PrepRung
Sign inSign up free

Class 9 Maths Chapter 5: I'm Up and Down, and Round and Round — Notes, Examples & Practice

Wheels, bangles, rangoli and the rim of a cup are all circles. This chapter proves the facts that make circles so useful: how chords relate to the centre, why exactly one circle passes through three points, why the angle at the centre is double the angle at the edge, and why opposite angles of a cyclic quadrilateral add up to 180°. Each result is proved from congruent triangles, so learn the reasoning — it is what the exam asks for.

1Circle vocabulary and symmetry

A circle is the set of all points in a plane at a fixed distance (the radius rr) from a fixed point (the centre O). A chord joins two points on the circle; a chord through the centre is a diameter, of length 2r2r, and it is the longest chord.

Part of the circle between two points is an arc (minor or major). The region between a chord and its arc is a segment; the region between two radii and their arc is a sector.

A circle has reflection symmetry in every diameter and rotational symmetry about its centre by any angle. These symmetries are behind many of the theorems below.

Key formulas & definitions
  • Diameter =2×= 2 \times radius; the diameter is the longest chord.
  • Equal circles (congruent circles) have equal radii.
  • Sector: between two radii. Segment: between a chord and its arc.
Common mistakes
  • Calling a sector a segment (and vice versa). Sector = 'slice of pizza'; segment = 'cut off by a straight chord'.

2The circle through three points

Through one point you can draw infinitely many circles; through two points, still infinitely many (their centres all lie on the perpendicular bisector of the segment joining them).

Through three non-collinear points A, B, C there is exactly one circle. Its centre must be equidistant from all three, so it lies on the perpendicular bisector of AB and of BC — and two such lines meet at just one point. This is the circumcentre of triangle ABC, and the circle is its circumcircle.

If the three points are collinear, the two perpendicular bisectors are parallel and never meet, so no circle passes through them.

Key formulas & definitions
  • Circumcentre = intersection of perpendicular bisectors of the sides.
  • Three non-collinear points determine exactly one circle.
Worked example 1

A broken circular plate leaves only an arc. How can you find the centre to restore it?

  1. Mark any three points A, B, C on the arc.
  2. Draw the perpendicular bisectors of chords AB and BC.
  3. Where they meet is the centre; its distance to A is the radius.

Answer: The intersection of the perpendicular bisectors of two chords.

Common mistakes
  • Using angle bisectors (which give the incentre) instead of perpendicular bisectors.

3Chords and the centre

The perpendicular from the centre to a chord bisects it. Proof: join O to both ends A, B of the chord and drop the perpendicular OM. Triangles OMA and OMB are right-angled with OA=OBOA = OB (radii) and common side OM, so they are congruent by RHS, giving AM=MBAM = MB. The converse also holds: the line from the centre to the midpoint of a chord is perpendicular to it.

Equal chords subtend equal angles at the centre (triangles OAB and OCD are congruent by SSS), and conversely equal central angles cut off equal chords.

Equal chords are equidistant from the centre, and the converse is true. Closer to the centre means longer: as the distance dd from the centre grows, the half-chord r2−d2\sqrt{r^2 - d^2} shrinks.

Key formulas & definitions
  • OM⊥AB⇒AM=MBOM \perp AB \Rightarrow AM = MB; and AM=MB⇒OM⊥ABAM = MB \Rightarrow OM \perp AB.
  • Chord length =2r2−d2= 2\sqrt{r^2 - d^2} where dd = distance from the centre.
  • Equal chords ⇔ equal central angles ⇔ equal distances from the centre.
Worked example 1

A chord of a circle of radius 13 cm is 5 cm from the centre. Find its length.

  1. Draw the perpendicular from O to the chord; it bisects the chord.
  2. Half-chord =132−52=169−25=144=12= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 cm.
  3. Chord =2×12=24= 2 \times 12 = 24 cm.

Answer: 24 cm.

Worked example 2

Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle of radius 5 cm. Find the distance between them.

  1. Distance of the 6 cm chord: 52−32=4\sqrt{5^2 - 3^2} = 4 cm.
  2. Distance of the 8 cm chord: 52−42=3\sqrt{5^2 - 4^2} = 3 cm.
  3. Opposite sides → add: 4+3=74 + 3 = 7 cm. (Same side would give 4−3=14 - 3 = 1 cm.)

Answer: 7 cm.

Trick / memory aid

Every chord problem hides a right triangle: radius (hypotenuse), distance from centre, half-chord. Draw it first.

Common mistakes
  • Using the full chord instead of half the chord in Pythagoras.
  • Forgetting to check whether two chords are on the same side or opposite sides of the centre.

