Class 9 Maths Ganita Manjari Chapter 5 Solutions (NCERT 2026–27) – I’m Up and Down, and Round and Round

These Class 9 Maths Ganita Manjari Chapter 5 solutions cover I’m Up and Down, and Round and Round (Circles) from the new NCF-2023 textbook (2026–27). Every exercise is solved step by step — with constructions, proofs and chord calculations — so you can understand each circle property and revise the whole chapter quickly.

Class: 9 Subject: Mathematics Book: Ganita Manjari (Part 1) Chapter: 5 Exercises: 5.1–5.6, End-of-Chapter Session: 2026–27
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Chapter 5 Overview

Chapter 5 of Ganita Manjari, I’m Up and Down, and Round and Round, is the circles chapter. It builds the key properties through constructions and proofs: the circumcircle of a triangle, chords and the angles they subtend at the centre, the perpendicular from the centre to a chord, the distance of chords from the centre, the angle subtended by an arc (centre angle = twice the angle on the circle), the angle in a semicircle, and cyclic quadrilaterals. The Class 9 Maths Ganita Manjari Chapter 5 solutions below work through every exercise step by step. Where a question depends on a textbook figure, the method and every value that follows from the given data are shown.

Key Concepts & Theorems

Circumcircle: the unique circle through the three vertices of a triangle; its centre (circumcentre) is where the perpendicular bisectors of the sides meet.

Chord properties: equal chords subtend equal angles at the centre (and conversely); equal chords are equidistant from the centre (and conversely); the longer chord is nearer the centre.

Perpendicular from centre: the line from the centre to the midpoint of a chord is perpendicular to it, and the perpendicular from the centre bisects the chord.

Angle at centre = 2 × angle on circle (same arc); all angles in the same segment are equal.

Angle in a semicircle = 90°.

Cyclic quadrilateral: a 4-gon whose vertices lie on a circle; opposite angles add up to 180° (and conversely).

Key Facts & Formulas

Chord length: for radius r and perpendicular distance d from the centre, chord = 2√(r2 − d2).

Distance from centre: d = √(r2 − (chord/2)2).

Diameter is the longest chord; the angle it subtends on the circle is 90°.

Arc angle: central angle = 2 × inscribed angle on the same arc.

Cyclic quadrilateral: opposite angles sum to 180°; an exterior angle = the interior opposite angle.

Exercise Set 5.1

1. Draw ΔABC with AB = 5 cm, ∡A = 70° and ∡B = 60°. Draw the circumcircle. Is the centre inside or outside the triangle?

SOLUTION Construct the triangle, then draw the perpendicular bisectors of any two sides; they meet at the circumcentre O. Draw the circle with centre O through A. Here ∡C = 180° − 70° − 60° = 50°, so all three angles are acute. For an acute triangle the circumcentre lies inside the triangle.

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

SOLUTION Construct the triangle and the perpendicular bisectors to locate O. Since ∡A = 100° is obtuse, the circumcentre lies outside the triangle (opposite the obtuse angle).

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

SOLUTION O is equidistant from the three vertices, so OA = OB = OC = R (the circumradius). By calculation: area (Heron, s = 10) = √(10·4·3·3) = 6√10 ≈ 18.97 cm2; R = (abc)/(4×area) = (6·7·7)/(4·6√10) ≈ 3.9 cm. The triangle is acute, so O is inside.

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

SOLUTION The smallest circle through A and B has AB as a diameter, so the least radius is AB/2.
Think, Draw and Infer A, B, C are three collinear points. Can you find P with PA = PB = PC? What about the perpendicular bisectors of AB and BC? Answer. No such P exists. The perpendicular bisectors of AB and BC are both perpendicular to the same line, so they are parallel and never meet — hence no point is equidistant from all three, and no circle can pass through three collinear points. A line can cut a circle in at most two points.

