Important Questions Class 9 Maths Chapter 5: I’m Up and Down, and Round and Round 2026-2027

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

This page gives Important Questions Class 9 Maths Chapter 5 with full answers. The chapter is about circles. The 12 questions here cover chords, the distance of a chord from the centre, the Baudhāyana–Pythagoras theorem, cyclic quadrilaterals, the angle in a semicircle and the symmetry of regular shapes.

Every answer is solved step by step in simple language. A figure is given with each question, so you can see the circle, the chords and the angles clearly. First try each question yourself, then check your working with the answer.

Q1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 that is "Chords of a circle having the same length are all at the same distance from the centre of the circle" must be true.

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

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

Let OM ⊥ AB and ON ⊥ CD.

We know that the perpendicular from the centre to a chord bisects the chord.

So, AM = MB = AB/2 and CN = ND = CD/2

Since AB = CD, we get: AM = CN

Also, OA = OC = radius of the circle.

Now apply the Baudhāyana–Pythagoras theorem in right triangles OMA and ONC:

In △OMA: OA² = OM² + AM²

In △ONC: OC² = ON² + CN²

Since OA = OC and AM = CN,

So, we have: OM² + AM² = ON² + CN²

Therefore, OM² = ON²

⇒ OM = ON

Hence, the equal chords AB and CD are equidistant from the centre.

Thus, Theorem 6 is proved.

Q2. Consider the given figure. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF. Solve using the Baudhāyana–Pythagoras theorem.

Answer: Given: CE ⊥ AB, CH ⊥ GF and CE = CH.

To Prove: AB = GF

Proof: Since CE is perpendicular to chord AB, it bisects AB.

So, E is the midpoint of AB.

⇒ AE = AB/2

Similarly, since CH is perpendicular to chord GF, it bisects GF.

So, H is the midpoint of GF.

⇒ GH = GF/2

Now apply the Baudhāyana–Pythagoras theorem in right triangles △CEA and △CHG.

In △CEA: CA² = CE² + AE²

In △CHG: CG² = CH² + GH²

But, CA = CG [Radii of same circle]

CE = CH [Given]

Therefore, CE² + AE² = CH² + GH²

⇒ AE² = GH² [Since CE = CH]

Hence, AE = GH

So, 2AE = 2GH

Therefore, AB = GF

Hence proved.

Q3. 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?

Answer: For a regular polygon, rotational symmetry occurs when it is rotated about its centre and fits exactly onto itself.

Thus, a regular n-sided polygon has rotational symmetry of order n and n lines of reflection symmetry.

Square (n = 4): rotational symmetry of order 4 (turns of 90°, 180°, 270° and 360°) and 4 lines of reflection symmetry.

Regular pentagon (n = 5): rotational symmetry of order 5 (turns of 72°, 144°, 216°, 288° and 360°) and 5 lines of reflection symmetry.

Regular hexagon (n = 6): rotational symmetry of order 6 (turns of 60°, 120°, 180°, 240°, 300° and 360°) and 6 lines of reflection symmetry.

Q4. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?

Answer: Join the centre O to the four vertices A, B, C and D.

Since all are radii of the same circle,

OA = OB = OC = OD

Therefore, the triangles △AOB, △BOC, △COD and △DOA are isosceles.

Let the equal base angles be represented as shown in the figure.

Then,

∠A = p + v

∠B = p + q

∠C = q + u

∠D = u + v

Now, the sum of the interior angles of quadrilateral ABCD = ∠A + ∠B + ∠C + ∠D = 360°

Therefore,

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

2p + 2q + 2u + 2v = 360°

2(p + q + u + v) = 360°

Hence,

p + q + u + v = 180° ...(1)

Now, consider the opposite angles A and C:

∠A + ∠C = (p + v) + (q + u)

= p + q + u + v

Using (1),

∠A + ∠C = 180°

Similarly,

∠B + ∠D = (p + q) + (u + v)

= p + q + u + v

= 180°

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

Q5. 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.

Answer: Let BC be the chord through A such that: BC ⊥ OA

Let PQ be any other chord passing through A.

Draw:

OM ⊥ PQ

Since M lies on PQ and A also lies on PQ, triangle OMA is right-angled at M.

Therefore, OA is the hypotenuse of △OMA.

Hence,

OA > OM

Thus, chord BC is farther from the centre O than any other chord PQ passing through A.

We know that the farther a chord is from the centre of a circle, the shorter it is.

Therefore,

BC < PQ

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

Hence proved.

Q6. 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.

Answer: Let the radius be r.

So, OA = OC = r.

For chord AB of length 10 cm:

Half of chord AB = 5 cm

Let its distance from centre OM = d₁

For chord CD of length 24 cm:

Half of chord CD = 12 cm

Let its distance from centre ON = d₂

Since the longer chord is closer to the centre:

d₁ − d₂ = 7 ........(1)

In triangle OAM, using Pythagoras theorem, we have

OA² = OM² + AM²

⇒ r² = d₁² + 5²

⇒ r² = d₁² + 25

In triangle OCN, using Pythagoras theorem, we have

OC² = ON² + CN²

⇒ r² = d₂² + 12²

⇒ r² = d₂² + 144

So, d₁² + 25 = d₂² + 144

⇒ d₁² − d₂² = 119

⇒ (d₁ − d₂)(d₁ + d₂) = 119

⇒ 7(d₁ + d₂) = 119 [Since d₁ − d₂ = 7]

⇒ d₁ + d₂ = 17 .....(2)

Now solving equations (1) and (2), we have

2d₁ = 24

⇒ d₁ = 12

Then d₂ = 5

Now, r² = d₁² + 5²

= 12² + 5²

= 144 + 25

= 169

⇒ r = 13 cm

Therefore, the radius of the circle is 13 cm.

