Class 9 Maths Ganita Manjari NCERT Solutions Chapter 5 - Up and Down, and Round and Round

Key Concepts & Theorems at a Glance

A circle is the set (locus) of all points in a plane at a fixed distance (the radius) from a fixed point (the centre).
A chord is a line segment joining two points on a circle; a diameter is a chord through the centre — the longest possible chord.
An arc is a connected portion of a circle between two points; the smaller piece is the minor arc, the larger is the major arc.
Points on the same circle are concyclic. A quadrilateral whose vertices are concyclic is a cyclic quadrilateral.

Theorem 1: A unique circle passes through three non-collinear points (its circumcircle; the centre is the circumcentre).

Theorem 2: Equal chords subtend equal angles at the centre.

Theorem 3: Chords subtending equal angles at the centre are equal.

Theorem 4: The line from the centre to the midpoint of a chord is perpendicular to the chord.

Theorem 5: The perpendicular from the centre to a chord bisects the chord.

Theorem 6: Equal chords are equidistant from the centre.

Theorem 7: Chords equidistant from the centre are equal.

Theorem 8: The longer of two chords is closer to the centre.

Theorem 9: The angle subtended by an arc at the centre is double the angle it subtends at any point on the circle outside the arc. Corollary: the angle in a semicircle is 90°.

Theorem 10: If AB subtends equal angles at C and D on the same side of AB, then A, B, C, D are concyclic.

Theorem 11: Opposite angles of a cyclic quadrilateral sum to 180°.

Theorem 12 (converse): If opposite angles of a quadrilateral sum to 180°, its vertices are concyclic.

\[ \text{chord length } = 2\sqrt{r^2-d^2}, \quad \text{where } r=\text{radius},\ d=\text{perpendicular distance from centre} \]

ActActivity: List some objects from nature that resemble a circle.

Many things in nature are circular or nearly circular in outline. Some examples:

The full moon and the sun;
ripples formed when a raindrop falls on still water;
the cross-section of a tree trunk (growth rings) or of a plant stem;
the centre/inflorescence of a sunflower;
a spider's orb web;
the pupil of an eye;
the cross-section of many fruits (e.g. an orange slice);
a rainbow (which is actually part of a circle);
the shape traced by the Earth's orbit around the Sun (approximately).

Examples: sun, moon, water ripples, tree-trunk rings, sunflower centre, spider webs, eye pupils, fruit cross-sections.

Think and Reflect

TRJamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?

Amina's suggestion uses the fact that a fold that makes the circular boundary overlap perfectly creates a crease along a diameter.

Fold the circular paper in half, so that the curved edge matches up with itself exactly on both sides. Open it out — the crease you see is a diameter of the circle.

Fold the paper again, in a different direction, so the boundary again overlaps perfectly, and open it out. This gives a second crease — another diameter.

Every diameter of a circle passes through the centre, so the point where the two creases (two different diameters) cross must be the centre of the circle.

Amina told her: fold the paper in half along the boundary twice, in two different directions. Each fold creates a crease that is a diameter, and the point where the two creases intersect is the centre.

Think and Reflect

TR1. 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?
2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
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?

1. A square has rotational symmetry at 90°, 180°, 270° and 360° (4 rotational symmetries — it looks the same after each quarter-turn), and 4 lines of reflection symmetry (the two diagonals, and the two lines joining midpoints of opposite sides).

A regular pentagon has 5 rotational symmetries (at multiples of \(360^\circ\div5=72^\circ\): 72°, 144°, 216°, 288°, 360°), and 5 lines of reflection symmetry (each joining a vertex to the midpoint of the opposite side).

A regular hexagon has 6 rotational symmetries (at multiples of \(360^\circ\div6=60^\circ\)), and 6 lines of reflection symmetry (3 through pairs of opposite vertices, and 3 through midpoints of opposite sides).

In general, a regular n-gon has n rotational symmetries and n lines of reflection symmetry. A circle is the limiting case — it has symmetry for every angle of rotation, and every diameter is a line of symmetry.

2. The longest chord of a circle is its diameter, so for radius 5, the longest chord is \(2\times5=10\) units. There is no smallest chord — chords can be drawn as short as we like (by choosing the two endpoints closer and closer together), so their length can get arbitrarily close to 0, but a chord needs two distinct points, so there is no chord of length exactly 0, and no "smallest" chord that isn't beaten by an even shorter one.

3. The locus of points equidistant from two given points A and B is the perpendicular bisector of segment AB — a straight line, not a circle. Using the hint: let M be the midpoint of AB, and let P be any point on the perpendicular bisector. In triangles PMA and PMB: PM is common, AM = BM (M is the midpoint), and \(\angle PMA=\angle PMB=90^\circ\). By SAS congruence, \(\triangle PMA\cong\triangle PMB\), so \(PA=PB\). This shows every point on the perpendicular bisector is equidistant from A and B. Combined with the fact (given) that every point equidistant from A and B lies on the perpendicular bisector, the locus is exactly the perpendicular bisector of AB.

Square: 4 rotational + 4 reflection symmetries. Pentagon: 5 + 5. Hexagon: 6 + 6. Longest chord in radius-5 circle = 10 units (the diameter); there is no smallest chord. Locus of points equidistant from two given points A, B = the perpendicular bisector of AB (a straight line).

Think and Reflect

TR1. How many circles pass through two points on a plane?
2. Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?
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?
4. As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?
5. You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?

Three of the infinitely many circles through fixed points A and B, centres sliding along the perpendicular bisector

1. Infinitely many circles pass through two given points A and B. Every point on the perpendicular bisector of AB can serve as the centre of one such circle (with radius equal to its distance to A, which equals its distance to B).

