NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down, and Round and Round 2026-27

Think and Reflect (NCERT Textbook Page No. 93)

Question 1. Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

Solution: Amina may have asked Jamuna to fold the circular paper exactly into half and make a crease. This crease represents a diameter of the circle. She should then unfold the paper and fold it again in another direction to form a second diameter. The point where the two creases intersect is the centre of the circular paper.

Think and Reflect (NCERT Textbook Page No. 94)

Question 1. What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?

Solution:

(a) Square

Image 1

A square has rotational symmetry of order 4. It matches its original position after rotations of 90°, 180°, 270°, and 360°.

It has 4 lines of reflection symmetry:

2 lines through the midpoints of opposite sides

2 diagonal lines through opposite vertices

(b) Regular Pentagon

Image 2

A regular pentagon has rotational symmetry of order 5. It matches its original position after rotations of 72°, 144°, 216°, 288°, and 360°.

It has 5 lines of reflection symmetry. Each line passes through a vertex and the midpoint of the opposite side.

(c) Regular Hexagon

Image 3

A regular hexagon has rotational symmetry of order 6. It matches its original position after rotations of 60°, 120°, 180°, 240°, 300°, and 360°.

It has 6 lines of reflection symmetry:

3 lines through pairs of opposite vertices

3 lines through the midpoints of opposite sides

Question 2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?

Solution:
The longest chord of a circle is its diameter.

Diameter = 2 × Radius

= 2 × 5

= 10 units

Therefore, the longest chord is 10 units long.

There is no smallest chord because two points on the circle can be chosen as close to each other as desired. The chord length can approach zero but cannot become zero for two distinct points.

Question 3. The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?
(Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)

Solution:
Let A and B be two fixed points. Any point P that is equally distant from A and B satisfies:

PA = PB

According to the perpendicular bisector theorem, every point lying on the perpendicular bisector of AB is equidistant from A and B. Conversely, every point equidistant from A and B lies on the perpendicular bisector of AB.

Image 4

Therefore, the locus of all points equidistant from two given points A and B is the perpendicular bisector of the line segment AB.

Think and Reflect (NCERT Textbook Page No. 95)

Question 1. How many circles pass through two points on a plane?

Solution: Infinitely many circles can pass through two given points on a plane. The centre of each circle can be chosen at any suitable point on the perpendicular bisector of the line segment joining the two points.

Question 2. Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?

Solution: 

No, circles of every possible radius cannot pass through A and B. The radius cannot be less than half the distance AB.

The smallest circle is formed when its centre is the midpoint of AB. Its radius is:

AB/2

There is no largest circle. As the centre moves farther away along the perpendicular bisector of AB, the radius continues to increase without any upper limit.

Question 3. As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?

Solution: The centre of every circle passing through A and B lies on the perpendicular bisector of AB. As the centre moves farther away from the midpoint of AB, its distance from A and B increases. Therefore, the radii of the circles increase.

Question 4. As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or. less curved?

Solution: As the centre moves farther away along the perpendicular bisector, the radius of the circle becomes larger. A circle with a larger radius appears flatter near A and B. Therefore, the circle will appear less curved.

Question 5. You are given two points, A and B, on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?

Solution:

(a) When A and B lie on the boundary:

Infinitely many squares can be drawn because their size, position, and orientation can be varied while keeping both A and B somewhere on the boundary.

(b) When A and B are corners:

Three squares can be drawn.

If A and B are adjacent vertices, two squares can be constructed on opposite sides of AB.

If A and B are opposite vertices, one square can be constructed by taking AB as its diagonal.

Think, Draw, and Infer (NCERT Textbook Page No. 98)

Question 1. A, B, and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B, and C, the perpendicular bisectors of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle into three distinct points?

Solution:
A point satisfying PA = PB must lie on the perpendicular bisector of AB. Similarly, a point satisfying PB = PC must lie on the perpendicular bisector of BC.

Since A, B, and C are collinear, both AB and BC lie on the same straight line. Their perpendicular bisectors are therefore perpendicular to the same line and are parallel to each other. As they do not meet, no point P can satisfy PA = PB = PC.

