NCERT Solutions Class 9 Maths Chapter 5 I M Up and Down and Round and Round

F R E E S T U D Y M AT E R I A L F O R E V E R Y S T U D E N T

C L A S S 9 · M AT H S

NCERT Solutions

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

NCERT Textbook — Ganita Manjari

BOOK PAGES SECTIONS QUESTIONS MEDIUM

92 – 117 17 64 English

Solutions, notes, sample papers & more at 75 pages

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

CLASS 9 · MATHS · GANITA MANJARI

NCERT Solutions — Chapter 5: I'm Up and Down, and
Round and Round
Chapter 5 builds the circle up from one idea — every point of a circle is the same distance from its centre
— and squeezes twelve theorems out of it. Chords, the angles they subtend, their distance from the centre,
arcs and cyclic quadrilaterals all follow from that single definition plus the congruence rules you already know.

TEXTBOOK BOOK PAGES

Ganita Manjari (Class 9) 92 – 117

SECTIONS QUESTIONS

17 64

MEDIUM

English

In-text Questions — Pages 92–93
Chapter opening

Q1 Can you recognise the origin of the shapes in Fig. 5.1?

Rain falling on still water Cross-section of a plant stem Sunflower head

Fig. 5.1, page 92 — redrawn sketch of the three photographs in the book.

All three pictures in Fig. 5.1 come from nature, and in each of them the shape is a circle.

Left: raindrops striking still water. Each drop sends out a ripple that spreads outward at the
same speed in every direction, so every point of the ripple stays at the same distance from
the point where the drop fell — a circle.
Middle: the cross-section of a tree trunk (a plant stem). The growth rings are nearly circular,
because the stem thickens outward by roughly the same amount all round.

Page 1 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Right: the head (inflorescence) of a sunflower. The florets are packed around a single centre,
and the outer boundary is a circle.

Why it happens: whenever something grows or spreads equally in all directions from
one point, the boundary it reaches is the set of points at a fixed distance from that
point. That is exactly the definition of a circle. The cave painters of Gudahandi drew
circles because nature kept showing them this shape.

Try This: the Moon and the Sun in Fig. 5.2 look circular too — they are spheres, and
the outline of a sphere seen from far away is a circle.

Q2 Activity: List some objects from nature that resemble a circle.

Any object that spreads out equally in all directions from a centre will look circular. A few from
around us:

Ripples made by a stone dropped into a pond.
The full Moon, and the Sun during a total solar eclipse.
The annual rings on the cut end of a log; the cross-section of a bamboo or a sugarcane stem.
The face of a sunflower; the flat top of a lotus seed-pod.
Cross-sections of fruits — orange, lemon, tomato.
A raindrop resting on a leaf; a soap bubble seen from above.
The pupil and iris of an eye; the outline of a bird's nest.

Check it yourself: for each object, ask "is there one point inside from which every
boundary point is the same distance?" If the answer is yes, the shape really is a
circle; if not (a mango, an egg), it is an oval, not a circle.

Think and Reflect — Page 93

Page 2 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Before Section 5.1 Definitions

THINK AND REFLECT

Q1 Jamuna 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 almost certainly told her: fold the paper in half twice; the point where the two creases
cross is the centre.
Here is the procedure and the reason behind each step.

1. Fold the circular paper so that the boundary falls exactly on itself. Open it. The crease is a
line of reflection symmetry of the circle, so it is a diameter — and every diameter passes
through the centre.
2. Fold again in a different direction so the boundary again falls on itself. This crease is a
second diameter.
3. Two different diameters both pass through the centre, and two distinct lines meet at only
one point. So their point of intersection is the centre.

crease 2

crease 1
centre

Two folds that make the boundary overlap give two diameters; they cross at the centre.

Page 3 of 75

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q2 What is the length of the longest chord in a circle of radius 5 units? Is there a
smallest chord?

The longest chord is the diameter.

Longest chord = 2 × radius = 2 × 5 = 10 units

Is there a smallest chord? No. There is no chord of least length.

Why it happens: a chord at distance d from the centre has length 2√(25 − d²), and d
can be any value with 0 ≤ d < 5. As d creeps up towards 5 the chord gets shorter and
shorter — 2√(25 − 24) = 2, then 2√(25 − 24.99) = 0.2, and so on — but it never actually
reaches 0, because at d = 5 the line only touches the circle at a single point and is no
longer a chord. So chords can be made as short as you please, and there is no
shortest one.

Tip: "longest" exists because d = 0 is allowed; "shortest" fails because d = 5 is not.

Q3 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? (Hint: We know
that any point that is equidistant from two given points A and B lies on the
perpendicular bisector of AB. Does this make the perpendicular bisector the locus?
For this, we have to show that all the points on the perpendicular bisector are
equidistant from A and B.)

The locus of points equidistant from two given points A and B is the perpendicular bisector of
AB.
A locus claim always needs two statements proved, not one:

1. Every point of the locus is on the line. Let P satisfy PA = PB, and let M be the midpoint of
AB. Then in ΔPMA and ΔPMB: PA = PB, AM = BM, PM common. By SSS, ΔPMA ≅ ΔPMB, so
∠PMA = ∠PMB. These are angles on a line, so each is 90°. Hence PM ⊥ AB and passes
through the midpoint — P lies on the perpendicular bisector.
2. Every point of the line is in the locus. Let Q be any point on the perpendicular bisector,
meeting AB at its midpoint M. In ΔQMA and ΔQMB: QM common, ∠QMA = ∠QMB = 90°, AM =
BM. By SAS, ΔQMA ≅ ΔQMB, so QA = QB.

Page 5 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

P is equidistant from A and B ⇔ P lies on the perpendicular bisector of AB

So the locus is exactly the perpendicular bisector of AB.

Why both halves matter: part 1 alone would allow the locus to be some part of the
line; part 2 alone would allow extra points off the line. Only together do they pin the
locus down exactly. The chapter uses this result immediately: the centres of all circles
through A and B are precisely the points of this line.

Think and Reflect — Page 95
Section 5.3 How Many Circles?

THINK AND REFLECT

Q1 How many circles pass through two points on a plane?

Infinitely many.

Why it happens: a circle through A and B must have its centre O at a point with OA =
OB, i.e. on the perpendicular bisector of AB. Conversely, every point O of that
perpendicular bisector gives a circle: take radius OA, and since OB = OA the circle
passes through B too. The perpendicular bisector contains infinitely many points, so
there are infinitely many such circles — one for each point of the line.

