CLASS 9 MATHEMATICS

I’m Up and Down, and
Round and Round

Chapter 5

Complete Page-wise Questions and Solutions

Ganita Manjari | Grade 9 | Part I

Source-based educational material for Class 9 students

This document includes the chapter’s introduction questions, activities, Think and Reflect tasks, all exercise sets, theorem-based prompts, and all 26 end-of-chapter exercises. Construction measurements may vary slightly.

Page 92 - Introduction

Question. Can you recognise the origin of the shapes in Fig. 5.1?

Answer: The first circular pattern comes from the cross-section of a plant stem or tree trunk. The second comes from the circular inflorescence of a sunflower. Both show how geometrical shapes occur in nature.

Page 93 - Introduction and Definitions

Question. What properties are common to all circles, big and small?

Answer: Every circle has a fixed point called its centre. All points lying on the circle are at the same distance from this centre. This fixed distance is called the radius of the circle.

Activity. List some objects from nature that resemble a circle.

Answer: Examples include:

1. The full Moon

2. The Sun

3. Circular ripples in water

4. A sunflower

5. The cross-section of a tree trunk or plant stem

6. The pupil of an eye

Think and Reflect. Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. Can you guess what Amina told her?

Answer: Amina probably suggested this method:

1. Fold the circular paper so that its two boundaries overlap exactly.

2. Open it. The crease formed is a diameter.

3. Fold it again in another direction and open it.

4. The point where the two diameter creases intersect is the centre.

Page 94 - Symmetries of a Circle

In-text Question. Can you tell whether the two points of a rotating wheel touching the ground at different times are the same point?

Answer: No. A circle looks the same after rotation through any angle. Therefore, without marking the wheel, we cannot determine whether the point touching the ground is the same or a different point.

In-text Question. Does a crease formed by folding a circular paper so that its boundaries overlap pass through the centre?

Answer: Yes. The crease passes through the centre and forms a diameter. Every diameter is a line of reflection symmetry of the circle.

Think and Reflect 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?

Answer: The symmetries are shown below. A rotation through 360° is the identity rotation.

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

Answer: The diameter is the longest chord of a circle.

Longest chord = 2r = 2 × 5 = 10 units

A chord joining two distinct points can be made arbitrarily small, so there is no smallest positive chord. If a degenerate chord with coinciding endpoints is allowed, its length is zero.

Think and Reflect 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?

Answer: The locus of points equidistant from two given points A and B is the perpendicular bisector of AB.

Let M be the midpoint of AB and P be any point on its perpendicular bisector. In right triangles PMA and PMB, AM = BM, PM is common, and both angles at M are 90°. By RHS congruence, PA = PB. Thus, every point on the perpendicular bisector is equidistant from A and B.

Page 95 - How Many Circles?

In-text Question. Given two points A and B on a plane, how many circles pass through them?

Answer: Infinitely many circles can pass through two given points. Their centres lie on the perpendicular bisector of AB, because every point on this line is equidistant from A and B.

Think and Reflect 1. How many circles pass through two points on a plane?

Answer: Infinitely many circles pass through two distinct points.

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

Answer: Circles of radius less than AB/2 cannot pass through both points. The smallest circle has the midpoint of AB as its centre and AB as its diameter.

r(minimum) = AB/2

There is no largest circle. As the centre moves farther away along the perpendicular bisector, the radius can increase without limit.

Think and Reflect 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?

Answer: The radii increase because the distance from the centre to both A and B becomes greater.

Think and Reflect 4. As you go along the perpendicular bisector, will the circle drawn through A and B appear more curved or less curved?

Answer: It will appear less curved because its radius increases. A circle with a larger radius has less curvature.

Think and Reflect 5. How many squares can be drawn with A and B on the boundary? How many squares can be drawn with A and B as corners?

Answer: Infinitely many squares can be drawn with A and B merely lying somewhere on their boundaries. Three squares can be drawn with A and B as vertices: two with AB as a side, one on each side of AB, and one with AB as a diagonal.

Pages 96-97 - Circles Through Three Points

In-text Question. How many circles can be drawn through three distinct points A, B and C? Is there always at least one such circle?

Answer: If A, B and C are collinear, no circle can pass through all three. If they are non-collinear, exactly one circle passes through them. Its centre is the unique intersection of the perpendicular bisectors of the segments joining the points.

