NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5: I’m Up and Down, and Round and Round by PW

Circles, chords, diameters, and angles subtended by arcs can be challenging to understand when you need to apply different theorems and solve geometry problems. Class 9 Maths Ganita Manjari Chapter 5, I’m Up and Down, and Round and Round, covers these concepts through definitions, geometric properties, and numerical questions.

NCERT Solutions for Chapter 5 provide step-by-step answers to help you understand the concepts and apply theorems correctly. The questions cover important definitions, calculations, and proofs, helping you practise problems and revise the chapter effectively.

NCERT Class 9 Maths Chapter 5 Solutions 

The following sections cover the questions and answers provided for Chapter 5, with each section including important concepts and worked solutions to support your revision.

01: Circles and Their Properties

Question 1: What is a circle?

Answer:

A circle is the set of all points in a plane that are at the same distance from a fixed point. That fixed point is called the centre, and the equal distance is called the radius.

Question 2: Define a chord and a diameter of a circle.

Answer:

• Chord: A line segment joining any two points on a circle.

• Diameter: A chord that passes through the centre of the circle. It is the longest chord, and its length is twice the radius.

Question 3: What are the rotational and line symmetries of a circle?

Answer:

• Rotational symmetry: A circle looks exactly the same after rotation through any angle. Thus, it has complete rotational symmetry.

• Line symmetry: Every diameter is an axis of reflection symmetry.

Question 4: How many circles can pass through three collinear points versus three non-collinear points?

Answer:

• Three collinear points: No circle can pass through three distinct collinear points.

• Three non-collinear points: Exactly one unique circle passes through three non-collinear points.

Question 5: State Theorem 1 of circles.

Answer:

Theorem 1: There is a unique circle passing through three non-collinear points.

01 PDF: Circles and Their Properties 

You can practise the questions and exercises covered in 01: Circles and Their Properties using the PDF below.

2: Circumcircle and Circumcentre

Question 1: What is the minimum radius of a circle passing through two given points A and B?

Answer:

The radius of the smallest circle passing through A and B is half the length of segment AB.

Minimum radius = AB ÷ 2

Question 2: Define circumcircle and circumcentre.

Answer:

• Circumcircle: The unique circle passing through all three vertices of a triangle.

• Circumcentre: The centre of the circumcircle.

Question 3: Where does the circumcentre lie for acute, obtuse, and right-angled triangles?

Answer:

• Acute triangle: The circumcentre lies inside the triangle.

• Obtuse triangle: The circumcentre lies outside the triangle.

• Right-angled triangle: The circumcentre lies at the midpoint of the hypotenuse.

Question 4: Why is it impossible for a circle to pass through three collinear points?

Answer:

For three collinear points A, B, and C, the perpendicular bisectors of AB and BC are parallel and never intersect. Since there is no common point of intersection to act as the centre, no circle can pass through three collinear points.

Question 5: Can a single straight line cut a circle at three distinct points?

Answer:

No. A straight line cannot cut a circle at more than two distinct points.

02 PDF: Circumcircle and Circumcentre 

You can practise the questions and exercises covered in 02: Circumcircle and Circumcentre using the PDF below.

03: Chords and Circle Theorems

Question 1: What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?

Answer:

The longest chord of a circle is its diameter.

Diameter = 2 × radius

Diameter = 2 × 5 = 10 units

There is no single smallest chord because chords can have lengths arbitrarily close to zero.

Question 2: State Theorem 2 and Theorem 3, the converse of Theorem 2.

Answer:

• Theorem 2: Equal chords of a circle subtend equal angles at the centre.

• Theorem 3: Chords of a circle that subtend equal angles at the centre are equal in length.

Question 3: Show that the triangle formed by a chord and the centre of the circle is an isosceles triangle.

Answer:

Let AB be a chord and O be the centre of the circle. In triangle OAB, OA = OB because both are radii of the same circle.

Since two sides of the triangle are equal, triangle OAB is an isosceles triangle.

Question 4: State Theorem 4 regarding the midpoint of a chord.

Answer:

Theorem 4: The line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.

Question 5: State Theorem 5 regarding a perpendicular from the centre to a chord.