Page 6 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

⊥ bisector of AB

A B

Every point of the perpendicular bisector of AB is the centre of a circle through A and B.

Q2 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?

No — not all radii are possible. Only radii of at least half of AB occur.

Let AB = c and let the centre O be at distance h from the midpoint M of AB.

Then OA² = OM² + MA², i.e. r² = h² + (c/2)²

So r = √(h² + c²/4) ≥ c/2 for every h ≥ 0.

Smallest radius = c/2 = ½ AB, obtained when h = 0, i.e. when the centre is the midpoint of AB.
For that circle AB is a diameter.
Largest radius: there is none. As h increases, r increases without bound, so the radii can be
made as large as we like but never reach a maximum.

Page 7 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Why it happens: AB is a chord of every such circle, and no chord can be longer than
the diameter. So 2r ≥ AB always, which is exactly r ≥ ½AB. Radii smaller than ½AB are
ruled out; every radius from ½AB upwards actually occurs.

Q3 As you move away from segment AB along its perpendicular bisector, do the radii of
the circles containing A and B increase or decrease?

The radii increase.

r = √(h² + (AB/2)²) where h = distance of the centre from the midpoint of AB

h ↑ ⇒ h² ↑ ⇒ h² + (AB/2)² ↑ ⇒ r ↑

Why it happens: A stays on the circle, so the radius is the distance OA. Moving O
further away from AB along the perpendicular bisector simply moves it further from
A as well — the right triangle OMA gets a longer vertical leg while its horizontal leg
MA stays fixed at ½AB, so the hypotenuse OA grows.

Q4 As you go along the perpendicular bisector, will the circle drawn from that point
through A and B appear more curved or less curved?

Less curved — the arc through A and B gets flatter and flatter.

Why it happens: curvature is decided by the radius: a small circle bends sharply, a
large circle bends gently. From Q3 the radius grows as you move out along the
perpendicular bisector, so near A and B the arc becomes flatter. In the limit, as the
centre races off to infinity, the arc through A and B straightens into the line AB itself.

Did you know? This is why the horizon looks straight. You are standing on a circle of
radius about 6400 km, so over a few metres the curvature is far too gentle to notice.

Page 8 of 75

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Why the sharp contrast: "on the boundary" leaves the size and the tilt free, so there
is a whole family of answers. "As corners" fixes one full side or one full diagonal, and
a square is completely determined once you know a side (plus which way to build it)
or a diagonal.

In-text Questions — Page 96
Section 5.3 How Many Circles?

Q1 What if A, B and C lie on a straight line, i.e., are collinear? Can you explain why, in
this case, there is no circle through A, B and C?

There is no circle at all through three collinear points.
Given: A, B, C on one line, all different. To show: no point O has OA = OB = OC.

Suppose such a circle existed, with centre O.

OA = OB ⇒ O lies on the perpendicular bisector of AB

OB = OC ⇒ O lies on the perpendicular bisector of BC

But AB and BC lie along the same line ℓ.

So both perpendicular bisectors are perpendicular to ℓ ⇒ they are parallel to each other.

They pass through different points (the midpoints of AB and of BC are different), so they are

distinct parallel lines — they never meet.

Hence no such point O exists. No circle passes through three collinear points.

Why it happens: Theorem 1 works because two perpendicular bisectors of a
genuine triangle are not parallel, so they cross at exactly one point. Collinearity
destroys precisely that step: the two bisectors become parallel, the meeting point
vanishes, and with it the centre.

Check it yourself: the same fact read the other way round says a straight line can
meet a circle in at most 2 points — never 3.

Page 10 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Exercise Set 5.1 — Page 98
Section 5.3 How Many Circles?

Q1 Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is
the centre inside or outside the triangle?

The centre lies inside the triangle.
Construction:

1. Draw AB = 5 cm.
2. At A draw a ray making 70° with AB; at B draw a ray making 60° with BA. They meet at C.
3. Construct the perpendicular bisectors of AB and of BC. They meet at O.
4. With centre O and radius OA, draw the circle. It passes through B and C as well.

∠C = 180° − 70° − 60° = 50°

Angles are 70°, 60°, 50° — all acute, so ΔABC is an acute-angled triangle.

For an acute-angled triangle the circumcentre lies inside the triangle.

Check it yourself: measure OA, OB, OC — all three should come out about 3.3 cm.
(Exactly, R = AB / (2 sin C) = 5 / (2 sin 50°) ≈ 3.26 cm.)

Why it happens: the circumcentre O sees side AB under the angle ∠AOB = 2∠C. O
falls inside the triangle exactly when all three of these central angles 2∠A, 2∠B, 2∠C
are less than 180° — that is, when every angle of the triangle is less than 90°.

Q2 Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is
the centre inside or outside the triangle?

The centre lies outside the triangle — on the far side of BC from A.
Construction: draw AB = 5 cm; at A draw a ray making 100° with AB and cut off AC = 4 cm on it;
join BC. Then draw the perpendicular bisectors of AB and AC; they meet at O, outside the
triangle. Draw the circle with centre O and radius OA.

Page 11 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

∠A = 100° > 90° ⇒ ΔABC is an obtuse-angled triangle.

For an obtuse-angled triangle the circumcentre lies outside the triangle,

beyond the side opposite the obtuse angle — here beyond BC.

Check it yourself: BC ≈ 6.9 cm and OA = OB = OC ≈ 3.5 cm.
(BC² = 5² + 4² − 2·5·4·cos 100° ≈ 47.9, so BC ≈ 6.92 cm; R = BC / (2 sin 100°) ≈ 3.52 cm.)

Why it happens: the central angle standing on BC is 2∠A = 200°, which is a reflex
angle. So A lies on the minor arc BC, and the centre is pushed across BC to the other
side, out of the triangle.

Q3 Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC.
Let the circumcentre be O. Measure OA, OB, OC.

OA = OB = OC ≈ 3.9 cm. All three must be equal — they are radii of the same circle.
Construction: draw AB = 6 cm; with centre A and radius 7 cm, and centre B and radius 7 cm,
draw arcs cutting at C. Construct the perpendicular bisectors of AB and BC; they meet at O.
Draw the circle with centre O, radius OA.

The triangle is isosceles (CA = CB = 7 cm), so it is symmetric about the perpendicular

bisector of AB.
Height from C to AB = √(7² − 3²) = √40 ≈ 6.32 cm

Area = ½ × 6 × 6.32 ≈ 18.97 cm²

R = (abc) / (4 × area) = (6 × 7 × 7) / (4 × 18.97) = 294 / 75.9 ≈ 3.87 cm

Tip: the angles here are about 64.6°, 64.6° and 50.8° — all acute — so O falls inside
the triangle, on the perpendicular bisector of AB, about 2.45 cm above AB.