Question. Why is there no circle passing through three distinct collinear points?

Answer: The perpendicular bisectors of the segments joining three collinear points are all perpendicular to the same line. Therefore, they are parallel and cannot meet at a common centre. Also, a straight line can intersect a circle at a maximum of two distinct points.

Question. Can more than one circle pass through three non-collinear points?

Answer: No. The perpendicular bisectors of two sides of the triangle formed by the points intersect at only one point. Therefore, there is only one possible centre and only one circle.

Page 98 - Exercise Set 5.1

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: Construction:

1. Draw AB = 5 cm.

2. At A, construct a ray making an angle of 70° with AB.

3. At B, construct a ray making an angle of 60° with BA. Let the rays meet at C.

4. Draw the perpendicular bisectors of any two sides and mark their intersection O.

5. With centre O and radius OA, draw the circumcircle.

∠C = 180° − 70° − 60° = 50°

All three angles are acute. Therefore, the circumcentre O lies inside the triangle.

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

Answer: Construction:

1. Draw AB = 5 cm.

2. At A, construct a ray making an angle of 100° with AB.

3. On this ray, mark C so that AC = 4 cm, and join BC.

4. Draw the perpendicular bisectors of any two sides and mark their intersection O.

5. With centre O and radius OA, draw the circumcircle.

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

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

Answer: Construction:

1. Draw AB = 6 cm.

2. With A and B as centres and radius 7 cm, draw arcs intersecting at C.

3. Join AC and BC.

4. Draw the perpendicular bisectors of two sides and mark their intersection O.

5. Draw the circumcircle with centre O and radius OA.

OA = OB = OC ≈ 3.9 cm

The three measurements are equal because O is the circumcentre. Slight variation may occur due to construction accuracy.

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

Answer: The smallest circle has AB as its diameter and the midpoint of AB as its centre.

r(minimum) = AB/2

Page 98 - Think, Draw and Infer

Question 1. A, B and C are three collinear points. Can you find P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Can a circle pass through three collinear points? Can a line cut a circle in three distinct points?

Answer: No point P can be equidistant from three distinct collinear points. The perpendicular bisectors of AB and BC are both perpendicular to the same line, so they are parallel and do not meet. Hence, no circle passes through three distinct collinear points, and a line can intersect a circle at a maximum of two points.

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

Answer: Yes. Rotating the triangle about the centre keeps its vertices on the same circle and produces congruent copies. Reflections in diameters also produce congruent triangles. Therefore, infinitely many congruent copies can share the circumcircle.

Pages 98-99 - In-text Questions on Chords

Question. After the wheel is rotated, what is common to the first and second positions of the thread?

Answer: The thread forms a chord in both positions. Since its length does not change, the two chords have equal lengths.

Question. What happens to the angle subtended by the chord at the centre after rotation?

Answer: The angle remains unchanged because the chord rotates rigidly about the centre. Thus, equal chords subtend equal angles at the centre.

Question. If two chords subtend equal angles at the centre, are the chords equal?

Answer: Yes. The triangles formed by joining their endpoints to the centre are congruent by SAS. Therefore, the corresponding chords are equal.

Page 100 - Exercise Set 5.2

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. Since OA and OB are radii, OA = OB. Therefore, ΔOAB has two equal sides and is isosceles.

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

Answer: Let AB and CD be equal chords. In triangles OAB and OCD, OA = OC and OB = OD because they are radii, while AB = CD is given. Therefore, by SSS congruence:

ΔOAB ≅ ΔOCD

Page 101 - Exercise Set 5.3

Question 1. Why does the perpendicular from the centre of a circle to a chord bisect the chord?

Answer: Let CM ⟂ AB. In right triangles CMA and CMB, CA = CB because they are radii, CM is common, and both angles at M are 90°. By RHS congruence, ΔCMA ≅ ΔCMB. Hence, AM = BM, so the perpendicular 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.

Answer: In an isosceles triangle, the altitude from A to base BC also bisects BC. It is therefore the perpendicular bisector of chord BC. The perpendicular bisector of every chord passes through the centre, so the altitude from A passes through the circle’s centre.