Answer:

Theorem 5: The perpendicular from the centre of a circle to a chord of the circle bisects the chord.

03 PDF: Chords and Circle Theorems 

You can practise the questions and exercises covered in 03: Chords and Circle Theorems using the PDF below.

04: Chord Lengths and Distances 

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?

Answer:

Using the Pythagorean theorem in the right triangle formed by the radius, the perpendicular distance, and half the chord:

Radius² = Distance² + (Half-chord)²

13² = 5² + B²

169 = 25 + B²

B² = 169 − 25 = 144

B = 12 cm

Total chord length = 2 × 12 = 24 cm

Question 2: The diameter of a circle is 26 cm. A chord of length 24 cm is drawn. Find the distance of the chord from the centre.

Answer:

Radius = 26 ÷ 2 = 13 cm

Half-chord = 24 ÷ 2 = 12 cm

Using the Pythagorean theorem:

Distance² = Radius² − (Half-chord)²

Distance = √(13² − 12²)

Distance = √(169 − 144)

Distance = √25 = 5 cm

Question 3: State Theorem 6 and Theorem 7.

Answer:

• Theorem 6: Chords of a circle having the same length are all at the same distance from the centre of the circle.

• Theorem 7: Chords of a circle that are equidistant from the centre have equal lengths.

Question 4: Two parallel chords AB = 6 cm and CD = 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Determine PQ, the distance between them.

Answer:

For chord AB:

Half-chord = 6 ÷ 2 = 3 cm

Distance from the centre = √(5² − 3²)

Distance = √(25 − 9) = 4 cm

For chord CD:

Half-chord = 8 ÷ 2 = 4 cm

Distance from the centre = √(5² − 4²)

Distance = √(25 − 16) = 3 cm

Since the chords lie on opposite sides of the centre:

PQ = 4 + 3 = 7 cm

Question 5: Two parallel chords AB = 6 cm and CD = 8 cm lie on the same side of the centre of a circle of radius 5 cm. Determine PQ.

Answer:

The distance of chord AB from the centre is 4 cm.

The distance of chord CD from the centre is 3 cm.

Since both chords lie on the same side of the centre:

PQ = 4 − 3 = 1 cm

04 PDF: Chord Lengths and Distances 

You can practise the questions and exercises covered in 04: Chord Lengths and Distances using the PDF below.

05: Equal Chords and Their Properties

Question 1: Explain using the Baudhāyana-Pythagoras theorem why two equal chords in a circle are equidistant from the centre (Theorem 6).

Answer:

Let OA = OC = r be radii, and let chords AB and CD be equal.

Since perpendiculars from the centre bisect the chords:

AM = AB ÷ 2

CN = CD ÷ 2

As AB = CD, we have AM = CN.

Applying the Pythagorean theorem:

OA² = AM² + OM²

OC² = CN² + ON²

Since OA² = OC² and AM² = CN², it follows that:

OM² = ON²

Therefore, OM = ON. Hence, the two equal chords are equidistant from the centre.

Question 2: Find the length of a chord of a circle where the radius is 7 cm and the perpendicular distance from the centre is 6 cm.

Answer:

Let half the chord length be x.

Using the Pythagorean theorem:

x = √(7² − 6²)

x = √(49 − 36)

x = √13 cm

Total chord length = 2x = 2√13 cm

Question 3: Express the length of a chord in terms of its perpendicular distance d from the centre and the radius r.

Answer:

The length of a chord is:

Length of chord = 2√(r² − d²)

Question 4: State Theorem 8 regarding chord length and its distance from the centre.

Answer:

Theorem 8: Let AB and DE be two chords of a circle with centre C. Suppose AB is longer than DE. Then the distance from C to AB is less than the distance from C to DE. In other words, longer chords lie closer to the centre.

Question 5: What happens to the length of a chord as its perpendicular distance from the centre increases?

Answer:

As the perpendicular distance of a chord from the centre increases, the length of the chord decreases.

05 PDF: Equal Chords and Their Properties 

You can practise the questions and exercises covered in 05: Equal Chords and Their Properties using the PDF below.

06: Parallel Chords and Angles

Question 1: 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 of the circle.

Answer:

Let x be the distance of the 24 cm chord from the centre. The half-lengths of the chords are 12 cm and 5 cm.

