Important Questions for Class 9 Maths (Ganita Manjari I) Chapter 5 I'm Up and Down, and Round and Round - 2026-27

Students can practise important questions class 9 Maths Chapter 5 to understand circle properties and geometric theorems. The questions are organised into MCQs, very short, short and long answer types, covering important concepts such as chords, arcs, circumcircles and angles in circles. They can also download the PDF to practise these questions and understand their solutions.

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round MCQ

Q1. How many circles can pass through three non-collinear points?

(a) No circle

(b) Exactly one circle

(c) Two circles

(d) Infinitely many circles

Answer: (b) Exactly one circle

Explanation: Three non-collinear points determine a unique circle. Its centre is the intersection of the perpendicular bisectors of the sides of the triangle formed by these points.

Q2. Where does the circumcentre of a right-angled triangle lie?

(a) Inside the triangle

(b) Outside the triangle

(c) At the midpoint of the hypotenuse

(d) At the midpoint of any side

Answer: (c) At the midpoint of the hypotenuse

Explanation: In a right-angled triangle, the circumcentre lies at the midpoint of the hypotenuse because this point is equidistant from all three vertices.

Q3. Two equal chords of a circle subtend which of the following angles at the centre?

(a) Unequal angles

(b) Equal angles

(c) Right angles only

(d) Straight angles

Answer: (b) Equal angles

Explanation: According to the equal chords theorem, equal chords of the same circle subtend equal angles at its centre.

Class 9 Maths Chapter 5 Very Short Answer Questions

Q4. What is the locus of points that are equidistant from a fixed point in a plane?

Answer: 

The locus is a circle. The fixed point is its centre, and the common distance is its radius.

Q5. What is the longest chord of a circle with a radius of 7 cm?

Answer:

The diameter is the longest chord of a circle.

Diameter = 2 × Radius

= 2 × 7 = 14 cm

Final Answer: 14 cm

Q6. What is the relationship between the perpendicular from the centre of a circle and a chord?

Answer: 

The perpendicular drawn from the centre of a circle to a chord bisects the chord into two equal parts.

Class 9 Maths Chapter 5 Short Answer Questions

Q7. A chord of a circle is 16 cm long, and the radius is 10 cm. Find the perpendicular distance of the chord from the centre.

Solution:

Let O be the centre, AB the chord and M its midpoint.

Given:

AB = 16 cm

OA = 10 cm

Since the perpendicular from the centre bisects the chord:

AM = 16 ÷ 2 = 8 cm

Using the Baudhāyana–Pythagoras theorem in right triangle OMA:

$$OA^2=OM^2+AM^2$$

$$10^2=OM^2+8^2$$

$$100=OM^2+64$$

$$OM^2=36$$

$$OM=6$$

Final Answer: The perpendicular distance is 6 cm.

Q8. In a circle with centre O, chord AB subtends an angle of 120° at the centre. Find the angle subtended by the same minor arc AB at a point P on the major arc.

Solution:

Given:

∠AOB = 120°

We know that the angle subtended by an arc at the centre is twice the angle subtended at a point on the circle outside that arc.

Therefore:

$$\angle APB=\frac{1}{2}\times\angle AOB$$

$$=\frac{1}{2}\times120^\circ$$

$$=60^\circ$$

Final Answer: ∠APB = 60°

Q9. Two chords AB and CD of a circle are equidistant from its centre. If AB = 12 cm, find CD and explain why.

Solution:

Given:

AB = 12 cm

Both chords are at equal perpendicular distances from the centre.

According to the theorem, chords of a circle that are equidistant from the centre are equal in length.

Therefore:

CD = AB = 12 cm

Final Answer: CD = 12 cm

Class 9 Maths Chapter 5 Long Answer Questions

Q10. Prove that the perpendicular drawn from the centre of a circle to a chord bisects the chord.

Solution:

Given: A circle with centre O and chord AB. OM is perpendicular to AB.

To Prove: AM = MB

Step 1: Join OA and OB.

OA = OB because both are radii of the same circle.

Step 2: Consider triangles OMA and OMB.

OA = OB (radii)

OM = OM (common side)

∠OMA = ∠OMB = 90°

Step 3: By the RHS congruence rule:

$$\triangle OMA\cong\triangle OMB$$

Step 4: By corresponding parts of congruent triangles:

AM = MB

Final Answer: The perpendicular from the centre of a circle to a chord bisects the chord. Hence proved.

Q11. Two parallel chords of lengths 6 cm and 8 cm lie on opposite sides of the centre of a circle of radius 5 cm. Find the distance between their midpoints.

Solution:

Step 1: Write the given information.

Radius = 5 cm

First chord = 6 cm

Second chord = 8 cm

Let O be the centre and M and N be the midpoints of the chords.

Step 2: Find the distance OM.

Half of the first chord = 6 ÷ 2 = 3 cm

Using the Baudhāyana–Pythagoras theorem:

$$OM^2+3^2=5^2$$

$$OM^2=25-9=16$$

$$OM=4$$ cm

Step 3: Find the distance ON.

Half of the second chord = 8 ÷ 2 = 4 cm

$$ON^2+4^2=5^2$$

$$ON^2=25-16=9$$

$$ON=3$$ cm

Step 4: Find the distance between the midpoints.

Since the chords are parallel and on opposite sides of the centre, their midpoints and the centre lie on the same perpendicular line.

MN = OM + ON

MN = 4 + 3

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

Q12. Prove that equal chords of a circle subtend equal angles at the centre.

Solution:

Given: A circle with centre O. AB and CD are two equal chords.

AB = CD

To Prove: ∠AOB = ∠COD

Step 1: Join OA, OB, OC and OD.

Since all are radii of the same circle:

OA = OC

OB = OD

Step 2: Consider triangles AOB and COD.

We have:

OA = OC (radii)

OB = OD (radii)

AB = CD (given)

Step 3: By the SSS congruence rule:

$$\triangle AOB\cong\triangle COD$$

Step 4: By corresponding parts of congruent triangles:

∠AOB = ∠COD

Final Answer: Equal chords of a circle subtend equal angles at its centre. Hence proved.

How to Solve Circle Geometry Questions in Class 9 Maths Chapter 5

While practising Chapter 5 Maths Class 9 Important Questions, students should focus on understanding the given diagram and identifying the correct circle property. The following methods can help students solve different types of geometry questions.

Quick Tip: Always draw a clear circle diagram, label the centre and mark the given angles or lengths before starting a solution. This helps you identify the correct theorem and avoid mistakes.

How Vedantu Helps Students Practise Class 9 Maths Chapter 5 Important Questions

Vedantu provides NCERT Class 9 Maths Chapter 5 important questions based on the concepts covered in I'm Up and Down and Round and Round. These questions help students understand circle properties, apply geometric theorems and practise different types of mathematical problems.

• Chapter-focused questions: Practise problems involving chords, arcs, circumcircles, perpendicular bisectors and concyclic points.

• Clear explanations: Understand the reasoning behind circle properties and geometric proofs.

• Different question types: Prepare for MCQs, very short, short and long answer questions.

• Diagram-based practice: Learn how to use circle diagrams to identify angles, chords and their relationships.

• Exam preparation: Improve accuracy and confidence in solving geometry questions through regular practice.

Students can also explore Class 9 Important Questions on Vedantu to practise questions from other subjects and strengthen their exam preparation.

Important Study Materials for Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round

CBSE Class 9 Maths Chapter-wise Important Questions

Other Important Related Links for CBSE Class 9 Maths