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

Answer: For the 6 cm chord:

d₁ = √(5² − 3²) = √16 = 4 cm

For the 8 cm chord:

d₂ = √(5² − 4²) = √9 = 3 cm

The chords are on opposite sides, so the distances are added:

Distance between midpoints = 4 + 3 = 7 cm

Page 102 - Activity on the Distance of Chords

Activity. Fold a circular paper to form a chord. Fold it again so that the endpoints of the chord meet. What do you observe?

Answer: The second crease passes through the centre, bisects the first crease or chord, and meets it at 90°. Therefore, it is the perpendicular bisector of the chord.

Question. How can another chord of the same length be drawn?

Answer: Set a compass equal to the first chord’s length. Choose another point on the circle and use the same opening to mark the second endpoint. The chord may also be copied on tracing paper and rotated about the centre.

Question. Is the new chord’s distance from the centre equal to that of the first chord?

Answer: Yes. Equal chords of a circle are equidistant from the centre.

Page 103 - In-text Reasoning

Question. Is checking several examples enough to prove that all equal chords are equidistant from the centre?

Answer: No. Examples can suggest a result but cannot prove that it is always true. A general geometrical proof using congruent triangles or the Baudhāyana-Pythagoras theorem is required.

Question. Why are equal chords equidistant from the centre?

Answer: Let equal chords AB and FG have perpendicular distances CE and CH from centre C. The perpendiculars bisect both chords, so AE = FH. Also, CA = CF are radii, and both triangles are right-angled. By RHS congruence, ΔCEA ≅ ΔCHF, so CE = CH.

Page 104 - Exercise Set 5.4

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

Answer: Let equal chords AB and FG be at distances CE and CH. Since the perpendiculars bisect the chords, AE = AB/2 and FH = FG/2. Thus, AE = FH. Using the Baudhāyana-Pythagoras theorem:

CE² = CA² − AE²

CH² = CF² − FH²

Since CA = CF and AE = FH, CE² = CH² and therefore CE = CH.

Question 2. In Fig. 5.15, CE ⟂ AB, CH ⟂ GF and CE = CH. Show that AB = GF.

Answer: The perpendiculars from the centre bisect the chords. Thus, AE = EB and GH = HF. In right triangles CEA and CHF, CA = CF are radii and CE = CH is given. By RHS congruence, AE = HF. Therefore, AB = 2AE and GF = 2HF, giving AB = GF.

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

Answer: In right triangles CEA and CHF:

CA² = CE² + EA²

CF² = CH² + HF²

Because CA = CF and CE = CH, EA² = HF², so EA = HF. Since the perpendiculars bisect the chords, AB = 2EA and GF = 2HF. Hence, AB = GF.

Pages 104-105 - Activity on Unequal Chords

Activity. Draw chords of various lengths and record their distances from the centre. What do you observe?

Answer: For example, in a circle of radius 5 cm:

The longer the chord, the closer it lies to the centre. Therefore, of two unequal chords, the shorter chord is farther from the centre.

Pages 105-106 - Exercise Set 5.5

Question 1. Find the length of a chord when the radius is 7 cm and the perpendicular distance from the centre is 6 cm.

Answer: Let M be the midpoint of chord AB. Then OA = 7 cm and OM = 6 cm.

AM² = OA² − OM² = 7² − 6² = 13

AB = 2AM = 2√13 cm ≈ 7.21 cm

Question 2. Explain why, if a chord is at distance d from the centre of a circle of radius r, its length is 2√(r² − d²).

Answer: Let OM ⟂ AB. Then M is the midpoint of AB. In right triangle OMA:

r² = d² + AM²

AM = √(r² − d²)

AB = 2AM = 2√(r² − d²)

Question 3. If the distance of chord AB from the centre is twice the distance of chord CD, can we conclude that CD = 2AB? Give reasons.

Answer: No. Chord length is not inversely proportional to its distance from the centre. If CD is at distance d, then AB is at distance 2d:

CD = 2√(r² − d²), AB = 2√(r² − 4d²)

For example, let r = 5 cm and d = 2 cm. Then CD = 2√21 cm, while AB = 6 cm. Clearly, CD ≠ 2AB.

Page 107 - Angles Subtended by an Arc

Exercise. Measure the angles subtended by arcs AKB and CLD at centre O in Fig. 5.19. State whether they are minor or major arcs.

Answer: Measurements from the printed figure are approximately:

Angle subtended by arc AKB: ∠AOB ≈ 100°

Since it is less than 180°, arc AKB is a minor arc.