2. No — not every radius is possible. The smallest circle through A and B has AB itself as a diameter, giving the minimum radius \(\dfrac{AB}{2}\) (this happens when the centre is the midpoint of AB). There is no largest circle — as the centre moves further along the perpendicular bisector, the radius keeps growing without any upper limit, so the radius can be made as large as we like.

3. The radii increase — the farther the centre is from AB along the perpendicular bisector, the greater its distance to A (and B), so the radius grows.

4. The circle appears less curved. A bigger circle (larger radius) curves more gently near any given arc — think of how the Earth's surface looks almost flat locally because its radius is so large. So as the radius increases, the circle looks less curved (flatter) near A and B.

5. If A and B are simply required to lie somewhere on the boundary of a square (not necessarily at corners), infinitely many squares work, since A and B could sit anywhere along the sides of squares of many different sizes and orientations.

If A and B must be corners of the square, there are two cases. If AB is a side of the square, exactly 2 squares can be drawn — one on each side of line AB (the square "grows" either upward or downward from AB, using AB as one edge). If AB is a diagonal of the square, exactly 1 square can be drawn — the diagonal's midpoint is the square's centre, and the other diagonal (same length, perpendicular to AB through that midpoint) is completely determined, so there is only one such square.

1: infinitely many. 2: smallest radius = \(\dfrac{AB}{2}\) (no largest — radius can grow without bound). 3: radii increase. 4: the circle appears less curved. 5: infinitely many squares with A, B merely on the boundary; exactly 3 squares with A, B as corners (2 with AB as a side, 1 with AB as a diagonal).

Exercise Set 5.1

1Draw \(\triangle ABC\) with AB = 5 cm, \(\angle A = 70^\circ\) and \(\angle B = 60^\circ\). Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?

Acute triangle ABC with circumcentre O inside the triangle

\(\angle C = 180^\circ - 70^\circ - 60^\circ = 50^\circ\)[angle sum of a triangle]

All three angles (70°, 60°, 50°) are less than 90°[△ABC is acute-angled]

Construction: draw AB = 5 cm; construct 70° at A, 60° at B

Rays from A, B meet at C, completing the triangle

Construct perpendicular bisectors of AB and BC — intersection is O

Draw the circle, centre O, radius OA

∠C = 50°, so △ABC is acute-angled — for an acute-angled triangle, the circumcentre lies inside the triangle.

2Draw \(\triangle ABC\) with AB = 5 cm, \(\angle A = 100^\circ\), AC = 4 cm. Draw the circumcircle of \(\triangle ABC\). Is the centre inside or outside the triangle?

Obtuse triangle ABC with circumcentre O outside the triangle

\(\angle A = 100^\circ > 90^\circ\)[△ABC is obtuse-angled at A]

Construction: draw AB = 5 cm; construct 100° at A; mark C with AC = 4 cm

Join BC to complete the triangle

Construct perpendicular bisectors of 2 sides — intersection is O

Draw the circle, centre O, radius OA

∠A = 100° is obtuse, so △ABC is obtuse-angled — for an obtuse-angled triangle, the circumcentre lies outside the triangle.

3Draw \(\triangle ABC\), with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of \(\triangle ABC\). Let the circumcentre be O. Measure OA, OB, OC.

Isosceles triangle ABC (BC = CA = 7 cm) with circumradii OA, OB, OC marked

Isosceles triangle, BC = CA = 7 cm; \(R=\dfrac{abc}{4\times\text{Area}}\)[a=7, b=7, c=6]

\(s=\dfrac{6+7+7}{2}=10\)[semi-perimeter]

Area \(=\sqrt{10(3)(3)(4)}=\sqrt{360}=6\sqrt{10}\approx18.97\) cm²[Heron's formula]

\(R=\dfrac{7\times7\times6}{4\times6\sqrt{10}}=\dfrac{49\sqrt{10}}{40}\approx3.87\) cm

OA, OB, OC are all radii of the same circumcircle[so they measure equal when you actually measure them]

OA = OB = OC ≈ 3.87 cm (the circumradius) — measuring confirms all three are equal, since O is equidistant from A, B and C by definition of the circumcentre.

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

The smallest circle through A, B is the one where AB is a diameter[a smaller circle can't reach both points]

The least possible radius is \(\dfrac{AB}{2}\) (half the distance between A and B).

Think, Draw and Infer

1A, B and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?

Collinear points A, B, C on line ℓ — the perpendicular bisectors of AB and BC are both ⟂ ℓ, hence parallel, and never meet

No such point P exists

A, B, C lie on a line ℓ

⟂ bisector of AB is ⟂ ℓ at the midpoint of AB

⟂ bisector of BC is ⟂ ℓ at the (different) midpoint of BC

Both bisectors are ⟂ the same line ℓ, so they are parallel to each other

Midpoints of AB, BC are different points on ℓ, so the two bisectors are also distinct

Distinct parallel lines never meet

A common centre P must lie on both bisectors — but they never meet

So no such P exists, and no circle passes through 3 collinear points

Any 3 points a line meets on a circle would have to be concyclic and collinear

This is impossible — so a line meets a circle in at most 2 points

No such P exists. The perpendicular bisectors of AB and BC are parallel (both perpendicular to the same line through A, B, C) and distinct, so they never meet — hence no circle passes through 3 collinear points, and a straight line can cut a circle in at most 2 points.

2The circumcircle of a given \(\triangle ABC\) is drawn. Can there be other triangles congruent to \(\triangle ABC\) that share the same circumcircle?

△ABC (teal) rotated about the circumcentre O to give a congruent △A′B′C′ (dashed purple) on the same circumcircle

Yes — rotate △ABC about circumcentre O by any angle

Rotation about the centre maps the circle to itself[new vertices still lie on the circle]

Rotation preserves distances[new triangle has the same side lengths]

Similarly, reflecting across any diameter also gives a congruent triangle

Yes — rotating or reflecting △ABC about its circumcentre produces infinitely many other triangles, all congruent to △ABC, sharing the same circumcircle.

