of AB.

Final answer:
The perpendicular bisector of a chord passes through the centre.

Question 6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB?

Answer:
A diameter subtends a right angle at any point on the circle.

Therefore:

∠ACB = 90°

Final answer:
∠ACB = 90°

Question 7. ABCD is a cyclic quadrilateral. If ∠A = 75°, find ∠C. If ∠B = 110°, find ∠D.

Answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

∠A + ∠C = 180°
75° + ∠C = 180°
∠C = 105°

Also:

∠B + ∠D = 180°
110° + ∠D = 180°
∠D = 70°

Final answer:
∠C = 105°, ∠D = 70°

Question 8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x - 20)°, find x and the measures of ∠P and ∠R.

Answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

∠P + ∠R = 180°

(2x + 10) + (3x - 20) = 180
5x - 10 = 180
5x = 190
x = 38

Now:

∠P = 2(38) + 10 = 86°
∠R = 3(38) - 20 = 94°

Final answer:
x = 38, ∠P = 86°, ∠R = 94°

Question 9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius.

Answer:
Chord = 16 cm
Half chord = 8 cm
Distance from centre = 6 cm

Radius² = 8² + 6²
= 64 + 36
= 100

Radius = 10 cm

Final answer:
10 cm

Question 10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Answer:
A cyclic quadrilateral with sides 5, 5, 12, 12 can be treated as an isosceles trapezium.

Let the parallel sides be 12 and 12, and the other two equal sides be 5 and 5. In this case, it forms a rectangle-like cyclic case with height 5.

Area = 12 × 5
= 60 square units

Final answer:
60 square units

Question 11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside or outside the quadrilateral?

Answer:
Use the angles of the cyclic quadrilateral.

If the cyclic quadrilateral is acute in a suitable sense and its diagonals are positioned so that the perpendicular bisectors meet inside, the circumcentre lies inside.

If the quadrilateral contains an obtuse angle or the perpendicular bisectors of two sides meet outside the quadrilateral, the circumcentre lies outside.

Best method:

1. Draw perpendicular bisectors of two sides.
2. Their point of intersection is the circumcentre.
3. Check whether this point lies inside or outside the quadrilateral.

Final answer:
Find the intersection of perpendicular bisectors of two sides and check its position.

Question 12. When two chords intersect, each 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.

Answer:
Let equal chords AB and CD intersect at P.

Since AB = CD:

AP + PB = CP + PD

For intersecting chords in a circle:

AP × PB = CP × PD

If two positive pairs have the same sum and same product, then the pairs are equal in corresponding order.

Therefore:

AP = CP and PB = PD

or

AP = PD and PB = CP

Final answer:
Equal intersecting chords are divided into equal corresponding segments.

Question 13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.

Answer:
Steps of construction:

1. Draw a line segment AB = 6 cm.
2. Find its midpoint M.
3. Draw a perpendicular line through M.
4. Mark O on this perpendicular such that OM = 3 cm.
5. Join OA.
6. With O as centre and OA as radius, draw a circle.

Since OM is perpendicular to AB and passes through the midpoint of AB, AB is a chord at distance 3 cm from centre O.

Final answer:
The required circle is obtained with centre O and radius OA.

Question 14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

Answer:
In a cyclic quadrilateral, opposite angles add up to 180°.

In a parallelogram, opposite angles are equal.

Let one pair of opposite angles be A and C.

Since it is cyclic:

A + C = 180°

Since it is a parallelogram:

A = C

Therefore:

2A = 180°
A = 90°

So all angles of the parallelogram are 90°.

Final answer:
A cyclic parallelogram must be a rectangle.

Question 15. Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Answer:
Let ABCD be a rectangle inscribed in a circle.

The diagonals AC and BD of a rectangle are equal and bisect each other.

Let their point of intersection be O.

Then:

OA = OB = OC = OD

So O is equidistant from all four vertices.

The centre of the circle is also the point equidistant from all points on the circle.

Final answer:
The intersection of the diagonals is the centre of the circle.

Question 16. Consider all chords of a circle of a fixed length. What shape is formed by the midpoints of all these chords?

Answer:
Equal chords are equidistant from the centre.

So, the midpoint of every chord of fixed length is at the same distance from the centre.

The locus of points at a fixed distance from a fixed point is a circle.

Final answer:
The midpoints form a circle with the same centre as the original circle.

Question 17. In a circle with centre O, chords AB and AC are congruent. Explain why the centre lies on the angle bisector of ∠BAC.

Answer:
Given:

AB = AC

Also:

OB = OC because both are radii.

OA is common.

So triangles OAB and OAC are congruent by SSS.

Therefore:

∠BAO = ∠OAC

This means AO bisects ∠BAC.

Final answer:
The centre O lies on the angle bisector of ∠BAC.

Question 18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre. The distance between the chords is 7 cm. Find the radius.

Answer:
Let radius be r.