Angle subtended by arc CLD: reflex ∠COD ≈ 204°

Since it is greater than 180°, arc CLD is a major arc. Measurements may differ slightly depending on the printed copy and protractor placement.

Activity. Draw a circle and chord AB. Fix arc AKB. Measure the angle subtended at the centre and at points P, Q and R on the circle outside the arc. What do you notice?

Answer: The angles at P, Q and R are equal, and each is half the angle at the centre:

∠APB = ∠AQB = ∠ARB = ½∠AOB

Thus, angles subtended by the same arc in the same segment are equal.

Pages 108-109 - Theorem 9

Question. Why is the angle subtended by an arc at the centre twice the angle subtended at a point on the circle outside the arc?

Answer: Let D be a point on the circle outside arc AFB. Join D to centre C and extend it to meet the circle at E. Since CB = CD, triangle BCD is isosceles, and the exterior-angle theorem gives ∠BCE = 2∠BDC. Similarly, since CA = CD, ∠ACE = 2∠ADC. Depending on E’s position, these angles are added or subtracted. In both cases:

∠BCA = 2∠BDA

Hence, the angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining circle.

Page 110 - Corollary

Question. Why is the angle subtended by a diameter at any point on the circle 90°?

Answer: A diameter divides the circle into two semicircles. The angle subtended by a semicircle at the centre is 180°. The angle at a point on the circle is half the central angle:

Angle in a semicircle = ½ × 180° = 90°

Pages 110-111 - Exercise Set 5.6

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

Answer: In triangle AOB, OA = OB = 12 cm and ∠AOB = 60°. The other two equal angles together measure 120°, so each is 60°. Therefore, triangle AOB is equilateral.

AB = 12 cm

Question 2(i). Are there points X and Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?

Answer: No. X and Y are in the same segment of chord AB. Angles subtended by the same chord in the same segment are equal, so ∠AXB = ∠AYB.

Question 2(ii). Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of AB?

Answer: Not always. Normally, angles formed on opposite sides of chord AB are supplementary. However, if AB is a diameter, each angle is 90°. Thus, points on opposite sides may also form equal angles, so equality alone does not prove that X and Y lie on the same side.

Question 2(iii). If ∠AXB = ∠AYB and X and Y do not lie on the given circle, does the circle through A, B and X also pass through Y?

Answer: Not necessarily. The concyclicity theorem also requires X and Y to lie on the same side of AB. If they are on the same side, A, B, X and Y are concyclic. If they are on opposite sides, the conclusion does not necessarily follow.

Question 3. Find x in Fig. 5.26.

Answer: ABCD is a cyclic quadrilateral, so its opposite angles are supplementary:

x + 100° = 180°

x = 80°

Pages 111-112 - Concyclicity of Points

Question. Why do four points A, B, C and D lie on a circle if ∠ACB = ∠ADB, where C and D lie on the same side of AB?

Answer: A unique circle can be drawn through non-collinear points A, B and C. Suppose D does not lie on it. If D lies outside, line AD meets the circle at E. Since C and E lie in the same segment, ∠ACB = ∠AEB. But ∠AEB is an exterior angle of triangle BED, so ∠AEB > ∠ADB, contradicting the given equality. A similar contradiction occurs if D lies inside. Therefore, D lies on the circle through A, B and C.

Pages 112-113 - Cyclic Quadrilaterals

Question. Why is the sum of two opposite angles of a cyclic quadrilateral 180°?

Answer: In cyclic quadrilateral ABCD, ∠BAD is half the central angle subtended by arc BCD, while ∠BCD is half the central angle subtended by the remaining arc BAD. Together, the two arcs form 360°. Therefore:

∠BAD + ∠BCD = ½ × 360° = 180°

Similarly, ∠ABC + ∠ADC = 180°

Exercise. A cyclic quadrilateral has ∠A = 80°, ∠B = 110°, ∠C = 100° and ∠D = 70°. Can such a quadrilateral be drawn? Explain.

Answer: Yes. Both pairs of opposite angles are supplementary:

∠A + ∠C = 80° + 100° = 180°

∠B + ∠D = 110° + 70° = 180°

Therefore, such a cyclic quadrilateral can be drawn.

Question. Why is a quadrilateral cyclic if two opposite angles add to 180°?