A circle cannot pass through three distinct collinear points because a straight line can intersect a circle at no more than two points. For the same reason, no line can cut a circle at three distinct points.

Question 2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?

Solution: Yes. The triangle can be rotated about the centre of the circumcircle or reflected across a line through the centre. Its side lengths and angles remain unchanged, so the new triangle is congruent to ΔABC and its vertices continue to lie on the same circumcircle.

Exercise (NCERT Textbook Page No. 107)

Question 1. A circle with centre O is drawn, and A, B, C, D are points on the circle (See Figure). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

Solution:

Image 5

Students should use a protractor to measure the angles subtended by arcs AKB and CLD at the centre O.

If the central angle is less than 180°, classify the corresponding arc as a minor arc. If the central angle is greater than 180°, classify it as a major arc.

Exercise (NCERT Textbook Page No. 113)

Question 1. A cyclic quadrilateral ABCD has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.

Solution:
In a cyclic quadrilateral, each pair of opposite angles must be supplementary.

Check the opposite angles:

∠A + ∠C = 80° + 100° = 180°

∠B + ∠D = 110° + 70° = 180°

Both pairs of opposite angles add up to 180°. Therefore, a cyclic quadrilateral with the given angle measurements can be drawn.

Class 9 Maths Ganita Manjari Chapter 5 Exercise Set 5.1 Solutions

Exercise Set 5.1 

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

Solution:
Given:

AB = 5 cm
∠A = 70°
∠B = 60°

Image 6

Steps of Construction:

1. Draw a line segment AB of length 5 cm.

2. At point A, construct an angle of 70° using a protractor.

3. At point B, construct an angle of 60°.

4. Let the two rays meet at point C. Join AC and BC to form ∆ABC.

5. Draw the perpendicular bisectors of any two sides of the triangle, such as AB and AC.

6. Mark their point of intersection as O. This point is the circumcentre of ∆ABC.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

Since all the angles of ∆ABC are less than 90°, it is an acute-angled triangle. Therefore, its circumcentre O lies inside the triangle.

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

Solution:

Image 7

Given:

AB = 5 cm
∠A = 100°
AC = 4 cm

Steps of Construction:

1. Draw a line segment AB of length 5 cm.

2. At point A, construct an angle of 100°.

3. With A as the centre and radius 4 cm, draw an arc cutting the second arm of the angle at C.

4. Join B and C to complete ∆ABC.

5. Draw the perpendicular bisectors of any two sides, such as AB and AC.

6. Let the perpendicular bisectors meet at O. This point is the circumcentre.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

Since ∠A = 100°, ∆ABC is an obtuse-angled triangle. Therefore, its circumcentre lies outside the triangle.

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

Solution:
Given:

AB = 6 cm
BC = 7 cm
CA = 7 cm

Steps of Construction:

1. Draw a line segment AB of length 6 cm.

2. With A as the centre and radius 7 cm, draw an arc.

3. With B as the centre and radius 7 cm, draw another arc intersecting the first arc at C.

4. Join AC and BC to form ∆ABC.

5. Draw the perpendicular bisectors of any two sides of the triangle.

6. Mark their point of intersection as O. This is the circumcentre.

7. With O as the centre and OA as the radius, draw a circle passing through A, B, and C.

The circumcentre is equally distant from all three vertices. Therefore:

OA = OB = OC

On measuring, each distance is approximately 3.87 cm. These equal distances represent the radius of the circumcircle.

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

Solution:
The centre of every circle passing through A and B lies on the perpendicular bisector of AB. The least possible radius is obtained when the centre is the midpoint of AB. In this case, AB becomes the diameter of the circle.

Therefore:

Least possible radius = AB/2

Hence, the smallest circle passing through A and B has a radius equal to half the length of AB.

Exercise Set 5.2 

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

Solution:
Given: AB is a chord of a circle with centre O.

To prove: ∆OAB is an isosceles triangle.

Proof:

Points A and B lie on the circumference of the circle. Therefore, OA and OB are radii of the same circle.