Exercise Set 5.2

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

SOLUTION Let AB be a chord and C the centre. Then CA and CB are both radii, so CA = CB. A triangle with two equal sides is isosceles. ✓

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

SOLUTION Both triangles have the two equal sides = radius r, and equal bases (given). By SSS congruence (r, r, equal base), the two triangles are congruent. ✓

Exercise Set 5.3

1. Explain why the converse of Theorem 4 is true: the perpendicular from the centre to a chord bisects the chord.

SOLUTION In ΔCMA and ΔCMB: CA = CB = r, CM is common, and ∡CMA = ∡CMB = 90°. By RHS congruence, ΔCMA ≅ ΔCMB, so AM = BM — the foot M is the midpoint, i.e. the perpendicular bisects the chord. ✓

2. An isosceles triangle ABC (AB = AC) is inscribed in a circle. Show that the altitude from A to BC passes through the centre.

SOLUTION Since AB = AC, A lies on the perpendicular bisector of BC. The centre O is also equidistant from B and C, so O lies on the perpendicular bisector of BC. The altitude from A to BC is that same perpendicular bisector, so it passes through O. ✓

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 the midpoints of the chords.

SOLUTION For the 6 cm chord: half = 3, distance = √(52 − 32) = √16 = 4 cm. For the 8 cm chord: half = 4, distance = √(52 − 42) = √9 = 3 cm. On opposite sides, the distance between the midpoints = 4 + 3 = 7 cm.

Exercise Set 5.4

1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 (equal chords are equidistant from the centre) is true.

SOLUTION If a chord has length 2ℓ and is at distance d from centre O (radius r), then r2 = d2 + ℓ2, so d = √(r2 − ℓ2). Two chords of equal length have the same ℓ, hence the same d — they are equidistant from the centre. ✓

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

SOLUTION CE and CH bisect AB and GF (perpendicular from centre). In ΔCEA and ΔCHG: CA = CG = r, CE = CH (given), ∡CEA = ∡CHG = 90°. By RHS, the triangles are congruent, so AE = GH; doubling gives AB = GF. ✓

3. Solve the previous question using the Baudhāyana–Pythagoras theorem.

SOLUTION AE = √(r2 − CE2) and GH = √(r2 − CH2). Since CE = CH, AE = GH, so AB = 2AE = 2GH = GF. ✓

Exercise Set 5.5

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

SOLUTION Chord = 2√(r2 − d2) = 2√(49 − 36) = 2√13 ≈ 7.21 cm.

2. Explain why: 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 The perpendicular from the centre bisects the chord. Half the chord, d and r form a right triangle with hypotenuse r, so (chord/2)2 + d2 = r2 ⇒ chord/2 = √(r2 − d2) ⇒ chord = 2√(r2 − d2). ✓

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

SOLUTION No. Chord length = 2√(r2 − d2) is not proportional to d. For example with r = 5: at d = 4 the chord is 6; at d = 2 the chord is 2√21 ≈ 9.17 — not 3 (= 6/2). So doubling the distance does not halve (or double) the chord.

Exercise Set 5.6

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

SOLUTION ΔOAB has OA = OB = 12 and ∡AOB = 60°, so the base angles are each 60° — the triangle is equilateral. Hence AB = 12 cm.

2. Let A and B be two points on a circle with centre O. (i) Are there points X, Y on the same side of AB with ∡AXB ≠ ∡AYB? (ii) Is it true that if ∡AXB = ∡AYB then X and Y lie on the same side? (iii) If ∡AXB = ∡AYB and X, Y are not on the circle, does the circle through A, B, X also pass through Y?

SOLUTION (i) No — all points on the same arc give the same angle (angles in the same segment are equal). (ii) Not necessarily — equal angles can occur on the same arc or on the corresponding arc of another position; equality of angle alone does not fix the side. (iii) Yes (when X, Y are on the same side of AB) — by the concyclicity theorem, A, B, X, Y then lie on one circle, so the circle through A, B, X passes through Y.

3. Find x in Fig. 5.26.

SOLUTION In Fig. 5.26, ABCD is a cyclic quadrilateral with the marked angle 100° at D and x at B. Opposite angles of a cyclic quadrilateral are supplementary, so x = 180° − 100° = 80°. (Reads the 100° and x positions from Fig. 5.26.)