Answer: Draw the circle through three vertices A, B and D. Suppose C does not lie on it, and let CD meet the circle at E. Since ABED is cyclic, ∠BAD + ∠BED = 180°. The given condition gives ∠BAD + ∠BCD = 180°, so ∠BED = ∠BCD. This is impossible if C lies inside or outside the circle because one of these angles becomes an exterior angle and must be greater. Thus, C lies on the circle, and the quadrilateral is cyclic.

Pages 114-116 - End-of-Chapter Exercises

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

Answer: The perpendicular from the centre bisects the chord.

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

Chord length = 2 × 12 = 24 cm

Question 2. An arc subtends an angle of 70° at the centre. What angle does it subtend at a point on the circle?

Answer: The angle at the circle is half the central angle:

½ × 70° = 35°

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

Answer: The radius is 13 cm and the half-chord is 12 cm.

d² + 12² = 13²

d = √(169 − 144) = 5 cm

Question 4. A circle has radius 15 cm. A chord is 9 cm from the centre. What is its length?

Answer: The perpendicular from the centre bisects the chord.

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

Chord length = 2 × 12 = 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 O the centre. OA = OB because both are radii, so O is equidistant from A and B. The locus of points equidistant from the endpoints of a segment is its perpendicular bisector. Therefore, O lies on the perpendicular bisector of AB.

Question 6. Diameter AB and point C lie on a circle. What is ∠ACB? Explain.

Answer: Diameter AB subtends 180° at the centre. The angle at a point on the circle is half of this:

∠ACB = ½ × 180° = 90°

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

Answer: Opposite angles of a cyclic quadrilateral are supplementary:

∠C = 180° − 75° = 105°

∠D = 180° − 110° = 70°

Question 8. PQRS is cyclic. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find x, ∠P and ∠R.

Answer: Opposite angles are supplementary:

(2x + 10) + (3x − 20) = 180

5x − 10 = 180 ⇒ x = 38

∠P = 2(38) + 10 = 86°

∠R = 3(38) − 20 = 94°

Question 9. A chord of length 16 cm is 6 cm from the centre. Find the radius.

Answer: The half-chord is 8 cm. By the Baudhāyana-Pythagoras theorem:

r² = 8² + 6² = 64 + 36 = 100

r = 10 cm

Question 10. A cyclic quadrilateral has consecutive side lengths 5, 5, 12 and 12 units. Find its area.

Answer: Let AB = BC = 5 and CD = DA = 12. The cyclic quadrilateral is a kite, so opposite angles A and C are equal. They are also supplementary, hence each is 90°. Diagonal BD divides the quadrilateral into two right triangles, each with perpendicular sides 5 and 12 units.

Area of each triangle = ½ × 5 × 12 = 30 square units

Total area = 30 + 30 = 60 square units

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

Answer: Draw the perpendicular bisectors of any two sides or diagonals. Their intersection is the circumcentre. If it lies inside the quadrilateral, the centre is inside; otherwise, it is outside. Using two long chords or diagonals usually gives a more accurate construction.

Question 12. Two equal chords intersect. Show that the segments of one chord are equal to the corresponding segments of the other chord.

Answer: Let equal chords AB and CD intersect at P. Draw OM ⟂ AB and ON ⟂ CD. Equal chords are equidistant from the centre, so OM = ON. Right triangles OMP and ONP are congruent by RHS, so OP bisects the angle between the chords. Reflection in OP maps one chord onto the other and preserves lengths. Therefore, after pairing corresponding endpoints:

PA = PC, PB = PD

The correspondence may be reversed depending on the endpoint labels.

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

Answer: Construction:

1. Draw AB = 6 cm.

2. Construct its midpoint M, so AM = BM = 3 cm.

3. At M, draw a perpendicular to AB.

4. Mark O on this perpendicular so that OM = 3 cm.

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

OA = √(OM² + AM²) = √(3² + 3²) = 3√2 cm

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

Answer: In a parallelogram, opposite angles are equal. In a cyclic quadrilateral, opposite angles are supplementary. Thus, if ∠A = ∠C and ∠A + ∠C = 180°, each equals 90°. Similarly, all four angles are right angles, so the parallelogram is a rectangle. Conversely, every rectangle is cyclic because each pair of opposite angles adds to 180°.

Question 15. Show that if a rectangle is inscribed in a circle, the intersection of its diagonals is the centre.