OA = OB

Thus, ∆OAB has two equal sides. Hence, ∆OAB is an isosceles triangle.

Therefore, the triangle formed by a chord and the centre of a circle is isosceles.

Image 8

Question 2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Solution:
Given: ∆OAB and ∆OCD are formed by chords AB and CD of the same circle with centre O, and:

AB = CD

To prove:

∆OAB ≅ ∆OCD

Proof:

In ∆OAB and ∆OCD:

OA = OC, as they are radii of the same circle.

OB = OD, as they are radii of the same circle.

AB = CD, as given.

Therefore, all three corresponding sides of the two triangles are equal.

Hence, by the SSS congruence criterion:

∆OAB ≅ ∆OCD

Therefore, two such isosceles triangles with equal bases are congruent.

Image 9

Exercise Set 5.3 

Question 1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use Figure. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM)

Solution:
Given: C is the centre of the circle, AB is a chord, and CM ⊥ AB.

Image 10

Therefore:

∠CMA = ∠CMB = 90°

To prove:

AM = BM

Proof:

Consider ∆CMA and ∆CMB.

CA = CB, as they are radii of the same circle.

CM = CM, as it is the common side.

∠CMA = ∠CMB = 90°, as given.

Therefore, by the RHS congruence criterion:

∆CMA ≅ ∆CMB

Hence, by corresponding parts of congruent triangles:

AM = BM

Therefore, the perpendicular drawn from the centre of a circle 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.

Solution:

Image 11

Given: ∆ABC is an isosceles triangle inscribed in a circle with centre O, where AB = AC. AM is the altitude drawn from A to BC.

To prove: AM passes through the centre O.

Proof:

Consider ∆AMB and ∆AMC.

AB = AC, as given.

AM = AM, as it is the common side.

∠AMB = ∠AMC = 90°, because AM is an altitude.

Therefore, by the RHS congruence criterion:

∆AMB ≅ ∆AMC

Hence:

BM = CM

Thus, M is the midpoint of chord BC.

The line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord. Therefore:

OM ⊥ BC

We also know that:

AM ⊥ BC

Since only one perpendicular can be drawn through point M to line BC, AM and OM must be the same straight line.

Therefore, A, O, and M are collinear. Hence, the altitude from A to BC passes through the centre O 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 of the circle is 5 cm, find the distance between the midpoints of the chords.

Solution:
Given:

Radius of the circle = 5 cm

Length of chord AB = 6 cm

Length of chord CD = 8 cm

Let N and M be the midpoints of chords AB and CD, respectively.

Image 12

Since the line joining the centre of a circle to the midpoint of a chord is perpendicular to the chord:

AN = AB/2 = 6/2 = 3 cm

In right-angled ∆ANO:

OA² = ON² + AN²

5² = ON² + 3²

25 = ON² + 9

ON² = 16

ON = 4 cm

Similarly:

CM = CD/2 = 8/2 = 4 cm

In right-angled ∆CMO:

OC² = OM² + CM²

5² = OM² + 4²

25 = OM² + 16

OM² = 9

OM = 3 cm

Since the chords lie on opposite sides of the centre:

MN = ON + OM

= 4 + 3

= 7 cm

Therefore, the distance between the midpoints of the two chords is 7 cm.

Exercise Set 5.4

Question 1. Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.

Solution:

Given: AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendicular to AB and CD, respectively.

Therefore:

OP ⊥ AB and OQ ⊥ CD

To prove:

OP = OQ

Proof:

A perpendicular drawn from the centre of a circle to a chord bisects the chord.

Therefore:

AP = AB/2

CQ = CD/2

Since AB = CD, we get:

AP = CQ

Also, OA and OC are radii of the same circle.

OA = OC = r

In right-angled ∆OPA, using the Baudhayana-Pythagoras theorem:

OA² = OP² + AP²

r² = OP² + AP² …(i)

In right-angled ∆OQC:

OC² = OQ² + CQ²

r² = OQ² + CQ² …(ii)

From equations (i) and (ii):

OP² + AP² = OQ² + CQ²

Since AP = CQ:

OP² = OQ²

Therefore:

OP = OQ

Hence, equal chords of a circle are equidistant from its centre.