Using the Pythagorean theorem:

r² = 12² + x²

r² = 5² + (x + 7)²

Equating the expressions:

144 + x² = 25 + (x + 7)²

144 + x² = 25 + x² + 14x + 49

144 = 74 + 14x

70 = 14x

x = 5 cm

Therefore:

r = √(12² + 5²)

r = √(144 + 25)

r = √169 = 13 cm

Question 2: Consider all chords of a circle of a fixed length. What path or shape is formed by the midpoints of all these chords?

Answer:

The locus formed by the midpoints of all equal-length chords is a concentric circle.

Question 3: State Theorem 9 regarding angles subtended by an arc.

Answer:

Theorem 9: The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at any point on the remaining part of the circle.

Question 4: What is the corollary of Theorem 9 regarding the angle subtended by a diameter?

Answer:

The angle subtended by a diameter at any point on the circle is 90°. An angle in a semicircle is a right angle.

Question 5: An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by it at the boundary of the circle?

Answer:

The angle at the boundary is half the angle at the centre.

Angle at the boundary = 70° ÷ 2 = 35°

06 PDF: Parallel Chords and Angles

You can practise the questions and exercises covered in 06: Parallel Chords and Angles using the PDF below.

08: Hexagons and Cyclic Quadrilaterals

Question 1: A regular hexagon is inscribed in a circle of radius r. Find the length of its side and its distance from the centre.

Answer:

The angle subtended by each side at the centre is:

360° ÷ 6 = 60°

The side length of the regular hexagon is r.

The distance of each side from the centre is (√3 ÷ 2)r.

Question 2: Justify the statement: “There is no chord of a circle that is longer than its diameter.”

Answer:

In triangle OCD, by the triangle inequality:

OC + OD > CD

Since OC and OD are radii:

r + r > CD

2r > CD

As 2r is the diameter, the diameter is longer than any chord that does not pass through the centre. A diameter itself is the longest chord of a circle.

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

Answer:

No. The relationship between chord length and its distance from the centre is non-linear.

Chord length = 2√(r² − d²)

Therefore, doubling the distance from the centre does not mean that the chord length will be halved, nor does halving the distance necessarily double the chord length.

Question 4: Prove that the exterior angle at any vertex of a cyclic quadrilateral is equal to the interior opposite angle.

Answer:

Let the opposite interior angle be x, and let the interior angle adjacent to the exterior angle z be y.

Opposite angles of a cyclic quadrilateral are supplementary:

x + y = 180°

The interior angle and its adjacent exterior angle form a linear pair:

y + z = 180°

Subtracting the second equation from the first gives:

x − z = 0°

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

Question 5: Show that a rectangle is the only parallelogram that can be inscribed in a circle.

Answer:

Let ABCD be a cyclic parallelogram.

Opposite angles of a parallelogram are equal:

∠B = ∠D

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

∠B + ∠D = 180°

Substituting ∠D = ∠B:

2∠B = 180°

∠B = 90°

A parallelogram with one right angle has all its angles equal to 90°. Therefore, it is a rectangle. Hence, a rectangle is the only parallelogram that can be inscribed in a circle.

07 PDF: Hexagons and Cyclic Quadrilaterals 

You can practise the questions and exercises covered in 07: Hexagons and Cyclic Quadrilaterals using the PDF below.

Why Choose NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 by PW?

NCERT Solutions for Class 9 Maths Chapter 5 by PW can help you revise circle properties and practise applying theorems to geometry problems.

• Understand key definitions: Revise terms such as radius, chord, diameter, circumcircle, and circumcentre.

• Learn circle theorems: Review relationships involving equal chords, perpendicular distances, and angles subtended by arcs.

• Practise numerical problems: Follow the calculations for finding chord lengths, distances from the centre, and the radius of a circle.

• Improve proof-writing skills: Study the reasoning used to prove results involving cyclic quadrilaterals and parallelograms.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5, I’m Up and Down, and Round and Round, help you revise circle properties, understand key theorems, and practise numerical problems and geometric proofs. Work through the questions and answers to strengthen your understanding of chords, angles, and cyclic quadrilaterals, and build confidence in solving chapter exercises.