Answer: Let diagonals AC and BD meet at P. The diagonals of a rectangle bisect each other, so PA = PC and PB = PD. They are also equal, AC = BD, so their halves are equal. Therefore, PA = PB = PC = PD. Thus, P is equidistant from all four vertices and is the circle’s centre.

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

Answer: All equal chords lie at the same perpendicular distance d from the centre. Their midpoints are the feet of these perpendiculars, so all midpoints lie on a circle concentric with the original circle.

Radius of midpoint circle: d = √(r² − (L/2)²)

For diameter-length chords, all midpoints coincide at the centre.

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

Answer: In triangles AOB and AOC, AB = AC is given, OB = OC are radii, and OA is common. By SSS congruence, ΔAOB ≅ ΔAOC. Therefore, ∠BAO = ∠OAC, so AO bisects ∠BAC.

Question 18. Parallel chords of lengths 10 cm and 24 cm are on the same side of the centre. Their separation is 7 cm. Find the radius.

Answer: The 24 cm chord is nearer the centre. Let its distance be x; the 10 cm chord is then x + 7 away.

r² = x² + 12²

r² = (x + 7)² + 5²

x² + 144 = x² + 14x + 74

14x = 70 ⇒ x = 5

r² = 5² + 12² = 169 ⇒ r = 13 cm

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

Answer: The six equal central angles measure 360°/6 = 60°. The triangle formed by two radii and one side is equilateral, so the side length equals r.

Side length = r

Distance = √(r² − (r/2)²) = (√3/2)r

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

Answer: Both angles subtend the same chord MP. Since O and N lie in the same segment of MP, ∠MOP = ∠MNP. Also, because MN is a diameter, the angles directly subtended by MN at O and P are right angles: ∠MON = ∠MPN = 90°.

Question 21. Explain why the exterior angle at any vertex of a cyclic quadrilateral equals its interior opposite angle.

Answer: Extend CD to E. Since ABCD is cyclic, ∠ABC + ∠ADC = 180°. Since ∠ADC and exterior angle ∠CDE form a linear pair, ∠ADC + ∠CDE = 180°. Therefore:

∠CDE = ∠ABC

Question 22. Justify: No chord of a circle is longer than its diameter.

Answer: For a chord of length L at distance d from the centre:

L = 2√(r² − d²)

Since d² ≥ 0, r² − d² ≤ r² and therefore L ≤ 2r. Equality occurs only when d = 0, meaning the chord passes through the centre and is a diameter.

Question 23. Let A be inside a circle with centre O. Show that the shortest chord through A is perpendicular to OA.

Answer: For any chord through A, let OM be its perpendicular distance from O. In right triangle OMA, OM ≤ OA. The greatest possible distance from O is therefore OA, achieved when M = A and the chord is perpendicular to OA. Since a chord farther from the centre is shorter, this perpendicular chord is the shortest through A.

Question 24. Use Fig. 5.30 to justify that the angle in a semicircle is 90°.

Answer: Let the angles at the diameter’s endpoints be a and b. The radii form two isosceles triangles, so the angle at the point on the semicircle is a + b. In the large triangle:

a + b + (a + b) = 180°

2(a + b) = 180° ⇒ a + b = 90°

Question 25. Chords CC′ and DD′ are perpendicular to diameter AB. Prove that MM′, joining the midpoints of CD and C′D′, is perpendicular to AB.

Answer: Because CC′ ⟂ AB, diameter AB bisects chord CC′, so C and C′ are reflections across AB. Similarly, D and D′ are reflections across AB. Reflection across AB maps segment CD onto C′D′ and maps midpoint M onto midpoint M′. Hence, AB is the perpendicular bisector of MM′, so MM′ ⟂ AB.

Question 26. Use Fig. 5.31 to justify that the sum of opposite angles of a cyclic quadrilateral is 180°.

Answer: Since OA = OB = OC = OD, the four triangles around O are isosceles. With the labels in the figure, ∠A = p + v and ∠C = q + u. The four central angles are 180° − 2p, 180° − 2q, 180° − 2u and 180° − 2v. Their sum is 360°:

720° − 2(p + q + u + v) = 360°

p + q + u + v = 180°

∠A + ∠C = (p + v) + (q + u) = 180°

Similarly, ∠B + ∠D = 180°. Thus, opposite angles are supplementary.