ActExercise: A circle with centre O is drawn, and A, B, C, D are points on the circle (Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

Fig. 5.19: circle with centre O; radii OA, OB (teal) and OC, OD (purple); K on arc AB, L on arc CD

This is a hands-on measuring activity — angle values depend on where you place A, B, C, D

Join OA, OB; measure \(\angle AOB\), swept along the same side as arc AKB (through K)

Compare to 180°: less than 180° → minor arc; more than 180° → major arc

Do the same for arc CLD: measure \(\angle COD\), swept through L, compare to 180°

Typically: if K is on the shorter path, AKB is minor; if L is on the longer path, CLD is major[check your own measurements against the 180° rule]

Method: measure the central angle swept along each arc; less than 180° → minor arc, more than 180° → major arc. (Exact classification of AKB and CLD depends on your own construction.)

Exercise Set 5.2

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

Chord AB with centre C — radii CA, CB and chord AB form △CAB

Given: AB a chord, centre C; join CA, CB

To Prove: \(\triangle CAB\) is isosceles

\(CA=CB\)[both radii of the same circle]

\(\therefore \triangle CAB\) is isosceles, with AB as its base[hence proved]

CA = CB (radii) ⟹ △CAB is isosceles.

2Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Equal chords AB, DE (AB = DE) with centre C — △CAB (teal) and △CDE (purple)

Given: AB, DE chords of the same circle, centre C; \(AB=DE\)

To Prove: \(\triangle CAB \cong \triangle CDE\)

In \(\triangle CAB\) and \(\triangle CDE\):

\(CA=CD\)[radii]

\(CB=CE\)[radii]

\(AB=DE\)[given]

\(\therefore \triangle CAB \cong \triangle CDE\)[SSS congruence, hence proved]

△CAB ≅ △CDE by SSS (CA = CD, CB = CE — both radii; AB = DE given).

Exercise Set 5.3

1Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that \(\angle CMA = \angle CMB = 90^\circ\). You need to show that AM = BM.)

Fig. 5.12: chord AB with midpoint M, centre C

Given: C is the centre; AB is a chord; \(CM \perp AB\)[so \(\angle CMA=\angle CMB=90^\circ\)]

To Prove: \(AM=BM\)

In \(\triangle CMA\) and \(\triangle CMB\):

\(CA=CB\)[radii of the same circle]

\(CM=CM\)[common side]

\(\angle CMA=\angle CMB=90^\circ\)[given]

\(\triangle CMA \cong \triangle CMB\)[RHS congruence]

\(\therefore AM=BM\)[corresponding parts of congruent triangles]

△CMA ≅ △CMB by RHS ⟹ AM = BM.

2An 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.

Isosceles △ABC (AB = AC) inscribed in a circle with centre O — altitude AM (dashed) passes through O

Given: \(\triangle ABC\) inscribed in a circle, centre O, \(AB=AC\); M is the midpoint of BC

To Prove: the altitude from A to BC passes through O

In \(\triangle ABM\) and \(\triangle ACM\):

\(AB=AC\)[given]

\(BM=CM\)[M is midpoint]

\(AM=AM\)[common]

\(\triangle ABM \cong \triangle ACM\)[SSS congruence]

\(\angle AMB=\angle AMC\)

\(\angle AMB+\angle AMC=180^\circ\)[linear pair]

\(\therefore \angle AMB=\angle AMC=90^\circ\)

AM ⟂ BC at its midpoint M — AM is the perpendicular bisector of BC

\(OB=OC=r\)[radii]

So O lies on the perpendicular bisector of BC — the same line AM[hence proved]

AM is the perpendicular bisector of BC (△ABM ≅ △ACM), and O also lies on it (OB = OC) — so the altitude from A passes through O.

3Two 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.

Two parallel chords on opposite sides of centre O — distance between midpoints = d₁ + d₂

6 cm chord: half-length = 3 cm

\(d_1=\sqrt{r^2-3^2}=\sqrt{16}=4\) cm[Pythagoras]

8 cm chord: half-length = 4 cm

\(d_2=\sqrt{r^2-4^2}=\sqrt{9}=3\) cm[Pythagoras]

Chords are on opposite sides of the centre[distances add]

Distance between midpoints \(=d_1+d_2=4+3=7\) cm

Distance between the midpoints = 7 cm.

Exercise Set 5.4

1Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

Equal chords AB (teal) and FG (purple) — the perpendiculars from centre C to each are equal in length

Given: AB, FG equal chords of a circle, centre C, radius r; \(AB=FG=2a\)

To Prove: AB and FG are equidistant from C

The perpendicular from C bisects each chord[perpendicular from centre to a chord bisects it]

For AB: right triangle with radius r, half-chord a, distance d

\(d^2+a^2=r^2 \Rightarrow d=\sqrt{r^2-a^2}\)[Pythagoras]

Same formula, same r and a, applies to FG[FG = 2a too]

\(\therefore\) both chords are at distance \(\sqrt{r^2-a^2}\) from C[hence proved]

Both distances equal √(r² − a²) for the same a (since AB = FG), so equal chords are equidistant from the centre.

2Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF.

Fig. 5.15: two chords AB, GF with perpendiculars CE, CH from centre C

Given: \(CE \perp AB\), \(CH \perp GF\), \(CE=CH\)

To Prove: \(AB=GF\)

E is the midpoint of AB, H is the midpoint of GF[perpendicular from centre bisects a chord]

In \(\triangle CEA\) and \(\triangle CHG\):

\(CA=CG\)[radii]

\(CE=CH\)[given]

\(\angle CEA=\angle CHG=90^\circ\)

\(\triangle CEA \cong \triangle CHG\)[RHS congruence]

\(EA=HG\)[corresponding parts of congruent triangles]

\(\therefore AB=2\,EA=2\,HG=GF\)[hence proved]

△CEA ≅ △CHG by RHS ⟹ EA = HG ⟹ AB = GF.