Page 12 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Why OA = OB = OC: that is what "circumcentre" means. O is on the perpendicular
bisector of AB (so OA = OB) and on the perpendicular bisector of BC (so OB = OC).
The two facts together force all three to be equal, which is exactly why a single circle
can pass through all three vertices.

Q4 What is the least possible radius of a circle through two points A and B?

The least possible radius is ½ AB — half the distance between the two points.

AB is a chord of any such circle.

No chord can exceed the diameter, so AB ≤ 2r, i.e. r ≥ AB/2.

The value r = AB/2 is actually reached: take the midpoint M of AB as centre.

Then MA = MB = AB/2, and the circle with centre M, radius AB/2, passes through both.

Least radius = AB/2, and for that circle AB is a diameter.

Why it happens: the centre must sit on the perpendicular bisector of AB, and its
distance to A is √(h² + (AB/2)²), where h is how far it has moved from the midpoint.
This is smallest when h = 0. Any move away from the midpoint only lengthens the
radius.

Think, Draw and Infer — Page 98

Page 13 of 75

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

Section 5.3 How Many Circles?
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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Both bisectors stand at right angles to the same line, so they are parallel and never meet — there is
no circumcentre.

No circle through collinear points: a circle needs a centre equidistant from all three, i.e.
exactly the point P we have just shown cannot exist.
No line cuts a circle in three points: if a line met a circle at three points, those three points
would be collinear and concyclic — impossible by the previous line. A line meets a circle in 0, 1 or
at most 2 points.

Why it happens: going from two points to three is what makes a circle unique — but
only if the third point steps off the line. Collinearity is exactly the degenerate case
where the third condition adds nothing new and the construction collapses.

Q2 The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent
to ΔABC that share the same circumcircle?

Yes — infinitely many.

Why it happens: a circle has complete rotational symmetry about its centre. Rotate
ΔABC about the circumcentre O through any angle θ. Distances from O do not
change, so the three image vertices A′, B′, C′ still lie on the same circle, and every
side length is preserved — so ΔA′B′C′ ≅ ΔABC and it has the same circumcircle. Since
θ can be any of infinitely many angles, there are infinitely many such triangles.

There is a second family too: reflect ΔABC in any diameter. The circle maps to itself and the
triangle maps to a congruent (mirror-image) triangle inscribed in it.

Page 15 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

O

Rotating an inscribed triangle about O gives a congruent triangle in the same circle.

Tip: the uniqueness in Theorem 1 runs the other way. Fix the three points and the
circle is unique. Fix the circle and there are endlessly many congruent triangles
inscribed in it.

Exercise Set 5.2 — Page 100
Section 5.4 Chords and the Angles They Subtend

Q1 Show that the triangle formed by a chord and the centre of the circle is isosceles.

Given: a circle with centre O and a chord AB. Join OA and OB. To show: ΔOAB is isosceles.

Page 16 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

OA = radius of the circle

OB = radius of the circle

∴ OA = OB

A triangle with two equal sides is isosceles, so ΔOAB is isosceles with base AB.

Consequence used again and again in this chapter: the base angles are equal, ∠OAB =
∠OBA.

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r r

A B

Two sides of ΔOAB are radii, so they are equal — the triangle is isosceles.

Why this is worth stating: every proof about chords in this chapter starts by joining
the ends of the chord to the centre. The instant you do that you own two equal sides
for free, and isosceles triangles hand you equal base angles. That is the engine
behind Theorems 2, 3, 4 and 9.

Page 17 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q2 Show that if two such isosceles triangles (occurring in the previous question) have
equal base length, they are congruent to each other.

Given: chords AB and DE of the same circle, centre C, with AB = DE. To show: ΔCAB ≅ ΔCDE.

CA = CB = r (radii)

CD = CE = r (radii)

⇒ CA = CD and CB = CE

AB = DE (given — the two bases are equal)

By the SSS congruence rule, ΔCAB ≅ ΔCDE
What follows at once: corresponding parts of congruent triangles are equal, so

∠ACB = ∠DCE — equal chords subtend equal angles at the centre (Theorem 2)

and the altitudes from C are equal — equal chords are equidistant from the centre

(Theorem 6)

Tip: if instead the two triangles have equal apex angles at C, use SAS (two radii and
the included angle) and you get AB = DE — that is Theorem 3, the converse.

Exercise Set 5.3 — Page 101

Page 18 of 75

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

Section 5.5 Midpoints and Perpendicular Bisectors of Chords
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Given: circle with centre C, chord AB, and CM ⊥ AB with M on AB. To show: AM = BM.
e m
g l as
a

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

In ΔCMA and ΔCMB:

∠CMA = ∠CMB = 90° (given — CM is perpendicular to AB)

CA = CB = r (radii — these are the hypotenuses)
CM = CM (common side)

By the RHS congruence rule, ΔCMA ≅ ΔCMB

∴ AM = BM (corresponding sides) — CM bisects the chord AB

Alternative, using Baudhāyana–Pythagoras:

AM² = CA² − CM² and BM² = CB² − CM²

CA = CB = r, and CM is the same in both

⇒ AM² = BM² ⇒ AM = BM (lengths are positive)

Why RHS and not SSS: we are given a right angle and the two hypotenuses, and we
do not yet know the third pair of sides — that is precisely the RHS situation. Note the
contrast with Theorem 4, where we were given the midpoint and had to produce the
right angle; there SAS was the right tool.

Q2 An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the
altitude from A to BC passes through the centre of the circle.

Given: ΔABC inscribed in a circle with centre O, and AB = AC. Let AD be the altitude from A to BC.
To show: O lies on AD.
Step 1 — the altitude from A is the perpendicular bisector of BC. In ΔABD and ΔACD: AB = AC
(given), AD common, ∠ADB = ∠ADC = 90°. By RHS, ΔABD ≅ ΔACD, so BD = DC. Thus AD is
perpendicular to BC and passes through its midpoint D — it is the perpendicular bisector of BC.
Step 2 — the perpendicular bisector of a chord passes through the centre. BC is a chord,
and OB = OC (radii), so O is equidistant from B and C. Every point equidistant from B and C lies
on the perpendicular bisector of BC. Hence O lies on that line.