Image 13

Question 2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Image 14

Solution:
Given: AB and GF are chords of a circle with centre C.

CE ⊥ AB
CH ⊥ GF
CE = CH

To prove:

AB = GF

Proof:

A perpendicular drawn from the centre of a circle to a chord bisects the chord.

Therefore:

AE = EB

and

GH = HF

Consider right-angled ∆CEA and ∆CHG.

CA = CG, as they are radii of the same circle.

CE = CH, as given.

∠CEA = ∠CHG = 90°

Therefore, by the RHS congruence criterion:

∆CEA ≅ ∆CHG

Hence, by CPCT:

AE = GH

Now:

AB = 2AE

and

GF = 2GH

Since AE = GH:

AB = GF

Therefore, the two chords are equal.

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

Image 15

Solution:
Given: AB and GF are chords of a circle with centre C.

CE ⊥ AB
CH ⊥ GF
CE = CH

To prove:

AB = GF

Proof:

Let:

CA = CG = r

because they are radii of the same circle.

Also, let:

CE = CH = d

In right-angled ∆CEA, using the Baudhayana-Pythagoras theorem:

CA² = CE² + AE²

AE² = CA² - CE²

AE² = r² - d² …(i)

In right-angled ∆CHG:

CG² = CH² + GH²

GH² = CG² - CH²

GH² = r² - d² …(ii)

From equations (i) and (ii):

AE² = GH²

Therefore:

AE = GH

A perpendicular drawn from the centre of a circle to a chord bisects the chord.

So:

AB = 2AE

and

GF = 2GH

Since AE = GH:

AB = GF

Hence proved.

Exercise Set 5.5 

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

Solution:

Image 16

Given:

Radius OA = 7 cm

Perpendicular distance OP = 6 cm

Let AB be the chord and P be its midpoint.

A perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore:

AP = PB

In right-angled ∆OPA, using the Baudhayana-Pythagoras theorem:

OA² = OP² + AP²

7² = 6² + AP²

49 = 36 + AP²

AP² = 13

AP = √13 cm

Since P is the midpoint of AB:

AB = 2AP

AB = 2√13 cm

AB ≈ 7.21 cm

Therefore, the length of the chord is 2√13 cm, or approximately 7.21 cm.

Question 2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√r2−d2.

Solution:

Image 17

Let AB be a chord of a circle with centre O. Let OM be perpendicular to AB.

Given:

OA = r

OM = d

Since the perpendicular from the centre of a circle to a chord bisects the chord:

AM = MB

In right-angled ∆OMA, using the Baudhayana-Pythagoras theorem:

OA² = OM² + AM²

r² = d² + AM²

AM² = r² - d²

AM = √(r² - d²)

Since M is the midpoint of AB:

AB = 2AM

Therefore:

AB = 2√(r² - d²)

Hence, the length of a chord at a perpendicular distance d from the centre of a circle of radius r is 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? Give reasons for your answer.

Solution:
No, we cannot conclude that CD = 2AB.

The length of a chord does not change in direct proportion to its distance from the centre. For a circle of radius r, the length of a chord at a perpendicular distance d from the centre is:

Chord length = 2√(r² - d²)

Let the distance of chord CD from the centre be x. Then the distance of chord AB from the centre is 2x.

Therefore:

AB = 2√(r² - 4x²)

and

CD = 2√(r² - x²)

The ratio of their lengths is:

CD/AB = √(r² - x²)/√(r² - 4x²)

This ratio depends on the values of r and x and is not always equal to 2.

Therefore, CD is longer than AB because it is closer to the centre, but we cannot conclude that CD = 2AB.

Exercise Set 5.6

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

Solution:

Image 18

OA and OB are radii of the same circle.

Therefore:

OA = OB = 12 cm

So, ∆AOB is an isosceles triangle.

Given:

∠AOB = 60°

Let:

∠OAB = ∠OBA = x

Using the angle-sum property of a triangle:

x + x + 60° = 180°

2x = 120°

x = 60°

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

Hence:

AB = OA = OB = 12 cm

Therefore, the length of chord AB is 12 cm.