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

Same configuration as Fig. 5.15 — right triangles CEA and CHG give the Pythagorean route to AB = GF

In right \(\triangle CEA\): \(CA^2=CE^2+EA^2\)[Pythagoras]

\(EA=\sqrt{r^2-CE^2}\)

In right \(\triangle CHG\): \(CG^2=CH^2+HG^2\)[Pythagoras]

\(HG=\sqrt{r^2-CH^2}\)

Since \(CE=CH\): \(EA=HG\)

\(\therefore AB=2\,EA=2\,HG=GF\)

EA = HG (since CE = CH), so AB = GF.

Exercise Set 5.5

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

Chord \(=2\sqrt{r^2-d^2}\), with \(r=7,\ d=6\)

\(=2\sqrt{49-36}=2\sqrt{13}\approx7.21\) cm

Chord length = 2√13 cm ≈ 7.21 cm.

2Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is \(2\sqrt{r^2-d^2}\).

Chord AB, centre C, foot of perpendicular M — right △CMA with CM = d, CA = r, MA = half the chord

Given: AB a chord, C the centre, \(CM \perp AB\), \(CM=d\), \(CA=r\)

M is the midpoint of AB[perpendicular from centre bisects a chord]

In right \(\triangle CMA\): \(CA^2=CM^2+MA^2\)[Pythagoras]

\(r^2=d^2+MA^2\)

\(MA=\sqrt{r^2-d^2}\)

\(AB=2\,MA=2\sqrt{r^2-d^2}\)[M is the midpoint, hence proved]

Chord = 2√(r² − d²), by the Baudhāyana–Pythagoras theorem in △CMA.

3*In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.

No — take a counterexample, \(r=5\)

Let distance of CD = 1, so distance of AB \(=2\times1=2\)[given ratio]

\(CD=2\sqrt{25-1^2}=2\sqrt{24}\approx9.80\)

\(AB=2\sqrt{25-2^2}=2\sqrt{21}\approx9.17\)

\(2\,AB\approx18.33\), but \(CD\approx9.80\ne 2\,AB\)

Chord \(=2\sqrt{r^2-d^2}\) is not linear in d[doubling d doesn't double or halve the chord]

No — CD ≠ 2AB in general, as shown by the counterexample (r = 5, distances 1 and 2 give CD ≈ 9.80, not 18.33).

Exercise Set 5.6

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

\(OA=OB=12\) cm[radii]

\(\angle AOB=60^\circ\)[given]

\(\angle OAB=\angle OBA=\dfrac{180^\circ-60^\circ}{2}=60^\circ\)[isosceles △OAB, base angles equal]

All three angles are 60° — △OAB is equilateral

\(AB=OA=OB=12\) cm

AB = 12 cm (△OAB is equilateral since it's isosceles with a 60° apex angle).

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

Chord AB with X, Y both on the lower arc — same-side points always subtend equal angles

(i) No

Angle subtended by AB at any point on the same arc = half the central angle of the far arc[angle at centre = 2 × angle at circumference]

This depends only on which arc the point lies on, not the specific point

So no such X, Y exist on the same side

(ii) Mostly yes, with one exception

Points on opposite arcs subtend supplementary angles[sum to 180°]

Two supplementary angles can be equal only if each is 90°

Exception: \(\angle AXB=\angle AYB=90^\circ\) — happens exactly when AB is a diameter[angle in a semicircle = 90°]

Then every point on the circle subtends 90°, so X, Y could be on opposite sides

(iii) Yes, provided X and Y are on the same side of AB

AB subtends equal angles at X, Y on the same side ⇒ A, B, X, Y are concyclic

So the circle through A, B, X automatically passes through Y too

(i) No — same-side points always subtend equal angles. (ii) Yes, unless the common angle is 90° (AB a diameter), in which case X, Y could be on either side. (iii) Yes — provided X, Y are on the same side of AB.

3Find x in Fig. 5.26.

Fig. 5.26: Cyclic quadrilateral ADCB with ∠D = 100°, ∠B = x

ADCB is a cyclic quadrilateral, D and B opposite vertices

\(\angle D + \angle B = 180^\circ\)[opposite angles of a cyclic quadrilateral sum to 180°]

\(100^\circ + x = 180^\circ\)

\(x = 80^\circ\)

x = 80°.

ExExercise: A cyclic quadrilateral has angles measuring \(\angle A = 80^\circ\), \(\angle B = 110^\circ\), \(\angle C = 100^\circ\), and \(\angle D = 70^\circ\). Can such a quadrilateral be drawn? Explain why or why not.

A quadrilateral is cyclic exactly when both pairs of opposite angles sum to 180°[converse: opposite angles summing to 180° ⇒ concyclic]

Pair A, C: \(\angle A+\angle C=80^\circ+100^\circ=180^\circ\)✓

Pair B, D: \(\angle B+\angle D=110^\circ+70^\circ=180^\circ\)✓

Sanity check: \(80+110+100+70=360^\circ\)[all 4 angles of any quadrilateral sum to 360°]

Yes, such a cyclic quadrilateral can be drawn — both pairs of opposite angles sum to 180°.

End-of-Chapter Exercises

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

Chord \(=2\sqrt{r^2-d^2}=2\sqrt{13^2-5^2}\)

\(=2\sqrt{144}=24\) cm

Chord length = 24 cm.

2An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

Angle at the centre = 2 × angle at a point on the circle

So angle at the circle = half the central angle

\(\dfrac{70^\circ}{2}=35^\circ\)

The angle subtended at a point on the circle is 35°.

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

Radius \(r=\dfrac{26}{2}=13\) cm

Half-chord \(=\dfrac{24}{2}=12\) cm

\(d=\sqrt{r^2-12^2}=\sqrt{169-144}=5\) cm[Pythagoras]

The distance from the centre to the chord is 5 cm.