Page 20 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Perpendicular bisector of BC = line AD (Step 1)

O lies on the perpendicular bisector of BC (Step 2)

∴ O lies on AD — the altitude from A passes through the centre

A

O

B D C

AD bisects BC at right angles, so it is the perpendicular bisector of the chord BC and must contain the
centre O.

Tip: for an isosceles triangle the altitude, the median, the angle bisector from the
apex and the perpendicular bisector of the base are all the same line — and here
that line is also a diameter of the circumcircle.

Page 21 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q3 Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a
circle. If the radius of the circle is 5 cm, find the distance between the midpoints of
the chords.

Distance = 7 cm.
Drop perpendiculars from the centre O to each chord. By Theorem 5 each perpendicular bisects
its chord, so its foot is the midpoint of that chord.

Chord AB = 6 cm ⇒ half-chord = 3 cm

d₁² = r² − 3² = 5² − 3² = 25 − 9 = 16 ⇒ d₁ = 4 cm

Chord PQ = 8 cm ⇒ half-chord = 4 cm

d₂² = r² − 4² = 5² − 4² = 25 − 16 = 9 ⇒ d₂ = 3 cm

The chords are on opposite sides of O, and both perpendiculars lie along the same line
through O.

Distance between midpoints = d₁ + d₂ = 4 + 3 = 7 cm

Page 22 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

6 cm

4

O

3

8 cm

Opposite sides of the centre: the two distances add. Note the longer chord (8 cm) is the nearer one (3
cm).

Tip: "opposite sides" ⇒ add the distances; "same side" ⇒ subtract them. Getting this
the wrong way round is the commonest slip in these questions.

Sanity check: Theorem 8 says the longer chord must be closer to the centre. Here
the 8 cm chord is 3 cm away and the 6 cm chord is 4 cm away — exactly as predicted.

In-text Questions — Page 102

Page 23 of 75

as e
a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

Section 5.6 Distance of Chords from the Centre
co m
e m.
m l as
.co
Activity: Take a paper circle. Fold the circle from the boundary, inwards. Open the
a g
emThe crease is now a chord (see Fig. 5.13 B). Now fold the paper again, so that
Q1

a s
fold.

a gl the end points of the chord meet. Open the fold (see Fig. 5.13 C). Measure the
lengths of the parts into which the chord is divided. Measure the angle between the

co m
creases. Measure the distance from the centre to the midpoint of the chord.

em . ag
g l as
a

co m
em.
m l as
m .co a g
l a se
a g
m a s
.co agl
Fig. 5.13 A Fig. 5.13 B Fig. 5.13 C

a s em
l fold is opened out, and the disc after the second fold is
agfirst
Fig. 5.13 A, B, C, page 102 — redrawn sketch of the paper-circle photographs: the plain
disc, the disc after the
opened out.

co m
m .
o m l a se
.c a g
m
se things come out of the folding, and each one is a theorem of this chapter in physical

g l a
a
Three
form.

se m
com
The two parts of the chord are equal. The second fold makes the endpoints of the chord
g l a
m . a
ase
coincide, so the crease is the perpendicular bisector of the chord — it cuts the chord exactly

agl
in half at its midpoint.
The angle between the two creases is 90°. Fold the paper along crease 2 and the chord

m
folds onto itself; the only way that happens is if crease 2 meets the chord at a right angle.

. co
m
The second crease passes through the centre, and the distance from the centre to the

o m l a se
midpoint of the chord is the distance from the centre to the chord. The second fold
.cmakes the boundary of the circle overlap itself, so crease a2 gis a diameter and contains the
m centre. "Distance from a point to a line" always means the perpendicular distance, and here
l a se
ag
c
that perpendicular is exactly the segment from the centre to the midpoint.

m .
m a s e
e m . co agl
g l as
a

com
m .
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.co


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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Crease 1 = chord AB Crease 2 = perpendicular bisector of AB

Crease 2 passes through the centre O, meets AB at M

AM = MB, ∠OMA = ∠OMB = 90°, and OM = distance from O to the chord

Why the paper knows this: a fold is a reflection. The second fold is a reflection that
swaps A and B and maps the circle to itself. A reflection that maps the circle to itself
must fix the centre, so the crease passes through O; a reflection that swaps A and B
must be the perpendicular bisector of AB. Those two statements together are
Theorems 4 and 5.

Q2 Now draw another chord of the same length. How will you do this? We will let you
figure this out yourself. Join the centre to the midpoint of the new chord and
measure its length. Is it the same as distance from the centre to the first chord?

How to draw a second chord of the same length: open a compass to the length of the first
chord AB. Put the compass point anywhere on the circle, at a point P, and cut the circle at Q.
Then PQ = AB. (Equivalently: trace the circle and the chord on tracing paper, then rotate the
tracing paper about the centre — the traced chord lands on a new chord of the same length.)
Yes — the distance is the same. The two equal chords are equidistant from the centre.

Let M, N be the midpoints of AB and PQ.

In ΔOMA and ΔONP: OA = OP = r, AM = ½AB = ½PQ = PN, ∠OMA = ∠ONP = 90°
By RHS congruence, ΔOMA ≅ ΔONP ⇒ OM = ON

Why rotation makes this obvious: the circle has complete rotational symmetry
about O. Rotating the chord AB about O gives a chord of the same length, and the
perpendicular from O rotates with it — so its length cannot change. The congruence
argument above is the same statement, written as a proof rather than an
observation. That distinction matters: examples suggest, proofs settle. This result is
Theorem 6.

Exercise Set 5.4 — Page 104

Page 25 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Section 5.6 Distance of Chords from the Centre

Q1 Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.

Theorem 6: chords of a circle having the same length are at the same distance from the centre.
Given: circle with centre C and radius r; chords AB and FG with AB = FG; E and H are the feet of
the perpendiculars from C to AB and FG. To show: CE = CH.

By Theorem 5 the perpendicular from the centre bisects the chord, so

AE = ½ AB and FH = ½ FG

AB = FG (given) ⇒ AE = FH

ΔCEA is right-angled at E, so by Baudhāyana–Pythagoras:
CE² = CA² − AE² = r² − AE²

Likewise CH² = CF² − FH² = r² − FH²

Since AE = FH, the two right-hand sides are equal:

CE² = CH² ⇒ CE = CH (lengths are positive)

Why this proof is worth having: the congruence proof in the text uses SSS or RHS;
this one turns the whole theorem into a single equation. Written as a formula it says
d = √(r² − (ℓ/2)²): the distance depends on the chord length ℓ and on nothing else.
Equal ℓ forces equal d — and, read backwards, it also gives Theorem 7 and Theorem
8 immediately.