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

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

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

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

Therefore:

∠AXB = ∠AYB

Hence, these two angles cannot be different.

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

Solution:
No, this is not always true.

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

∠AXB = ∠AYB = 90°

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

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

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

∠AXB = ∠AYB

then A, B, X, and Y are concyclic.

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

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

Question 3. Find x in Fig. 5.26.

Image 19

Solution:
Given:

∠ADC = 100°

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

Therefore:

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

The complete angle around the centre is 360°.

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

360° - 200° = 160°

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

x = 160°/2

x = 80°

Therefore, x = 80°.

End of Chapter Exercises Solutions

Question 1. 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:

Image 20

Given:

Radius OA = 13 cm

Perpendicular distance OP = 5 cm

Let P be the midpoint of chord AB.

A perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore:

AP = PB

In right-angled ∆OPA, using the Baudhayana-Pythagoras theorem:

OA² = OP² + AP²

13² = 5² + AP²

169 = 25 + AP²

AP² = 144

AP = 12 cm

Therefore:

AB = 2AP

= 2 × 12

= 24 cm

Hence, the length of the chord is 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?

Solution:
The angle subtended by an arc at the centre is twice the angle subtended by the same arc at a point on the remaining part of the circle.

Therefore:

Angle at the circumference = 70°/2

= 35°

Hence, the angle subtended by the arc at a point on the remaining part of the circle is 35°.

Question 3. 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:

Image 21

Given:

Diameter = 26 cm

Chord AB = 24 cm

Radius OA = 26/2 = 13 cm

Let OP be the perpendicular distance from the centre to chord AB.

A perpendicular from the centre to a chord bisects the chord.

Therefore:

AP = 24/2 = 12 cm

In right-angled ∆OPA:

OA² = OP² + AP²

13² = OP² + 12²

169 = OP² + 144

OP² = 25

OP = 5 cm

Therefore, the distance from the centre to the chord is 5 cm.

Question 4. 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:

Image 22

Given:

Radius OA = 15 cm

Perpendicular distance OP = 9 cm

Let P be the midpoint of chord AB.

In right-angled ∆OPA:

OA² = OP² + AP²

15² = 9² + AP²

225 = 81 + AP²

AP² = 144

AP = 12 cm

Since P is the midpoint of AB:

AB = 2AP

= 2 × 12

= 24 cm

Therefore, the length of the chord is 24 cm.

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

Solution:

Image 23

Given: AB is a chord of a circle with centre O. Its perpendicular bisector meets AB at M.

To prove: The perpendicular bisector of AB passes through O.

Proof:

OA and OB are radii of the same circle.

Therefore:

OA = OB

This means that O is equidistant from A and B.

The locus of all points equidistant from the endpoints of a line segment is its perpendicular bisector.

Therefore, O lies on the perpendicular bisector of AB.

Hence, the perpendicular bisector of a chord passes through the centre of the circle.

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

Solution:
AB is a diameter of the circle, and C is a point on its circumference.

The angle subtended by a diameter at any point on the circumference is a right angle.

Therefore:

∠ACB = 90°

Hence, the measure of ∠ACB is 90°.

Question 7. 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:
Opposite angles of a cyclic quadrilateral are supplementary.

Therefore:

∠A + ∠C = 180°

75° + ∠C = 180°

∠C = 105°

Similarly:

∠B + ∠D = 180°

110° + ∠D = 180°

∠D = 70°

Therefore:

∠C = 105°

∠D = 70°

Question 8. 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:
Since PQRS is a cyclic quadrilateral, its opposite angles are supplementary.

Therefore:

∠P + ∠R = 180°

(2x + 10) + (3x - 20) = 180

5x - 10 = 180

5x = 190

x = 38

Now:

∠P = 2x + 10

= 2(38) + 10

= 86°

Also:

∠R = 3x - 20

= 3(38) - 20

= 94°

Therefore:

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 of the circle.

Solution:

Image 24

Given:

Chord AB = 16 cm

Perpendicular distance OP = 6 cm

The perpendicular from the centre bisects the chord.