Class 9 Maths Ganita Manjari Chapter 5 Solutions — End-of-Chapter Exercises

1. A chord is 5 cm from the centre. If the radius is 13 cm, find the length of the chord.

SOLUTIONChord = 2√(132 − 52) = 2√144 = 24 cm.

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

SOLUTIONAngle on the circle = half the central angle = 70° ÷ 2 = 35°.

3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn. Find the distance from the centre to the chord.

SOLUTIONr = 13, half-chord = 12; distance = √(132 − 122) = √25 = 5 cm.

4. A circle has radius 15 cm. A chord is at distance 9 cm from the centre. Find the chord length.

SOLUTIONChord = 2√(152 − 92) = 2√144 = 24 cm.

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

SOLUTION Every point on the perpendicular bisector of a chord is equidistant from its two endpoints. The centre O is equidistant from the endpoints (both are radii), so O lies on the perpendicular bisector. ✓

6. AB is a diameter and C is on the circumference. What is ∡ACB?

SOLUTIONThe angle in a semicircle is a right angle, so ∡ACB = 90°.

7. ABCD is a cyclic quadrilateral. If ∡A = 75°, find ∡C. If ∡B = 110°, find ∡D.

SOLUTIONOpposite angles are supplementary: ∡C = 180° − 75° = 105°; ∡D = 180° − 110° = 70°.

8. PQRS is cyclic with ∡P = (2x + 10)° and ∡R = (3x − 20)°. Find x, ∡P and ∡R.

SOLUTION ∡P + ∡R = 180° ⇒ (2x + 10) + (3x − 20) = 180 ⇒ 5x − 10 = 180 ⇒ x = 38. ∡P = 2(38) + 10 = 86°, ∡R = 3(38) − 20 = 94° (86 + 94 = 180 ✓).

9. The distance of a chord of length 16 cm from the centre is 6 cm. Find the radius.

SOLUTIONhalf-chord = 8; r = √(82 + 62) = √100 = 10 cm.

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

SOLUTION With adjacent sides 5, 5, 12, 12 the quadrilateral is a right kite whose diagonal = √(52 + 122) = 13 (a diameter). It splits into two 5–12–13 right triangles, so area = 2 × (½ × 5 × 12) = 60 sq units. (Brahmagupta’s formula gives the same: √(12·12·5·5) = 60.)

*11. For a cyclic quadrilateral, how can we tell whether the centre of the circumcircle lies inside or outside it?

SOLUTION The centre lies inside the quadrilateral if and only if every interior angle is less than 90°… more precisely, the centre is inside when no side subtends a reflex arc — equivalently, when all four arcs are less than a semicircle (no angle of the quadrilateral is obtuse enough to put the centre across a side). The quick test: drop perpendicular bisectors of the sides; they meet at the centre — check whether that point falls inside the quadrilateral.

*12. When two equal chords intersect, show that the segments of one chord equal the corresponding segments of the other.

SOLUTION Equal chords are equidistant from the centre. Drop perpendiculars from O to both chords; they bisect the chords into equal halves (equal chords ⇒ equal halves). The intersection point is at equal perpendicular distances, so by symmetry the corresponding segments are equal. ✓

*13. Draw a circle in which a chord of length 6 cm stands at a distance of 3 cm from the centre.

SOLUTION The radius satisfies r = √(32 + 32) = √18 = 3√2 ≈ 4.24 cm. Construction: draw a line, mark the chord’s midpoint M, erect a perpendicular, mark O at 3 cm from M; with centre O and radius 3√2 cm draw the circle — it passes through both ends of the 6 cm chord.

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

SOLUTION An inscribed parallelogram is cyclic, so opposite angles are supplementary; but in a parallelogram opposite angles are equal. Equal and supplementary ⇒ each is 90°. A parallelogram with right angles is a rectangle. ✓

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

SOLUTION A rectangle’s diagonals are equal and bisect each other; each diagonal is a diameter (it subtends a right angle at the opposite vertices). Their common midpoint is therefore the centre of the circle. ✓

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

SOLUTION Every such chord is the same distance d = √(r2 − (L/2)2) from the centre, so its midpoint is at that fixed distance. The midpoints form a concentric circle of radius d.