Page 26 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q2 Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE =
CH, show that AB = GF.

B E A

G

C

H

F

Fig. 5.15, page 104 — circle with centre C; AB and GF are chords, and CE and CH are
drawn from C to E on AB and H on GF.

(In Fig. 5.15, C is the centre; AB and GF are chords, E lies on AB and H lies on GF. The condition is
that CH is perpendicular to GF — the chord — as the figure shows.)
Given: CE ⊥ AB, CH ⊥ GF, CE = CH. To show: AB = GF.

Page 27 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

In ΔCEA and ΔCHF:

∠CEA = ∠CHF = 90° (given)

CA = CF = r (radii — the hypotenuses)
CE = CH (given)

By the RHS congruence rule, ΔCEA ≅ ΔCHF

⇒ AE = FH (corresponding sides)

By Theorem 5, E and H are the midpoints of AB and GF, so

AB = 2 AE and GF = 2 FH

∴ AB = 2 AE = 2 FH = GF

Why the midpoint step is essential: the congruence only gives us equal half-
chords. It is Theorem 5 — the perpendicular from the centre bisects the chord —
that turns AE = FH into AB = GF. This exercise proves Theorem 7, the converse of
Theorem 6.

Note: the printed question says "CH is perpendicular to GH". Since H lies on the
chord GF, the segment intended is GF; Fig. 5.15 marks the right angle at H on the
chord GF.

Q3 Solve the previous question using the Baudhāyana–Pythagoras theorem.

Given: CE ⊥ AB, CH ⊥ GF, CE = CH, with C the centre and radius r. To show: AB = GF.

Page 28 of 75

as e
a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
em.
ΔCEA is right-angled at E: AE² = CA² − CE² = r² − CE²
m l as
.co
ΔCHF is right-angled at H: FH² = CF² − CH² = r² − CH²
m a g
l a se
g
aGiven CE = CH, so CE² = CH², hence

co m
. ag
AE² = FH² ⇒ AE = FH
em
g l as
By Theorem 5, E and H bisect the chords: a

co m
m.
AB = 2 AE and GF = 2 FH

m as e
.co l
∴ AB = GF
a g
se m
g l a
a Tip: the whole of Sections 5.6 and 5.6.1 sits inside one identity — (½ chord)² + d² =

s
r². Fix r. Then d decides the chord and the chord decides d. Equal chords ⇒ equal d
m a
.co agl
(Theorem 6); equal d ⇒ equal chords (Theorem 7); bigger chord ⇒ smaller d

se m
a
(Theorem 8).

a g l

co m
In-text Questions — Page 104
m .
m as e
.co
Section 5.6.1 Which of the two unequal chords is farther from the centre?
a g l
a s em
a gl Q1 Activity: Draw a circle. Draw chords of various lengths. Drop a perpendicular to each
chord from the centre. Record the length of the chord and its distance from the
se m
com
centre in a table. (Table 1: Length of Chord / Distance from Centre.) What do you
g l a
m . a
ase
observe?

agl

c
Observation: the longer the chord, the smaller its distance from the centre.
. om
s e m own circle and
Here is the table filled in for a circle of radius 5 cm (do the same for your

. com
compare).
a gla
a s em
agl LENGTH OF CHORD (CM) 10 9 8 6 4 2
c
m .
m a s e
agl
DISTANCE FROM CENTRE (CM)
co
0 2.18 3 4 4.58 4.90

m .
as e
a g l

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m .
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.co


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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Each entry comes from the same right triangle:

d = √(r² − (½ chord)²) = √(25 − (½ chord)²)

e.g. chord 8 ⇒ d = √(25 − 16) = 3; chord 6 ⇒ d = √(25 − 9) = 4

Why it happens: r is fixed, so d² + (½ chord)² is a constant (= 25 here). If one of the
two squares grows the other must shrink. So a longer chord forces a smaller
distance — which is exactly Theorem 8.

Check the two extremes: the longest chord is the diameter, 10 cm, at distance 0 —
it passes through the centre. Push the chord outwards and it shrinks towards a
single point, with distance approaching the full radius, 5 cm.

Exercise Set 5.5 — Pages 105–106
Section 5.6.1 Which of the two unequal chords is farther from the centre?

Q1 Find the length of the chord of a circle where the radius is 7 cm and perpendicular
distance is 6 cm.

Chord = 2√13 cm ≈ 7.2 cm.

Let AB be the chord, O the centre, M the foot of the perpendicular from O.

By Theorem 5, M is the midpoint of AB, and ΔOMA is right-angled at M.

AM² = OA² − OM² = 7² − 6² = 49 − 36 = 13

AM = √13 cm

AB = 2 × AM = 2√13 cm ≈ 7.21 cm

Check it yourself: the answer must be less than the diameter 14 cm — and it is.
Also, 6 cm is a fairly large distance for a radius of 7 cm, so a shortish chord is exactly
what we should expect.

Page 30 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q2 Explain 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√(r² − d²).

Given: circle with centre O, radius r; chord AB whose perpendicular distance from O is d. To
show: AB = 2√(r² − d²).

Let M be the foot of the perpendicular from O to AB, so OM = d and ∠OMA = 90°.

By Theorem 5 (the perpendicular from the centre bisects the chord), AM = MB = ½ AB.

In the right triangle OMA, by Baudhāyana–Pythagoras:

OA² = OM² + AM²

r² = d² + AM²

AM² = r² − d²

AM = √(r² − d²)

AB = 2 AM = 2√(r² − d²)

Page 31 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

O
r
d

A √(r² − d²) M B

One right triangle — radius, distance, half-chord — carries every chord calculation in the chapter.

Reading the formula: put d = 0 and you get AB = 2r, the diameter — the longest
chord. Push d up towards r and the chord shrinks towards 0. And since r² − d²
decreases as d increases, a bigger distance always means a shorter chord, which is
Theorem 8 in one line.

Q3 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, we cannot. The conclusion CD = 2 AB is false in general; it holds only in one very special
case.

Page 32 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Let the distance of CD from the centre be d, so the distance of AB is 2d.

AB = 2√(r² − 4d²) and CD = 2√(r² − d²)

Take r = 10 cm and d = 3 cm:

CD = 2√(100 − 9) = 2√91 ≈ 19.08 cm

AB = 2√(100 − 36) = 2√64 = 16 cm

CD / AB ≈ 1.19, not 2.