Q7. 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".

Answer: Given: AB = AC

Join OA, OB and OC.

In △AOB and △AOC, we have

OB = OC [Both are radii]

OA = OA [Common]

AB = AC [Given]

Therefore, by SSS congruence

△AOB ≅ △AOC

So, ∠BAO = ∠OAC

Hence, AO bisects ∠BAC.

Therefore, the centre O lies on the angle bisector of ∠BAC.

Q8. 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 the midpoint M of AB.

3. Draw a perpendicular line to AB at M.

4. On this perpendicular, mark a point O such that OM = 3 cm.

5. Join OA or OB.

6. With centre O and radius OA, draw a circle.

Q9. 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.

Answer: Let chords AB and CD intersect at point P inside the circle, and O be the centre.

From the figure, ON ⊥ CD and OM ⊥ AB.

Equal chords ⇒ equal distances from centre

Given: AB = CD .........(1)

To Prove: PB = PD and AP = CP.

So, the perpendicular distances from the centre are equal.

So, OM = ON

In △OPM and △OPN, we have

OP = OP [common side]

OM = ON [Proved above]

∠OMP = ∠ONP = 90° [Perpendiculars]

So, by RHS congruency

△OPM ≅ △OPN

So, PM = PN [CPCT] ................(2)

Now, OM ⊥ AB, so AM = MB [Perpendicular from the centre bisects the chord]

Similarly, ON ⊥ CD, so CN = ND

Since, AB = CD [From (1)]

Therefore, BM = DN [BM = (1/2)AB and DN = (1/2)CD] ............(3)

Adding (2) and (3), we get

PM + BM = PN + DN

⇒ PB = PD ............(4)

Subtracting (4) from (1), we get

AB − PB = CD − PD

⇒ AP = CP

Hence proved.

Q10. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.

Answer: Since AB is the diameter of the circle, A, O and B lie on the same straight line.

Therefore, ∠AOB = 180°

Now, by the theorem:

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

Here, arc AB subtends ∠AOB at the centre O and ∠ACB at point C on the circle.

Therefore, ∠AOB = 2∠ACB

Substituting,

180° = 2∠ACB

∠ACB = 180°/2

∠ACB = 90°

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

Q11. 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.

Answer: Given: Radius of the circle OA = OC = 5 cm

Lengths of the two chords are AB = 6 cm and CD = 8 cm

The perpendiculars from the centre O to the chords AB and CD bisect the chords at M and N respectively.

So, half-lengths are:

For the 6 cm chord: AM = 3 cm

For the 8 cm chord: CN = 4 cm

Let the distances of the chords from the centre be d₁ and d₂.

In triangle OAM, using Pythagoras theorem:

OM² + AM² = OA²

⇒ d₁² + 3² = 5²

⇒ d₁² + 9 = 25

⇒ d₁² = 16

⇒ d₁ = 4 cm

Similarly, in triangle OCN, using Pythagoras theorem:

ON² + CN² = OC²

⇒ d₂² + 4² = 5²

⇒ d₂² + 16 = 25

⇒ d₂² = 9

⇒ d₂ = 3 cm

Since the chords are on opposite sides of the centre and their midpoints M and N lie on the same straight line through O, the distance between their midpoints = d₁ + d₂ = 4 + 3 = 7 cm

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

Q12. 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: Given: △ABC is inscribed in a circle and AB = AC.

Since AB = AC, △ABC is isosceles with base BC.

Let AD be the altitude from A to BC.

Then AD ⊥ BC

In △ABD and △ACD,

AB = AC [Given]

AD = AD [Common to both triangles]

∠ADB = ∠ADC [Each 90°, as AD ⊥ BC]

Therefore, by RHS congruence

△ABD ≅ △ACD

Hence, BD = DC [CPCT]

So, AD is bisector of BC.

Since AD is perpendicular to BC and also bisects BC, AD is the perpendicular bisector of chord BC.

Therefore, AD passes through the centre of the circle.

Hence, the altitude from A to BC passes through the centre of the circle.

Hence proved.

Related Study Material on Extramarks

• Important Questions for Class 9 Maths Chapter 1 The Use of Coordinates
• Important Questions for Class 9 Maths Chapter 2 Introduction to Linear Polynomials
• Important Questions for Class 9 Maths Chapter 3 The World of Numbers
• Important Questions for Class 9 Maths Chapter 4 Exploring Algebraic Identities
• Important Questions for Class 9 Maths Chapter 6 Measuring Space: Perimeter and Area
• Important Questions for Class 9 Maths Chapter 7 Introduction to Probability
• Important Questions for Class 9 Maths Chapter 8 Sequences and Progressions
• Important Questions for Class 9 Maths Chapter 9 Propositions
• Important Questions for Class 9 Maths Chapter 10 Data
• Important Questions for Class 9 Maths Chapter 11 Algorithms
• Important Questions for Class 9 Maths Chapter 12 Quadrilaterals
• Important Questions for Class 9 Maths Chapter 13 Two Variables, One Line
• Important Questions for Class 9 Maths Chapter 14 Math of Space: Surface Area and Volume