4A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?

Chord \(=2\sqrt{r^2-d^2}=2\sqrt{15^2-9^2}\)

\(=2\sqrt{144}=24\) cm

Chord length = 24 cm.

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

Chord AB with radii OA, OB (dashed) — the perpendicular bisector of AB (gold) passes through O

Given: AB a chord of a circle, centre O

To Prove: O lies on the perpendicular bisector of AB

\(OA=OB\)[radii]

So O is equidistant from A and B

The perpendicular bisector of AB is the locus of points equidistant from A, B

\(\therefore\) O lies on the perpendicular bisector of AB[hence proved]

OA = OB (radii) ⟹ O lies on the perpendicular bisector of AB.

6The diameter of a circle is AB. Point C is on the circumference. What is the measure of the \(\angle ACB\)? Explain your reasoning.

AB is a diameter, C is on the circle — ∠ACB is always 90°

The arc from A to B (not through C) subtends 180° at the centre[AB is a diameter — that arc is half the circle]

Angle at the centre = 2 × angle at a point on the circle

Angle at C \(=\dfrac{180^\circ}{2}=90^\circ\)

∠ACB = 90° — the angle in a semicircle is always a right angle.

7ABCD is a cyclic quadrilateral inscribed in a circle. If \(\angle A\) measures 75°, what is the measure of \(\angle C\)? If \(\angle B\) measures 110°, what is the measure of \(\angle D\)?

Opposite angles of a cyclic quadrilateral sum to 180°

\(\angle A+\angle C=180^\circ \Rightarrow \angle C=180^\circ-75^\circ=105^\circ\)

\(\angle B+\angle D=180^\circ \Rightarrow \angle D=180^\circ-110^\circ=70^\circ\)

∠C = 105°, ∠D = 70°.

8Quadrilateral PQRS is inscribed in a circle. If \(\angle P = (2x+10)^\circ\) and \(\angle R = (3x-20)^\circ\), find the value of x and the measures of \(\angle P\) and \(\angle R\).

P, R opposite vertices: \(\angle P+\angle R=180^\circ\)[opposite angles of a cyclic quadrilateral sum to 180°]

\((2x+10)+(3x-20)=180\)

\(5x-10=180 \Rightarrow 5x=190 \Rightarrow x=38\)

\(\angle P=2(38)+10=86^\circ\)

\(\angle R=3(38)-20=94^\circ\)

Check: \(86^\circ+94^\circ=180^\circ\)✓

x = 38, ∠P = 86°, ∠R = 94°.

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

Half-chord \(=\dfrac{16}{2}=8\) cm

\(r=\sqrt{8^2+6^2}=\sqrt{100}=10\) cm[Pythagoras]

The radius of the circle is 10 cm.

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

Kite ABCD: AB = DA = 5, BC = CD = 12, right angles at B and D, AC = 13 is a diameter

Sides 5, 5, 12, 12 in kite order — this is a kite[two pairs of adjacent equal sides]

By the kite's symmetry, the two angles where a 5-side meets a 12-side are equal

These are opposite angles of the cyclic quadrilateral, so they sum to 180°[opposite angles of a cyclic quadrilateral sum to 180°]

Equal and summing to 180° ⇒ each is 90°

Kite = 2 congruent right triangles (legs 5, 12), joined on a common hypotenuse

Hypotenuse \(=\sqrt{5^2+12^2}=\sqrt{169}=13\)[Pythagoras]

This diagonal is a diameter of the circumcircle[angle in a semicircle = 90°, matching the right angles found]

Area of one right triangle \(=\dfrac12\times5\times12=30\) sq units

Total \(=2\times30=60\) sq units

Area = 60 square units.

11*Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?

Construct the perpendicular bisectors of any 2 sides (say AB, CD)[ruler and compass]

Their intersection is the circumcentre O[O is equidistant from A, B and from C, D]

Check by inspection whether O lies inside or outside the quadrilateral's boundary

Construct the perpendicular bisectors of any two sides; their intersection gives O without drawing the circle. Then check whether O lies inside or outside the quadrilateral's boundary.

12*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.

Equal chords AB, CD (AB = CD) meeting at P — the theorem shows AP = CP and PB = PD

Given: Chords AB, CD of a circle (centre O) meet at P, \(AB=CD\); M, N are midpoints of AB, CD

To Prove: \(AP=CP\) and \(PB=PD\)

\(OM=ON\)[equal chords are equidistant from the centre]

In \(\triangle OMP\) and \(\triangle ONP\):

\(OP=OP\)[common]

\(OM=ON\)[shown above]

\(\angle OMP=\angle ONP=90^\circ\)[perpendicular from centre to a chord]

\(\triangle OMP \cong \triangle ONP\)[RHS congruence]

\(MP=NP\)

\(AM=\dfrac{AB}{2}=\dfrac{CD}{2}=CN\)[M, N are midpoints, AB = CD]

\(AP=AM+MP=CN+NP=CP\), and similarly \(PB=PD\)[hence proved]

13*Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)

Chord AB = 6 cm at perpendicular distance OM = 3 cm from centre O (radius = 3√2 cm)

Chord \(=2\sqrt{r^2-d^2}\), with chord = 6, \(d=3\)

\(6=2\sqrt{r^2-9} \Rightarrow 3=\sqrt{r^2-9}\)

\(r^2=18 \Rightarrow r=3\sqrt2\approx4.24\) cm

Construction: Draw \(AB=6\) cm

Bisect AB at M, draw \(MO\perp AB\) with \(OM=3\) cm

With centre O, radius \(OA=\sqrt{OM^2+MA^2}=3\sqrt2\) cm, draw the circle[Pythagoras]

Required radius = 3√2 cm ≈ 4.24 cm. Draw AB = 6 cm, erect a perpendicular at its midpoint M with OM = 3 cm, then draw the circle centred at O with radius OA = 3√2 cm.