4Angles subtended by arcs

The angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle. Proof idea: join the point P on the circle through O to the far side. Each of the two triangles formed is isosceles (two radii), and the exterior angle at O of each triangle is twice its base angle; adding (or subtracting) gives ∠AOB=2∠APB\angle AOB = 2\angle APB.

Two consequences follow at once. Angles in the same segment are equal — every point on the same arc sees the chord at the same angle, since each is half the same central angle.

The angle in a semicircle is a right angle: a diameter subtends 180∘180^\circ at the centre, so it subtends 90∘90^\circ at every point on the circle.

Key formulas & definitions
  • ∠AOB=2∠APB\angle AOB = 2\angle APB (P on the major arc).
  • Angles in the same segment are equal.
  • Angle in a semicircle =90∘= 90^\circ.
Worked example 1

O is the centre, and ∠AOB=110∘\angle AOB = 110^\circ. P is on the major arc and Q on the minor arc. Find ∠APB\angle APB and ∠AQB\angle AQB.

  1. ∠APB=110∘2=55∘\angle APB = \dfrac{110^\circ}{2} = 55^\circ.
  2. The reflex angle AOB =360∘−110∘=250∘= 360^\circ - 110^\circ = 250^\circ, and Q sees the major arc, so ∠AQB=125∘\angle AQB = 125^\circ.
  3. Check: APBQ is cyclic, and 55∘+125∘=180∘55^\circ + 125^\circ = 180^\circ ✓.

Answer: 55∘55^\circ and 125∘125^\circ.

Trick / memory aid

“Centre is double” — the angle at the centre is always the bigger one, twice the angle at the edge standing on the same arc.

Common mistakes
  • Halving the wrong arc — make sure both angles stand on the same arc (open towards the same chord).
  • Using the angle-in-a-semicircle result when the chord is not a diameter.

5Cyclic quadrilaterals

Points that lie on one circle are concyclic. A quadrilateral whose four vertices lie on a circle is cyclic.

**Opposite angles of a cyclic quadrilateral add up to 180∘180^\circ.** Opposite angles stand on the two arcs that together make the whole circle, so their central angles add to 360∘360^\circ and the angles themselves to 180∘180^\circ. The converse is a useful test: if a pair of opposite angles is supplementary, the quadrilateral is cyclic.

Consequence: an exterior angle of a cyclic quadrilateral equals the interior opposite angle. A rectangle is always cyclic; a parallelogram that is cyclic must be a rectangle.

Key formulas & definitions
  • Cyclic ABCD: ∠A+∠C=180∘\angle A + \angle C = 180^\circ, ∠B+∠D=180∘\angle B + \angle D = 180^\circ.
  • Exterior angle = interior opposite angle.
  • Converse: opposite angles supplementary ⇒ cyclic.
Worked example 1

ABCD is cyclic with ∠A=(2x+10)∘\angle A = (2x + 10)^\circ and ∠C=(3x−5)∘\angle C = (3x - 5)^\circ. Find both angles.

  1. 2x+10+3x−5=180⇒5x=175⇒x=352x + 10 + 3x - 5 = 180 \Rightarrow 5x = 175 \Rightarrow x = 35.
  2. ∠A=80∘\angle A = 80^\circ, ∠C=100∘\angle C = 100^\circ.

Answer: 80∘80^\circ and 100∘100^\circ.

Worked example 2

Prove that a cyclic parallelogram is a rectangle.

  1. In a parallelogram, opposite angles are equal: ∠A=∠C\angle A = \angle C.
  2. Cyclic: ∠A+∠C=180∘\angle A + \angle C = 180^\circ, so 2∠A=180∘2\angle A = 180^\circ and ∠A=90∘\angle A = 90^\circ.
  3. A parallelogram with one right angle is a rectangle.

Answer: Proved.

Common mistakes
  • Adding adjacent angles to 180° — it is the opposite angles.
  • Assuming a quadrilateral is cyclic without checking (or being told).

Important questions

Try each question on paper first, then tap it to check the answer.

Basic Recall and direct use

The radius of a circle is 5 cm and a chord is 8 cm long. Find the distance of the chord from the centre.Answer
52−42=3\sqrt{5^2 - 4^2} = 3 cm.
AB is a diameter and C is a point on the circle with ∠CAB=35∘\angle CAB = 35^\circ. Find ∠CBA\angle CBA.Answer
∠ACB=90∘\angle ACB = 90^\circ (angle in a semicircle). So ∠CBA=180∘−90∘−35∘=55∘\angle CBA = 180^\circ - 90^\circ - 35^\circ = 55^\circ.
In cyclic quadrilateral PQRS, ∠P=72∘\angle P = 72^\circ and ∠Q=95∘\angle Q = 95^\circ. Find ∠R\angle R and ∠S\angle S.Answer
∠R=180∘−72∘=108∘\angle R = 180^\circ - 72^\circ = 108^\circ; ∠S=180∘−95∘=85∘\angle S = 180^\circ - 95^\circ = 85^\circ.