Therefore:

AP = 16/2 = 8 cm

In right-angled ∆OPA:

OA² = OP² + AP²

r² = 6² + 8²

r² = 36 + 64

r² = 100

r = 10 cm

Therefore, the radius of the circle is 10 cm.

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

Solution:

Image 25

Let ABCD be the cyclic quadrilateral such that:

AB = BC = 5 units

CD = DA = 12 units

Join diagonal BD.

In ∆ABD and ∆CBD:

AB = CB

AD = CD

BD = BD

Therefore:

∆ABD ≅ ∆CBD by the SSS congruence criterion.

Hence:

∠BAD = ∠BCD

Since ABCD is cyclic, opposite angles are supplementary.

∠BAD + ∠BCD = 180°

As the two angles are equal:

∠BAD = ∠BCD = 90°

Therefore, ∆ABD and ∆CBD are right-angled triangles with perpendicular sides 5 units and 12 units.

Area of ∆ABD:

= 1/2 × 5 × 12

= 30 square units

Similarly, area of ∆CBD = 30 square units.

Total area of ABCD:

= 30 + 30

= 60 square units

Therefore, the area of the cyclic quadrilateral is 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 the quadrilateral or outside? What is the best way of finding out?

Solution: Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Since these sides are chords of its circumcircle, their perpendicular bisectors pass through the circumcentre.

The point where the two perpendicular bisectors intersect is the centre of the circumcircle.

Now observe the position of this point:

If the point lies within the quadrilateral, the circumcentre is inside.

If the point lies beyond its boundary, the circumcentre is outside.

If it lies on one of the sides, the circumcentre lies on the quadrilateral.

Therefore, constructing the perpendicular bisectors of two sides is the simplest and most reliable method.

Question 12. 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:
Let equal chords AB and CD intersect at E.

Image 26

Since:

AB = CD

and equal chords of a circle are equidistant from its centre, draw perpendiculars OM and ON from the centre O to AB and CD.

Then:

OM = ON

Also, perpendiculars from the centre bisect the chords.

Therefore:

AM = MB

CN = ND

Since AB = CD:

AM = CN

and:

MB = ND

Now consider right-angled ∆OME and ∆ONE.

OM = ON

OE = OE

∠OME = ∠ONE = 90°

Therefore:

∆OME ≅ ∆ONE by the RHS congruence criterion.

Hence:

ME = NE

Using the arrangement shown in the figure:

AE = AM + ME

CE = CN + NE

Since AM = CN and ME = NE:

AE = CE

Similarly:

BE = MB - ME

DE = ND - NE

Since MB = ND and ME = NE:

BE = DE

Therefore, the corresponding segments of the two equal intersecting chords are equal.

Question 13. 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?)

Solution:

Image 27

Steps of Construction:

1. Draw a line segment AB of length 6 cm.

2. Construct the perpendicular bisector of AB. Let it meet AB at P.

3. Since the centre is 3 cm away from the chord, mark a point O on the perpendicular bisector such that:

OP = 3 cm

4. Join OA and OB.

5. With O as the centre and OA as the radius, draw a circle.

The circle passes through A and B, and the perpendicular distance of chord AB from the centre O is 3 cm.

Therefore, this is the required circle.

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

Solution:

Image 28

Let parallelogram ABCD be inscribed in a circle.

Since ABCD is cyclic, its opposite angles are supplementary.

Therefore:

∠A + ∠C = 180°

Since ABCD is also a parallelogram, opposite angles are equal.

Therefore:

∠A = ∠C

So:

2∠A = 180°

∠A = 90°

Similarly, all the other angles are also 90°.

A parallelogram with four right angles is a rectangle.

Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.

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.

Solution:

Image 29

Let rectangle ABCD be inscribed in a circle. Let its diagonals AC and BD intersect at O.

In a rectangle, the diagonals bisect each other.

Therefore:

AO = OC

BO = OD

The diagonals of a rectangle are also equal.

AC = BD

Hence, their halves are equal:

AO = BO = CO = DO

Thus, point O is equidistant from all four vertices A, B, C, and D.