When does CD = 2 AB actually happen?

2√(r² − d²) = 2 × 2√(r² − 4d²)

r² − d² = 4(r² − 4d²)

r² − d² = 4r² − 16d²

15d² = 3r² ⇒ r² = 5d², i.e. r = d√5

Check with d = 2, r = 2√5: CD = 2√(20 − 4) = 8, AB = 2√(20 − 16) = 4 ✓

Why the guess fails: distance and chord length are not proportional to each other.
They are tied by d² + (½ chord)² = r², a squared relation, so doubling one quantity
does not halve or double the other. All we may safely say from Theorem 8 is the
direction: since AB is farther from the centre, AB is the shorter chord, so CD > AB.
Whether CD is 1.1 times AB or exactly 2 times AB depends on r.

In-text Questions — Page 107

Page 33 of 75

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a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

Section 5.7 Angles Subtended by an Arc
co m
e m.
m l as
.co a g
Exercise: A circle with centre O is drawn, and A, B, C, D are points on the circle (see
em5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If
Q1

a s
Fig.

a gl 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.
com
e m . ag
g l as
a C

co m
B em.
com g l as
m . a
ase
agl L

m a s
m .co agl
l a se
a g O
K
co m
m .
m as e
.co a g l
se m
g l a
a
se m
com g l a
A m . a
ase
agl
D
co m
m .
m l a se
.co ag at O.
Fig. 5.19, page 107 — circle with centre O; A, B, C, D, K and L lie on the circle. The two

se m shaded regions are the angles

g l a
a c
m .
a s e
. c om agl
s e m5.19 with a protractor:
a
Measuring the two angles in Fig.

agl

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m .
m ase
.co


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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

ARC ANGLE SWEPT AT O LESS / GREATER THAN 180° TYPE

arc AKB about 100° less than 180° minor arc

arc CLD about 205° (reflex) greater than 180° major arc

How to read the figure correctly. The angle an arc subtends is the angle swept as the radius
turns from one end of the arc to the other along that arc.

For arc AKB you turn from OA to OB passing through OK. K sits between A and B on the short
way round, so you sweep the ordinary (non-reflex) angle AOB — about 100°.
For arc CLD you must go from OC to OD passing through OL. L lies on the long way round, so
you sweep the reflex angle COD — about 205°. The direct angle COD is about 155°, but that
belongs to the other arc from C to D, the one not containing L.

Angle for arc CLD + angle for the other arc from C to D = 360°

205° + 155° = 360° ✓

Why the naming letter matters: "arc CD" is ambiguous — there are two arcs
joining C and D. Writing CLD names the middle point L and so fixes which of the two
you mean. This is exactly why the book insists on three letters for an arc.

Q2 Activity: Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K
between A, B on the circle. Measure the angle subtended at the centre by arc AKB.
Take three points P, Q, R on the circle outside arc AKB. Measure the angles
subtended by arc AKB at points P, Q, R. What do you notice?

Two things emerge, and together they are Theorem 9.

The three angles ∠APB, ∠AQB and ∠ARB come out equal to one another — it makes no
difference where on the circle (outside arc AKB) the point is taken.
Each of them is exactly half the angle that arc AKB subtends at the centre.

Sample reading (draw your own and check):

angle at the centre, ∠AOB = 110°

∠APB = ∠AQB = ∠ARB = 55° = ½ × 110°

Page 35 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Why it must be so: join P to the centre O and extend PO to meet the circle again.
That splits ∠APB into two pieces, and each piece sits in an isosceles triangle whose
two equal sides are radii. The exterior-angle theorem then doubles each piece at the
centre. Adding the two pieces gives ∠AOB = 2∠APB — and the argument never used
where P was, only that it lies on the circle outside the arc. That is why every position
of P gives the same answer.

Try This: now take a point F inside the circle and a point G outside it, and measure
∠AFB and ∠AGB. You will find ∠AGB < ∠APB < ∠AFB. The constancy is special to
points that lie on the circle — see Fig. 5.25.

Exercise Set 5.6 — Pages 110–111
Section 5.7.1 Angle subtended by an arc at a point on the circle outside the arc

Q1 In 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?

AB = 12 cm — the chord equals the radius.

In ΔOAB: OA = OB = 12 cm (radii) ⇒ the triangle is isosceles

∠OAB = ∠OBA (base angles of an isosceles triangle)

∠OAB + ∠OBA + 60° = 180°

2 ∠OAB = 120° ⇒ ∠OAB = ∠OBA = 60°

All three angles are 60°, so ΔOAB is equilateral

∴ AB = OA = OB = 12 cm

Check with the chord formula: the distance of AB from O is d = 12 cos 30° = 6√3, so
AB = 2√(144 − 108) = 2√36 = 12 cm ✓

Page 36 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Worth remembering: a central angle of 60° always gives a chord equal to the
radius. That is why a regular hexagon inscribed in a circle has side equal to the
radius — six 60° slices fill up 360°.

Q2 Let 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 ∠AXB is different from ∠AYB? (ii) Is it true
that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle? (iii) If ∠AXB =
∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also
pass through Y?

(i) No. If X and Y are both on the circle and on the same side of the chord AB, they lie on the
same arc, so ∠AXB = ∠AYB always.

Both X and Y lie outside the same arc AB, so by Theorem 9

∠AXB = ½ (angle subtended by that arc at O) = ∠AYB

There is no freedom left — the angle cannot vary.

(ii) No, this is not always true. Equal angles do not force X and Y onto the same side of AB.

If X is on the major arc and Y on the minor arc, then AXBY is a cyclic 4-gon, so

∠AXB + ∠AYB = 180° (Theorem 11)

These two can be equal only when each is 90°, i.e. when AB is a diameter.

Counterexample: let AB be a diameter. Take X above AB and Y below it.

∠AXB = ∠AYB = 90° (angle in a semicircle), yet X and Y are on opposite sides.

So the correct statement is: if ∠AXB = ∠AYB and AB is not a diameter, then X and Y must be on
the same side; if AB is a diameter, they need not be.
(iii) Yes, provided X and Y lie on the same side of the line AB. This is exactly Theorem 10.

Page 37 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

A, B, X are non-collinear ⇒ a circle passes through A, B, X (Theorem 1)

∠AXB = ∠AYB and X, Y are on the same side of AB

⇒ by Theorem 10, A, B, X, Y are concyclic
⇒ the circle through A, B, X also passes through Y

Why the "same side" condition cannot be dropped: reflect Y in the line AB to get
Y′. Then ∠AY′B = ∠AYB, but Y′ generally lies on a different circle through A and B.
Theorem 10's proof works by ruling out "Y inside" and "Y outside" using the exterior-
angle theorem, and that argument needs Y on the same side as X.