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

Cyclic parallelogram ABCD — the proof forces every angle to 90°, so it must be a rectangle

Given: ABCD is a cyclic parallelogram

To Prove: ABCD is a rectangle

\(\angle A=\angle C\), \(\angle B=\angle D\)[opposite angles of a parallelogram are equal]

\(\angle A+\angle C=180^\circ\)[ABCD is cyclic — opposite angles sum to 180°]

\(2\angle A=180^\circ \Rightarrow \angle A=90^\circ\)

Similarly \(\angle B=90^\circ\)

All angles = 90° ⇒ ABCD is a rectangle[hence proved]

∠A = ∠C (parallelogram) and ∠A + ∠C = 180° (cyclic) together force ∠A = 90°, and similarly all angles = 90° — so ABCD must be a rectangle.

15*Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.

Rectangle ABCD inscribed in a circle — diagonals AC, BD (dashed) intersect at the centre O

Given: ABCD is a rectangle inscribed in a circle

To Prove: diagonals AC, BD meet at the centre

\(\angle ABC=90^\circ\) is the angle subtended by AC at B[angle of a rectangle]

\(\therefore\) AC is a diameter[an angle of 90° on the circle means the chord is a diameter]

Similarly \(\angle BAD=90^\circ\) is subtended by BD at A[angle of a rectangle]

\(\therefore\) BD is also a diameter

Two diameters of a circle meet only at the centre

\(\therefore\) AC and BD meet at the centre[hence proved]

Both diagonals are diameters (each subtends 90° at the opposite vertex), and diameters meet only at the centre.

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

Three chords of equal length (gold), each with its midpoint (green dots) — all lying on a smaller concentric circle (dashed)

Let the circle have radius r, centre O; all chords have fixed length L

Every chord of length L is at the same distance \(d=\sqrt{r^2-\left(\tfrac{L}{2}\right)^2}\) from O[equal chords are equidistant from the centre]

The midpoint of each chord is the foot of the perpendicular from O[perpendicular from centre to a chord meets it at the midpoint]

So every midpoint is at distance d from O

As the chord's direction varies, its midpoint sweeps out every point at distance d from O

The midpoints form a circle, concentric with the original, of radius \(\sqrt{r^2-\left(\dfrac{L}{2}\right)^2}\).

17*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 \(\angle BAC\)".

Congruent chords AB = AC, with M, N the feet of the perpendiculars from O — △OMA ≅ △ONA

Given: \(AB=AC\), chords of a circle, centre O; M, N midpoints of AB, AC

To Prove: AO bisects \(\angle BAC\)

\(OM=ON\)[equal chords are equidistant from the centre]

\(OM\perp AB\), \(ON\perp AC\)[perpendicular from centre to a chord]

In \(\triangle OMA\) and \(\triangle ONA\):

\(OA=OA\)[common]

\(OM=ON\)[shown above]

\(\angle OMA=\angle ONA=90^\circ\)

\(\triangle OMA \cong \triangle ONA\)[RHS congruence]

\(\angle OAB=\angle OAC\)

\(\therefore\) AO bisects \(\angle BAC\)[hence proved]

△OMA ≅ △ONA (RHS) ⟹ ∠OAB = ∠OAC ⟹ AO bisects ∠BAC.

18Two 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.

Both chords on the same side of O — the 24 cm chord is closer, the 10 cm chord is farther, 7 cm apart

Half-chords: \(\dfrac{10}{2}=5\), \(\dfrac{24}{2}=12\)

Distance to 10 cm chord: \(d_1=\sqrt{r^2-25}\); to 24 cm chord: \(d_2=\sqrt{r^2-144}\)[Pythagoras]

The longer chord (24 cm) is closer to the centre[longer chord ⇒ closer to centre]

Same side, so distance between them = \(d_1-d_2=7\)

\(\sqrt{r^2-25}-\sqrt{r^2-144}=7\)

\(\sqrt{r^2-25}=7+\sqrt{r^2-144}\)

Squaring: \(r^2-25=49+14\sqrt{r^2-144}+r^2-144\)

\(-25=-95+14\sqrt{r^2-144}\)

\(70=14\sqrt{r^2-144} \Rightarrow 5=\sqrt{r^2-144}\)

\(25=r^2-144 \Rightarrow r^2=169 \Rightarrow r=13\) cm

Check: \(d_1=\sqrt{144}=12\), \(d_2=\sqrt{25}=5\), \(d_1-d_2=7\)✓

The radius of the circle is 13 cm.

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

Regular hexagon inscribed in a circle — split into 6 equilateral triangles (one shaded)

Each side subtends a central angle of \(\dfrac{360^\circ}{6}=60^\circ\)

In \(\triangle OAB\): \(OA=OB=r\), \(\angle AOB=60^\circ\)[O centre, A, B adjacent vertices]

Base angles \(=\dfrac{180^\circ-60^\circ}{2}=60^\circ\) — the triangle is equilateral

\(\therefore AB=r\)

Distance from centre: \(d=\sqrt{r^2-(\text{side}/2)^2}\)

\(d=\sqrt{r^2-\dfrac{r^2}{4}}=\dfrac{\sqrt3}{2}r\)

Side length = r. Distance from centre to each side = \(\dfrac{\sqrt3}{2}r\).

20A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about \(\angle MOP\) and \(\angle MNP\)? Explain your reasoning.

MNOP inscribed with MN a diameter; ∠MOP and ∠MNP both subtend chord MP (purple) from the same arc

\(\angle MOP\) and \(\angle MNP\) are both angles subtended by MP, at O and N

In cyclic order M, N, O, P, both N and O lie on the same arc of chord MP

\(\therefore \angle MOP=\angle MNP\)[angles subtended by the same chord from the same arc are equal]

∠MOP = ∠MNP, since both are angles subtended by MP from points on the same arc.

21Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., \(\angle CDE = \angle ABC\), where E is a point on the extension of side AD beyond D).

Cyclic quadrilateral ABCD with side AD extended to E — ∠CDE is the exterior angle at D

A, D, E are collinear, so \(\angle CDE\) and \(\angle CDA\) are supplementary[angles on a straight line]

\(\angle CDE+\angle CDA=180^\circ\) …(1)

ABCD is cyclic, so \(\angle CDA+\angle ABC=180^\circ\) …(2)[opposite angles of a cyclic quadrilateral sum to 180°]

From (1), (2): both \(\angle CDE\) and \(\angle ABC\) equal \(180^\circ-\angle CDA\)

\(\therefore \angle CDE=\angle ABC\)

∠CDE = 180° − ∠CDA (angles on a line) = ∠ABC (since ∠CDA + ∠ABC = 180° too) — the exterior angle at any vertex of a cyclic quadrilateral equals the interior angle at the opposite vertex.

22*"There is no chord of a circle that is longer than its diameter." How do you justify this statement?

Chord \(=2\sqrt{r^2-d^2}\), where \(d\ge0\)

Maximised when \(d=0\), giving chord \(=2\sqrt{r^2}=2r\)[the diameter]

Since d can't be negative, no chord exceeds 2r

Chord = 2√(r² − d²) is maximised at d = 0, giving 2r (the diameter) — so no chord can be longer than the diameter.

23*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.

Point A inside the circle — the chord through A perpendicular to OA (teal) is shorter than any other chord through A (dashed purple)

Let \(OA=k\)

Any chord through A has distance \(d'\le OA=k\) from O[equality only when the chord ⟂ OA at A]

Chord \(=2\sqrt{r^2-d'^2}\), which decreases as \(d'\) increases

Chord is shortest when \(d'\) is largest, i.e. \(d'=k\)

This happens exactly when the chord ⟂ OA at A

The chord ⟂ OA at A has the largest possible distance from O (= OA), hence gives the shortest chord, with length 2√(r² − OA²).

24How would you use the following figure to justify the statement that the angle in a semicircle is 90°?

Fig. 5.30: Angle in a semicircle, base angles a and b

Join OA (dashed)[OA and the base radii are all radii — 2 isosceles triangles]

Base angles a and b as marked

Angle at A \(=a+b\)

Angle sum of the big triangle: \(a+b+(a+b)=180^\circ\)

\(2(a+b)=180^\circ \Rightarrow a+b=90^\circ\)

\(\therefore\) angle at A \(=90^\circ\)

a + b = 90° (angle sum of the big triangle), and the angle at A equals a + b — so the angle in a semicircle is 90°.

25*In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C'D' is perpendicular to AB.

CC′ and DD′ ⟂ diameter AB; CD and C′D′ (dashed) have midpoints M, M′ — segment MM′ (red) is vertical, i.e. ⟂ AB

Given: \(CC'\perp AB\), \(DD'\perp AB\), AB a diameter; M, M′ midpoints of CD, C′D′

To Prove: \(MM'\perp AB\)

AB bisects \(CC'\) — so C, C′ are reflections across AB[AB passes through the centre, ⟂ to CC′ — bisects it]

Similarly D, D′ are reflections of each other across AB

So CD reflects to C′D′ across AB

Hence their midpoints M, M′ are also reflections across AB

A point and its reflection are joined by a segment ⟂ the mirror line

\(\therefore MM'\perp AB\)[hence proved]

C, C′ and D, D′ are reflections across AB, so M, M′ are reflections across AB too — hence MM′ ⟂ AB.

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

Fig. 5.31: Cyclic quadrilateral ABCD with centre O joined to all vertices

Join O to A, B, C, D — 4 isosceles triangles[OA = OB = OC = OD = radii]

Let base angles: \(\angle OAB=\angle OBA=p\); \(\angle OBC=\angle OCB=q\); \(\angle OCD=\angle ODC=u\); \(\angle ODA=\angle OAD=v\)

\(\angle A=v+p,\ \angle B=p+q,\ \angle C=q+u,\ \angle D=u+v\)

Sum of angles of a quadrilateral \(=360^\circ\)

\((v+p)+(p+q)+(q+u)+(u+v)=360^\circ\)

\(2(p+q+u+v)=360^\circ \Rightarrow p+q+u+v=180^\circ\)

\(\therefore \angle A+\angle C=(v+p)+(q+u)=180^\circ\)

and \(\angle B+\angle D=(p+q)+(u+v)=180^\circ\)

Both opposite angle pairs equal p + q + u + v = 180° — an alternative proof that opposite angles of a cyclic quadrilateral sum to 180°.

Extra Practice Questions

Seven extra questions in the style of the textbook's own exercises, for independent practice once you've gone through the solved questions above. Each one is shown open with its full working, since a diagram is part of the answer here — cover the solution with your hand and attempt it on paper first.

1A chord of a circle is 16 cm long and is at a distance of 6 cm from the centre. Find the radius of the circle.

Rough sketch — chord of length 16 cm, 6 cm from centre O.

Perpendicular from centre bisects the chord[perpendicular from centre to a chord bisects it]

Half the chord \(=\dfrac{16}{2}=8\) cm

In the right triangle formed: \(r^2=8^2+6^2\)[Baudhāyana–Pythagoras theorem]

\(r^2=64+36=100\)

\(r=\sqrt{100}=10\) cm

Radius of the circle = 10 cm.

2A chord AB of a circle subtends an angle of 70° at the centre O. Find the angle subtended by AB at a point on the major arc.

Rough sketch — chord AB subtends 70° at centre O, and ∠APB at a point P on the major arc.