The point equidistant from all points on a circle is its centre.

Therefore, the point of intersection of the diagonals of the rectangle is the centre of the circle.

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

Solution:
All chords of equal length in a circle are at the same perpendicular distance from the centre.

Therefore, the midpoint of every fixed-length chord remains at the same distance from the centre.

A set of points lying at a fixed distance from one point forms a circle.

Hence, the midpoints of all chords of a fixed length form a smaller circle concentric with the original circle.

Question 17. 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:

Image 30

Given:

AB = AC

To prove:

∠BAO = ∠OAC

Consider ∆AOB and ∆AOC.

OA = OA, common side

OB = OC, radii of the same circle

AB = AC, given

Therefore:

∆AOB ≅ ∆AOC by the SSS congruence criterion.

Hence, by CPCT:

∠BAO = ∠OAC

Therefore, AO bisects ∠BAC.

Thus, 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 of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Solution:

Image 31

Let the perpendicular distance from the centre O to the longer chord of length 24 cm be x cm.

Since the shorter chord is farther from the centre, its distance from O is:

x + 7 cm

Half of the 24 cm chord:

24/2 = 12 cm

Half of the 10 cm chord:

10/2 = 5 cm

Let the radius be r.

For the 24 cm chord, using the Baudhayana-Pythagoras theorem:

r² = x² + 12²

r² = x² + 144 …(i)

For the 10 cm chord:

r² = (x + 7)² + 5²

r² = x² + 14x + 49 + 25

r² = x² + 14x + 74 …(ii)

Equating equations (i) and (ii):

x² + 144 = x² + 14x + 74

144 - 74 = 14x

70 = 14x

x = 5 cm

Substitute x = 5 in equation (i):

r² = 5² + 144

r² = 25 + 144

r² = 169

r = 13 cm

Therefore, the radius of the circle is 13 cm.

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:

Image 32

Let ABCDEF be a regular hexagon inscribed in a circle with centre O and radius r.

The six equal sides subtend six equal angles at the centre.

Therefore, each central angle is:

360°/6 = 60°

Consider ∆AOB.

OA = OB = r

and:

∠AOB = 60°

The remaining two angles are equal and their sum is:

180° - 60° = 120°

Therefore, each is 60°.

Thus, ∆AOB is equilateral.

Hence:

AB = OA = OB = r

Therefore, the length of each side of the regular hexagon is r.

Now, let M be the midpoint of AB. Then:

AM = r/2

Also:

OM ⊥ AB

In right-angled ∆OMA:

OA² = OM² + AM²

r² = OM² + (r/2)²

OM² = r² - r²/4

OM² = 3r²/4

OM = r√3/2

Therefore:

Length of each side = r

Distance of each side from the centre = r√3/2

Question 20. 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:

Image 33

Both ∠MOP and ∠MNP are angles subtended by the same chord MP at points O and N on the same segment of the circle.

Angles subtended by the same chord in the same segment are equal.

Therefore:

∠MOP = ∠MNP

Question 21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).

Solution:
The intended exterior angle at vertex D is the angle formed by extending one side of the quadrilateral.

Since ABCD is cyclic:

∠ABC + ∠ADC = 180°

The exterior angle at D and ∠ADC form a linear pair.

Therefore:

Exterior angle at D + ∠ADC = 180°

Since both expressions are equal to 180°:

Exterior angle at D = ∠ABC

Hence, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.

Question 22. “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?

Solution:

Image 34

Let AB be any chord of a circle with centre O and radius r.

Join OA and OB.

In ∆AOB:

OA = OB = r

By the triangle inequality:

AB < OA + OB

AB < r + r

AB < 2r

Here, 2r is the diameter of the circle.

If chord AB passes through the centre, then AB itself is a diameter and:

AB = 2r

Therefore, no chord can be longer than the diameter. The diameter is the longest chord of a circle.

Question 23. 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:

Image 35

Let PQ be the chord passing through A such that:

OA ⊥ PQ

Consider any other chord RS passing through A. Draw:

OM ⊥ RS

Since M lies on chord RS and A also lies on RS, ∆OMA is right-angled at M.