*17. In a circle with centre O, chords AB and AC are congruent. Explain why the centre lies on the bisector of ∡BAC.

SOLUTION Equal chords are equidistant from O, so O is equidistant from lines AB and AC. The locus of points equidistant from two lines through A is the bisector of the angle between them, so O lies on the bisector of ∡BAC. ✓

18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre, 7 cm apart. Find the radius.

SOLUTION Distances from centre: √(r2 − 25) and √(r2 − 144). Same side ⇒ difference = 7: √(r2 − 25) − √(r2 − 144) = 7. Multiplying by the conjugate: (119)/(sum) = 7 ⇒ sum = 17. With difference 7: the two distances are 12 and 5, so r2 − 25 = 144 ⇒ r = 13 cm.

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

SOLUTION Each side subtends 60° at the centre, giving an equilateral triangle, so the side = r. The distance of a side from the centre (apothem) = √(r2 − (r/2)2) = (√3/2)r.

20. MNOP is inscribed in a circle with MN a diameter. What can you say about the angles subtended by MN at the other vertices?

SOLUTION Since MN is a diameter, the angle it subtends at any other point of the circle is 90° (angle in a semicircle). So ∡MPN = 90° and ∡MON = 90°.

21. For cyclic quadrilateral ABCD, explain why the exterior angle at a vertex equals the interior opposite angle (e.g. ∡CDE = ∡ABC).

SOLUTION ∡CDE + ∡ADC = 180° (straight line) and ∡ABC + ∡ADC = 180° (opposite angles of a cyclic quad). Subtracting gives ∡CDE = ∡ABC. ✓

*22. “There is no chord longer than the diameter.” Justify this.

SOLUTION Chord length = 2√(r2 − d2) is largest when d = 0, giving 2r = the diameter. Any chord with d > 0 is shorter, so the diameter is the longest chord. ✓

*23. For a point A inside a circle (centre O), show that the shortest chord through A is perpendicular to OA.

SOLUTION A chord through A is shortest when its distance from O is greatest (since length = 2√(r2 − d2) decreases as d grows). For chords through A, the perpendicular distance from O is largest, equal to OA, exactly when the chord is perpendicular to OA. So that chord is the shortest. ✓

24. How would you use Fig. 5.30 to justify that the angle in a semicircle is 90°?

SOLUTION Join the vertex to the centre O. This creates two isosceles triangles (OA = OB = the radius = OC). Their base angles a and b satisfy 2a + 2b = 180°, so a + b = 90° — and a + b is exactly the angle in the semicircle. Hence it is 90°. ✓

*25. Two chords CC′ and DD′ are perpendicular to a diameter AB. Prove the segment MM′ joining the midpoints of CD and C′D′ is perpendicular to AB.

SOLUTION The diameter AB is an axis of symmetry. Chords perpendicular to AB are bisected by AB, so C, C′ (and D, D′) are symmetric about AB. The midpoints M of CD and M′ of C′D′ therefore lie on lines perpendicular to AB at equal positions, and MM′ runs parallel to those chords — i.e. perpendicular to AB. ✓

*26. How would you use Fig. 5.31 to justify that opposite angles of a cyclic quadrilateral sum to 180°?

SOLUTION Join each vertex to the centre O. Each angle of the quadrilateral is half the central angle on the arc it stands on (angle at centre = 2 × angle on circle). Two opposite angles together are half of the full 360° rotation at O, i.e. ½ × 360° = 180°. ✓

Common Mistakes to Avoid

Watch out for these

  • Using chord/2 as r in the Pythagoras step — the hypotenuse is the radius r, not the chord.
  • Forgetting the angle at the centre is double (not equal to) the angle on the circle.
  • Thinking opposite angles of a cyclic quadrilateral are equal — they are supplementary (sum 180°).
  • Assuming chord length is proportional to distance from the centre — it follows 2√(r2 − d2).
  • For parallel chords, remember to add the distances if on opposite sides and subtract if on the same side.
  • Placing the circumcentre inside an obtuse triangle — it lies outside.