Angle at the centre = 2 × angle at the circle (same arc)[angle subtended by an arc at the centre is double the angle at the circle]

\(\angle AOB=2\times\angle APB\)

\(70^\circ=2\times\angle APB\)

\(\angle APB=35^\circ\)

Angle subtended by AB at a point on the major arc = 35°.

3AB is a diameter of a circle with centre O. C is a point on the circle such that AC = 8 cm and BC = 6 cm. Find the radius of the circle.

Rough sketch — AB is a diameter; C lies on the circle with AC = 8 cm, BC = 6 cm.

AB is a diameter \(\Rightarrow \angle ACB=90^\circ\)[angle in a semicircle is a right angle]

In right triangle ACB: \(AB^2=AC^2+BC^2\)[Baudhāyana–Pythagoras theorem]

\(AB^2=8^2+6^2=64+36=100\)

\(AB=10\) cm

Radius \(=\dfrac{AB}{2}=\dfrac{10}{2}=5\) cm[radius = half the diameter]

Radius of the circle = 5 cm.

4PQRS is a cyclic quadrilateral in which ∠P = 3x and ∠R = (2x + 10)°. Find the value of x and the measure of ∠P.

Rough sketch — cyclic quadrilateral PQRS with ∠P = 3x, ∠R = (2x+10)°.

Opposite angles of a cyclic quadrilateral sum to 180°[cyclic quadrilateral property]

\(\angle P+\angle R=180^\circ\)

\(3x+(2x+10)=180\)

\(5x+10=180 \Rightarrow 5x=170\)

\(x=34\)

\(\angle P=3x=3\times34=102^\circ\)

x = 34; ∠P = 102°.

5Two circles of radii 10 cm and 8 cm intersect at two points, and the distance between their centres is 12 cm. Find the length of the common chord.

Rough sketch — circles of radii 10 cm, 8 cm, centres 12 cm apart, common chord AB.

Let the common chord AB meet \(O_1O_2\) at M, and let \(O_1M=d\), so \(O_2M=12-d\)[centre line is the perpendicular bisector of the common chord]

In right triangle \(O_1MA\): \(AM^2=10^2-d^2\)[Baudhāyana–Pythagoras theorem]

In right triangle \(O_2MA\): \(AM^2=8^2-(12-d)^2\)

\(100-d^2=64-(144-24d+d^2)\)[equating the two expressions for AM²]

\(100-d^2=64-144+24d-d^2\)

\(100=-80+24d \Rightarrow 24d=180\)

\(d=7.5\) cm

\(AM^2=10^2-7.5^2=100-56.25=43.75\)

\(AM=\sqrt{43.75}\approx6.61\) cm

\(AB=2\times AM\approx13.23\) cm[centre line bisects the common chord]

Length of the common chord ≈ 13.23 cm.

6In a circle, chord AB = 30 cm is at a distance of 8 cm from the centre O. Another chord CD of the same circle is at a distance of 15 cm from O. Which chord is longer, and what is the length of CD?

Rough sketch — chord AB (8 cm from O) and chord CD (15 cm from O), same circle.

First find the radius using chord AB[AB = 30 cm, 8 cm from O]

Half of AB \(=15\) cm

\(r^2=15^2+8^2=225+64=289\)[Baudhāyana–Pythagoras theorem]

\(r=17\) cm

Now find CD using the same radius, distance 15 cm

Half of CD \(=\sqrt{r^2-15^2}=\sqrt{289-225}=\sqrt{64}=8\) cm

\(CD=16\) cm

Since \(CD(16)[the chord closer to the centre is the longer chord]

AB is the longer chord; CD = 16 cm.

7Find the radius of the circumcircle of a right-angled triangle whose legs are 9 cm and 12 cm.

Rough sketch — right triangle ABC, right angle at A, legs 9 cm and 12 cm.

Find the hypotenuse: \(BC^2=AB^2+AC^2=9^2+12^2\)[Baudhāyana–Pythagoras theorem]

\(BC^2=81+144=225\)

\(BC=15\) cm

The right angle at A subtends BC \(\Rightarrow\) BC must be a diameter of the circumcircle[converse: a 90° angle on the circle is subtended only by a diameter]

Circumradius \(=\dfrac{BC}{2}=\dfrac{15}{2}=7.5\) cm

Radius of the circumcircle = 7.5 cm.

Frequently Asked Questions

Through 1 point, infinitely many circles can pass (any radius, any centre at that distance). Through 2 points, infinitely many circles can pass, with centres lying on the perpendicular bisector of the segment joining them. Through 3 non-collinear points, exactly one circle can pass — its circumcircle. Through 3 collinear points, no circle can pass.

Chord length = 2√(r² − d²), where r is the radius and d is the perpendicular distance from the centre to the chord. This comes directly from the Baudhāyana–Pythagoras theorem applied to the right triangle formed by the radius, the distance, and half the chord.

The angle subtended by an arc at the centre of a circle is always double the angle it subtends at any point on the circle outside that arc. A special case: the angle subtended by a diameter (which corresponds to a 180° arc at the centre) at any point on the circle is always 90°.

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a single circle. Its defining property is that each pair of opposite angles sums to 180° — and conversely, if a quadrilateral's opposite angles sum to 180°, it must be cyclic.

The longer of two chords is always closer to the centre. The diameter (the longest possible chord) passes through the centre, so its distance from the centre is 0.

Chapter 5I'm Up and Down, and Round and Round

Key Concepts & Theorems at a Glance

Think and Reflect

Think and Reflect

Think and Reflect

Exercise Set 5.1

Think, Draw and Infer

Exercise Set 5.2

Exercise Set 5.3

Exercise Set 5.4

Exercise Set 5.5

Exercise Set 5.6

End-of-Chapter Exercises

Extra Practice Questions

Frequently Asked Questions