Therefore, OA is the hypotenuse.

Hence:

OA > OM

The perpendicular distance of chord PQ from the centre is OA, while the perpendicular distance of chord RS is OM.

Therefore, chord PQ is farther from the centre than chord RS.

Among chords of the same circle, the chord farther from the centre is shorter.

Hence:

PQ < RS

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

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

Image 36

Solution:

Image 37

Let BC be a diameter of the circle with centre O, and let A be a point on the semicircle.

Join OA.

Since OA, OB, and OC are radii:

OA = OB = OC

In ∆AOB:

OA = OB

Therefore:

∠OAB = ∠ABO = a

In ∆AOC:

OA = OC

Therefore:

∠OAC = ∠ACO = b

Now:

∠BAC = a + b

Since B, O, and C lie on a straight line:

∠AOB + ∠AOC = 180°

In ∆AOB:

∠AOB = 180° - 2a

In ∆AOC:

∠AOC = 180° - 2b

Therefore:

(180° - 2a) + (180° - 2b) = 180°

360° - 2a - 2b = 180°

2a + 2b = 180°

a + b = 90°

Therefore:

∠BAC = 90°

Hence, the angle in a semicircle is a right angle.

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 the chords CD and C’ D’ is perpendicular to AB.

Solution:

Image 38

Given:

CC’ ⊥ AB

DD’ ⊥ AB

Therefore:

CC’ ∥ DD’

Consider quadrilateral CDD’C’. Since CC’ and DD’ are parallel, it is a trapezium.

M and M’ are the midpoints of the non-parallel sides CD and C’D’, respectively.

By the midpoint theorem for a trapezium, the line segment joining the midpoints of its non-parallel sides is parallel to its parallel sides.

Therefore:

MM’ ∥ CC’

Since:

CC’ ⊥ AB

and MM’ is parallel to CC’, we get:

MM’ ⊥ AB

Hence proved.

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

Image 39

Solution:
Let O be the centre of the circle containing cyclic quadrilateral ABCD.

Join OA, OB, OC, and OD.

Since all are radii:

OA = OB = OC = OD

Therefore, the four triangles formed are isosceles.

Let:

∠OAB = ∠OBA = p

∠OBC = ∠OCB = q

∠OCD = ∠ODC = u

∠ODA = ∠OAD = v

Now:

∠A = p + v

∠B = p + q

∠C = q + u

∠D = u + v

The sum of the angles of quadrilateral ABCD is 360°.

Therefore:

(p + v) + (p + q) + (q + u) + (u + v) = 360°

2p + 2q + 2u + 2v = 360°

p + q + u + v = 180°

Now:

∠A + ∠C

= (p + v) + (q + u)

= p + q + u + v

= 180°

Similarly:

∠B + ∠D

= (p + q) + (u + v)

= 180°

Therefore, the sum of each pair of opposite angles of a cyclic quadrilateral is 180°.

Test Yourself on Class 9 Maths Chapter 5 I’m Up and Down, and Round and Round

Before closing this chapter, check if you can answer these quickly: What is the angle in a semicircle? How long is a chord 5 cm from the centre of a circle of radius 13 cm? Why is a rectangle the only parallelogram that fits inside a circle? If any takes more than a few seconds, revisit the solved theorems above or revise with Vedantu's Chapter 5 Revision Notes.

How Vedantu’s NCERT Solutions for Chapter 5 I’m Up and Down, and Round and Round Helps You Score in Class 9 Exams?

Circles carries strong weightage in Class 9 Maths, mixing one-mark angle questions with full proof-based theorem questions. The chord-distance problems are reliable scoring questions once you master r² = d² + (chord/2)². After these solutions, attempt Vedantu's Sample Papers and Important Questions to see how often cyclic quadrilaterals and the angle-in-a-semicircle result appear.

Access Exercise Wise NCERT Solutions for Chapter 5 Maths Class 9

CBSE Class 9 Maths Chapter 5 Other Study Materials

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Additional Study Materials for Class 9 Maths