Page 38 of 75

as e
a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
m.
Find x in Fig. 5.26.
e
Q3

m l as
m .co a g
l a se D
a g

. com ag
asem 100°
agl

co m
em.
Am l as C
m .co a g
l a se
a g
m a s
m .co agl
l a se
a g
x
co m
m .
m as e
.co a g l
se m
g l a B
a
se m
o m l a
m .cB lie on one circle; the angle marked at D is 100° and the ag
se angle marked at B is x.
Fig. 5.26, page 111 — A, D, C and

l a
ag

co m
m .
e

m l as
.co g
x = 80°.
a
emIn Fig. 5.26 the four points A, D, C, B lie on a circle, so ADCB is a cyclic quadrilateral. The angle
a s
agl 100° is marked at D and the angle x is marked at B, and D and B are opposite vertices.
.c
s e m
m a
e m . co agl
g l as
a

co m
m .
m ase
.co


a g l Page 39 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

By Theorem 11, opposite angles of a cyclic quadrilateral add up to 180°:

∠ADC + ∠ABC = 180°

100° + x = 180°
x = 180° − 100° = 80°

D

100°

A C

x

B
ADCB is cyclic, so the angles at the opposite vertices D and B add to 180°.

Why opposite angles add to 180°: ∠ADC stands on the arc ABC and ∠ABC stands
on the arc ADC. Together those two arcs make up the whole circle, so the two central
angles add to 360°. Each inscribed angle is half its central angle, so the two inscribed
angles add to ½ × 360° = 180°.

Page 40 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Check it yourself: the other pair must also add to 180°. So ∠DAB + ∠DCB = 180°
here as well, whatever those two individual values happen to be.

In-text Questions — Page 113
Section 5.8 Concyclicity of Points

Q1 Exercise: A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°,
and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.

Yes, such a quadrilateral can be drawn. Both tests are satisfied.

Test 1 — angle sum of a quadrilateral:

80° + 110° + 100° + 70° = 360° ✓

Test 2 — opposite angles of a cyclic quadrilateral:

∠A + ∠C = 80° + 100° = 180° ✓

∠B + ∠D = 110° + 70° = 180° ✓

Both pairs of opposite angles add to 180°, so by Theorem 12 the four vertices must lie on a
circle. Such a cyclic quadrilateral exists.

Why the second test is the decisive one: every quadrilateral, cyclic or not, has
angle sum 360°, so Test 1 alone proves nothing about the circle. It is Theorem 12 —
opposite angles supplementary ⇒ concyclic — that does the real work. Notice also
that once ∠A + ∠C = 180° holds, ∠B + ∠D = 360° − 180° = 180° follows
automatically; the two conditions are not independent.

Try This: change ∠C to 95° and keep the rest so the sum stays 360° (take ∠D = 75°).
Now ∠A + ∠C = 175° ≠ 180°, so a quadrilateral with those angles exists — but no
circle can pass through all four of its vertices.

End-of-Chapter Exercises — Pages 114–116

Page 41 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Whole chapter

Q1 In 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 = 24 cm.

Half-chord² = r² − d² = 13² − 5² = 169 − 25 = 144

Half-chord = 12 cm

Chord = 2 × 12 = 24 cm

Tip: (5, 12, 13) is a Baudhāyana triple, so no square roots are needed. Keep (3,4,5),
(5,12,13), (8,15,17) and (7,24,25) at your fingertips — the questions in this exercise
are built out of them.

Q2 An 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?

35°, at any point of the circle lying outside that arc.

By Theorem 9: angle at the centre = 2 × angle at a point of the circle outside the arc
70° = 2 × angle at the point

Angle at the point = 70° ÷ 2 = 35°

Why the answer does not depend on the point: the halving argument uses only
the fact that the point is on the circle and outside the arc — it never uses where on
the circle. So every such point gives 35°; this is why all angles in the same segment
are equal.

Careful: a point on the other side — i.e. on the 70° arc itself — sees the chord under
180° − 35° = 145°, not 35°. The theorem is about points outside the arc.

Page 42 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q3 The 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.

Distance = 5 cm.

Radius r = 26 ÷ 2 = 13 cm

Half-chord = 24 ÷ 2 = 12 cm (Theorem 5: the perpendicular from the centre bisects the

chord)

d² = r² − (half-chord)² = 13² − 12² = 169 − 144 = 25

d = 5 cm

Tip: this is Q1 run backwards — same circle, same chord, the unknown moved to the
other side of the equation.

Q4 A 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 = 24 cm.

Chord = 2√(r² − d²)

= 2√(15² − 9²)

= 2√(225 − 81)

= 2√144 = 2 × 12 = 24 cm

Check it yourself: (9, 12, 15) is just (3, 4, 5) scaled by 3. And 24 cm < 30 cm, the
diameter — as every chord must be.

Page 43 of 75

as e
a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
m.
Prove that the perpendicular bisector of a chord passes through the centre of the
e
Q5

m l as
.co
circle.

a g
sem
g l a
a

Given: a circle with centre O and a chord AB. To show: the perpendicular bisector of AB passes

m
through O.
. co ag
em
g l as
a
OA = OB = r (both are radii)

So O is equidistant from A and B.

co m
m.
Every point equidistant from A and B lies on the perpendicular bisector of AB.

m as e
.co
∴ O lies on the perpendicular bisector of AB.
a g l
a s em
a l locus step, written out. Let M be the midpoint of AB and join OM. In ΔOMA and ΔOMB: OA
gThe
= OB (radii), AM = BM (M is the midpoint), OM common. By SSS, ΔOMA ≅ ΔOMB, so ∠OMA =
a s
com it — OM is the perpendicular bisector, and it agl
∠OMB. These are angles on a straight line, so ∠OMA + ∠OMB = 180°, giving ∠OMA = ∠OMB =
90°. Hence OM is perpendicular to AB and .bisects
a s em
agl
passes through O.