Practice MCQs & Assertion–Reason

1. The longest chord of a circle is its:

(a) radius    (b) diameter    (c) tangent    (d) arc

2. The angle in a semicircle is:

(a) 45°    (b) 60°    (c) 90°    (d) 180°

3. The angle subtended by an arc at the centre is ___ the angle it subtends on the circle:

(a) equal to    (b) half    (c) double    (d) one-third

4. The perpendicular from the centre of a circle to a chord:

(a) bisects the chord    (b) trisects it    (c) is parallel to it    (d) equals the radius

5. Opposite angles of a cyclic quadrilateral add up to:

(a) 90°    (b) 180°    (c) 270°    (d) 360°

6. A chord of length 24 cm in a circle of radius 13 cm is at distance:

(a) 5 cm    (b) 7 cm    (c) 12 cm    (d) 10 cm

7. The number of non-collinear points needed to determine a unique circle is:

(a) 1    (b) 2    (c) 3    (d) 4

8. Equal chords of a circle are:

(a) equidistant from the centre    (b) of different distances    (c) always diameters    (d) perpendicular

9. The side of a regular hexagon inscribed in a circle of radius r is:

(a) r/2    (b) r    (c) r√3    (d) 2r

10. In a cyclic quadrilateral, if ∠A = 85°, then ∠C =

(a) 85°    (b) 95°    (c) 105°    (d) 180°

Answer key: 1-(b), 2-(c), 3-(c), 4-(a), 5-(b), 6-(a), 7-(c), 8-(a), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: The angle in a semicircle is a right angle.

Reason: A diameter subtends 180° at the centre, and the angle on the circle is half of it.

A-R 2. Assertion: Equal chords of a circle are equidistant from the centre.

Reason: The triangles formed by equal chords and the radii are congruent.

A-R 3. Assertion: The perpendicular from the centre to a chord bisects the chord.

Reason: The two triangles formed are congruent by RHS.

A-R 4. Assertion: The angle at the centre is half the angle on the circle for the same arc.

Reason: The angle subtended by an arc at the centre is twice the angle at a point on the circle.

A-R 5. Assertion: The diameter is the longest chord of a circle.

Reason: Chord length 2√(r2 − d2) is greatest when d = 0.

Answer key: 1-(A), 2-(A), 3-(A), 4-(D), 5-(A).

Quick Revision Summary

  • Three non-collinear points determine a unique circle (the circumcircle of their triangle).
  • Equal chords subtend equal central angles and are equidistant from the centre (and conversely); the longer chord is nearer the centre.
  • The perpendicular from the centre bisects a chord; chord = 2√(r2 − d2).
  • Angle at centre = 2 × angle on the circle (same arc); angles in the same segment are equal.
  • The angle in a semicircle is 90°; the diameter is the longest chord.
  • Opposite angles of a cyclic quadrilateral are supplementary; an exterior angle equals the interior opposite angle.

How to score full marks in this chapter

Always draw a clean labelled figure and write “Given / To show” before a proof. For chord problems, form the right triangle of half-chord, distance and radius. State the theorem you use (e.g. “angle in a semicircle”) before applying it, and keep surd answers exact unless a decimal is asked.

Frequently Asked Questions

What is Class 9 Maths Ganita Manjari Chapter 5 about?

It is the circles chapter — chords and the angles they subtend, perpendiculars from the centre, distances of chords, the angle subtended by an arc, the angle in a semicircle, and cyclic quadrilaterals.

What is the formula for the length of a chord?

Chord = 2√(r2 − d2), where r is the radius and d is the perpendicular distance of the chord from the centre.

What is the angle in a semicircle?

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

Are these Class 9 Maths Ganita Manjari Chapter 5 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Manjari textbook for 2026–27.

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