For chord 10 cm:

Half chord = 5 cm
Distance from centre = √(r² - 25)

For chord 24 cm:

Half chord = 12 cm
Distance from centre = √(r² - 144)

Since the longer chord is closer to the centre, the distance between the chords is:

√(r² - 25) - √(r² - 144) = 7

Try r = 13:

√(169 - 25) - √(169 - 144)
= √144 - √25
= 12 - 5
= 7

Final answer:
Radius = 13 cm

Question 19. A regular hexagon is inscribed in a circle of radius r. Find the length of each side of the hexagon and the distance of each side from the centre.

Answer:
A regular hexagon divides the circle into 6 equal central angles.

Each central angle = 360°/6 = 60°

Each side of the hexagon is a chord subtending 60° at the centre.

The triangle formed by two radii and one side is equilateral.

So:

Side length = r

Distance from centre to each side:

Draw perpendicular from centre to a side. It bisects the side.

Half side = r/2

Distance d:

d² + (r/2)² = r²
d² = r² - r²/4
= 3r²/4

d = (√3/2)r

Final answer:
Side length = r, distance of each side from centre = (√3/2)r.

Question 20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP?

Answer:
Since MN is a diameter, the angle subtended by MN at any point on the circle is 90°.

If O lies on the circle, then:

∠MON or ∠MPN-type angles subtended by diameter MN are 90° depending on the vertex.

For the given angles, both are angles in the cyclic quadrilateral related to arcs and opposite angles.

Since MNOP is cyclic:

∠MOP + ∠MNP = 180°

Final answer:
∠MOP and ∠MNP are supplementary. Their sum is 180°.

Question 21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle.

Answer:
Let exterior angle at D be ∠CDE, where E lies on extension of CD.

Since ABCD is cyclic:

∠ADC + ∠ABC = 180°

Also, ∠ADC + ∠CDE = 180° because they form a linear pair.

Therefore:

∠CDE = ∠ABC

Final answer:
The exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Question 22. There is no chord of a circle longer than its diameter. Justify this statement.

Answer:
The diameter passes through the centre and is the longest possible chord.

Any other chord lies at a positive distance from the centre. Its length is:

2√(r² - d²)

where d > 0.

Since r² - d² < r²:

2√(r² - d²) < 2r

But diameter = 2r.

Final answer:
No chord can be longer than the diameter.

Question 23. Let A be any point within a given circle with centre O. Show that the shortest chord passing through A is perpendicular to OA.

Answer:
All chords passing through A have different distances from O.

The chord perpendicular to OA at A has distance OA from the centre. This is the greatest possible perpendicular distance among chords through A.

The farther a chord is from the centre, the shorter it is.

Therefore, the chord through A perpendicular to OA is the shortest.

Final answer:
The shortest chord through A is perpendicular to OA.

Question 24. How would you use the given figure to justify that the angle in a semicircle is 90°?

Answer:
Let AB be a diameter and O be the centre.

The arc AB is a semicircle, so the angle subtended by arc AB at the centre is 180°.

The angle subtended by the same arc at any point on the circle is half the central angle.

So:

Angle in semicircle = 180°/2
= 90°

Final answer:
The angle in a semicircle is 90°.

Question 25. In a circle, two chords CC′ and DD′ are drawn perpendicular to a diameter AB. Prove that the segment MM′ joining the midpoints of CD and C′D′ is perpendicular to AB.

Answer:
Since CC′ and DD′ are perpendicular to diameter AB, they are parallel chords.

The midpoints of corresponding segments formed symmetrically about AB lie on a line parallel to the chords and perpendicular to AB.

By symmetry of the circle about the diameter AB, the midpoint M of CD and midpoint M′ of C′D′ are mirror-related across the diameter.

Therefore, MM′ is perpendicular to AB.

Final answer:
MM′ is perpendicular to AB.

Question 26. How would you use the figure to justify that the sum of opposite angles of a cyclic quadrilateral is 180°?

Answer:
Let ABCD be a cyclic quadrilateral with centre O.

Angle ∠A subtends arc BCD.
Angle ∠C subtends arc BAD.

The two arcs together make the full circle, whose angle at the centre is 360°.

Since angle at the circle is half the corresponding angle at the centre:

∠A + ∠C = 1/2 × 360°
= 180°

Similarly:

∠B + ∠D = 180°

Final answer:
Opposite angles of a cyclic quadrilateral add up to 180°.

Topics Covered in NCERT Solutions for Class 9 Maths Chapter 5

1. Circle
2. Centre
3. Radius
4. Chord
5. Diameter
6. Locus of points
7. Reflection symmetry of circle
8. Rotational symmetry of circle
9. Circles through two points
10. Perpendicular bisector
11. Circumcircle
12. Circumcentre
13. Equal chords
14. Angle subtended by a chord
15. Perpendicular from centre to chord
16. Distance of chord from centre
17. Longer and shorter chords
18. Arc of a circle
19. Major arc and minor arc
20. Angle subtended by an arc
21. Angle in a semicircle
22. Concyclic points
23. Cyclic quadrilateral
24. Opposite angles of cyclic quadrilateral

Important Formulas and Theorems in NCERT Solutions for Class 9 Maths Chapter 5

NCERT Class 9 Maths Ganita Manjari 2026 Chapter Solutions