Medium Two or three steps

A, B, C are on a circle with centre O and ∠ABC=48∘\angle ABC = 48^\circ (B on the major arc). Find ∠AOC\angle AOC and ∠OAC\angle OAC.Answer
∠AOC=2×48∘=96∘\angle AOC = 2 \times 48^\circ = 96^\circ. Triangle OAC is isosceles (OA=OCOA = OC), so ∠OAC=180∘−96∘2=42∘\angle OAC = \dfrac{180^\circ - 96^\circ}{2} = 42^\circ.
Two concentric circles have centre O. A line cuts the outer circle at A and D and the inner circle at B and C (in the order A, B, C, D). Prove that AB=CDAB = CD.Answer
Draw OM⊥ADOM \perp AD. The perpendicular from the centre bisects every chord on that line, so AM=MDAM = MD (outer circle) and BM=MCBM = MC (inner circle). Subtracting, AM−BM=MD−MCAM - BM = MD - MC, i.e. AB=CDAB = CD.
Find the radius of a circle in which a chord of 24 cm is 5 cm from the centre.Answer
r=122+52=169=13r = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 cm.

Advanced Reasoning and HOTS

ABCD is a cyclic quadrilateral whose diagonals meet at E. If ∠DBC=70∘\angle DBC = 70^\circ and ∠BAC=30∘\angle BAC = 30^\circ, find ∠BCD\angle BCD. If also AB=BCAB = BC, find ∠ECD\angle ECD.Answer
∠CAD=∠CBD=70∘\angle CAD = \angle CBD = 70^\circ (same segment, on chord CD). So ∠BAD=30∘+70∘=100∘\angle BAD = 30^\circ + 70^\circ = 100^\circ and ∠BCD=180∘−100∘=80∘\angle BCD = 180^\circ - 100^\circ = 80^\circ. If AB=BCAB = BC, then ∠BCA=∠BAC=30∘\angle BCA = \angle BAC = 30^\circ, so ∠ECD=80∘−30∘=50∘\angle ECD = 80^\circ - 30^\circ = 50^\circ.
Prove that the circle drawn with any equal side of an isosceles triangle as diameter bisects the base.Answer
Let AB=ACAB = AC and draw the circle on AB as diameter, cutting BC at D. Then ∠ADB=90∘\angle ADB = 90^\circ (angle in a semicircle), so AD ⊥ BC. In triangles ADB and ADC: AB=ACAB = AC, AD common, right angles at D → congruent (RHS). Hence BD=DCBD = DC.

Self-check quiz

Five quick questions. Tap an option to see if you are right.

The longest chord of a circle is
If a chord subtends 80∘80^\circ at the centre, the angle it subtends at a point on the major arc is
How many circles pass through three non-collinear points?
In a cyclic quadrilateral, one angle is 65∘65^\circ. Its opposite angle is
A chord 6 cm long is drawn in a circle of radius 5 cm. Its distance from the centre is

FAQ

Why is the angle in a semicircle 90°?

The diameter makes a straight angle (180∘180^\circ) at the centre, and the angle at the circumference is half of the central angle on the same arc, so it is 90∘90^\circ.

Is every quadrilateral cyclic?

No. Only those whose opposite angles add to 180∘180^\circ. Squares, rectangles and isosceles trapeziums are cyclic; a general parallelogram or rhombus is not.

Which circle theorems need proofs in the exam?

Usually: perpendicular from the centre bisects a chord, equal chords are equidistant from the centre, angle at the centre is twice the angle at the circumference, and opposite angles of a cyclic quadrilateral are supplementary.

Book references

  • NCERT Ganita Manjari Part I, Chapter 5: I'm Up and Down, and Round and RoundRead the chapter and do every exercise set first.
  • Mathematics for Class 9 — R.D. Sharma (Dhanpat Rai)Extra graded practice; its chapter numbers follow the old syllabus, so match by topic.
  • Mathematics for Class 9 — R.S. Aggarwal (Bharati Bhawan)Good for drill on basics; match by topic.
  • Exemplar Problems in Mathematics, Class IX — NCERTHarder, concept-testing questions (written for the old book).

These notes, examples and questions are written by PrepRung in our own words to help you study the NCERT Ganita Manjari book. NCERT and CBSE are not affiliated with PrepRung. Always study from your textbook too.