Why this small result matters so much: it is the reason the circumcentre
co m
construction works. Given three points, each side gives a perpendicular bisector that
m .
o m l a se
.c recipe for finding the centre of a given circle: draw gany two chords, bisect
must contain the centre; two of them therefore pin the centre down. It is also the

m a
asethem at right angles, and the creases cross at the centre.
practical

agl
se m
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m . a
ase
The diameter of a circle is AB. Point C is on the circumference. What is the measure

agl
Q6
of the ∠ACB? Explain your reasoning.

co m
.

em
as
∠ACB = 90° — the angle in a semicircle is a right angle.
m l
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ag
.c
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e m . co agl
g l as
a

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Let O be the centre, so A, O, B are collinear and AB is a diameter.

Take the arc from A to B not containing C. The angle it sweeps at O is the straight angle:

∠AOB = 180°

C lies on the circle, outside that arc, so by Theorem 9

∠ACB = ½ ∠AOB = ½ × 180° = 90°

A second proof, using isosceles triangles. Join OC. Then OA = OC = OB = r, so ΔOAC and ΔOBC
are both isosceles.

Let ∠OAC = ∠OCA = p and ∠OBC = ∠OCB = q

Angle sum of ΔABC: p + q + ∠ACB = 180°, and ∠ACB = p + q

So 2(p + q) = 180° ⇒ p + q = 90° ⇒ ∠ACB = 90°

Did you know? The converse is also true and is used constantly: if ∠ACB = 90° for a
point C on a circle through A and B, then AB must be a diameter. That is how Q15
below is solved.

Q7 ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the
measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?

∠C = 105° and ∠D = 70°.

By Theorem 11, opposite angles of a cyclic quadrilateral are supplementary.

∠A + ∠C = 180° ⇒ 75° + ∠C = 180° ⇒ ∠C = 105°

∠B + ∠D = 180° ⇒ 110° + ∠D = 180° ⇒ ∠D = 70°

Check it yourself: 75° + 110° + 105° + 70° = 360° ✓ — the angle sum of the
quadrilateral comes out right, as it must.

Page 45 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q8 Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find
the value of x and the measures of ∠P and ∠R.

x = 38, ∠P = 86° and ∠R = 94°.

In the quadrilateral PQRS, P and R are opposite vertices, so

∠P + ∠R = 180°

(2x + 10) + (3x − 20) = 180

5x − 10 = 180

5x = 190
x = 38

∠P = 2(38) + 10 = 76 + 10 = 86°

∠R = 3(38) − 20 = 114 − 20 = 94°

Check it yourself: 86° + 94° = 180° ✓, and both angles are positive and less than
180°, so the quadrilateral is genuine.

Q9 The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the
radius of the circle.

Radius = 10 cm.

Half-chord = 16 ÷ 2 = 8 cm

r² = d² + (half-chord)² = 6² + 8² = 36 + 64 = 100

r = 10 cm

Tip: (6, 8, 10) is (3, 4, 5) doubled. Notice the pattern across Q1, Q3, Q4 and Q9: one
single relation, r² = d² + (half-chord)², with a different letter unknown each time.

Page 46 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Q10 A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Area = 60 square units.
Let the quadrilateral be ABCD with AB = BC = 5 and CD = DA = 12 — the two equal pairs
adjacent, so it is a kite. Join the diagonal BD.

The kite is symmetric about BD, so ∠A = ∠C.

ABCD is cyclic, so ∠A + ∠C = 180° (Theorem 11)

⇒ 2∠A = 180° ⇒ ∠A = ∠C = 90°

So ΔABD and ΔCBD are right-angled at A and at C.

BD² = AB² + AD² = 5² + 12² = 25 + 144 = 169 ⇒ BD = 13

Area = area ΔABD + area ΔCBD

= ½ × 5 × 12 + ½ × 5 × 12

= 30 + 30 = 60 square units

Page 47 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

B
5
A

13
12

C

D
The kite is cyclic, so the two equal angles at A and C must each be 90° — and BD becomes a diameter.

If instead the sides alternate 5, 12, 5, 12: opposite sides are equal, so the quadrilateral is a
parallelogram; a cyclic parallelogram is a rectangle (Q14), and the area is 5 × 12 = 60 square
units again.

Why both arrangements give 60: in each case the figure splits into two right
triangles with legs 5 and 12 and hypotenuse 13. In fact BD = 13 is a diameter in both
cases, so the circumradius is 6.5 units either way.

Page 48 of 75

as e
a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
m.
Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find
e
Q11

om
out whether the centre of the circumcircle lies inside the quadrilateral or outside?
What.cis g l as
e m
the best way of finding out? a
ag las

m
Best method: draw one diagonal and look at the two triangles it makes.
. co ag
se m
A diagonal, say AC, cuts the cyclic quadrilateral ABCD into ΔABC and ΔACD. Both triangles are

l a
aga triangle's circumcentre sits:
inscribed in the same circle, so the circumcentre of the quadrilateral is the circumcentre of each
of them. And we already know where

co m
m.
TRIANGLE ABC OR ACD CIRCUMCENTRE LIES CONCLUSION FOR ABCD

m as e
.co
one of them is acute-angled inside that triangle
a g l
inside the quadrilateral

se m
g l a on the diagonal AC (AC is a diameter)
a
one of them is right-angled at the midpoint of its hypotenuse

s
both are obtuse-angled outside both triangles outside the quadrilateral

om a
e
. c
m side of the quadrilateral cuts off an arc. The centre agl
s
An equivalent test using the sides. Each

a g
lies inside exactly when no side cutsla off an arc bigger than a semicircle. Since an inscribed
angle is half its arc, this becomes a test you can carry out with a protractor:

c o m
Look at the angle each side subtends at one of the two opposite vertices:m .
o m l a se
.c (on side AB), ∠BDC (on side BC), ∠CAD (on side CD), a g∠DBA (on side DA)
m
ase
∠ACB

agl
all four less than 90° ⇒ centre inside
se m
omside (that side is a diameter)
one of them equal to 90° ⇒ centre on.cthat g l a
em outside a
a
one of them greater than 90°l⇒ scentre
ag

co m
Why one angle bigger than 90° throws the centre out: if ∠ACB > 90°, then the arc
m .
o m l a se
AB not containing C is more than a semicircle, so C and D are both squeezed onto an
g The centre lies on the
.carc smaller than a semicircle, on one side of the chordaAB.
se m other side of AB — outside the quadrilateral. It is the same mechanism as an obtuse-
g l a
a c
.
angled triangle pushing its circumcentre outside.

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e m . co agl
g l as
a

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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Tip: in a rectangle the centre is at the meeting point of the diagonals (Q15); in a very
"flat" cyclic quadrilateral, all four vertices bunched on one arc, the centre falls well
outside.

Q12 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.

Given: chords AB and CD of a circle with centre O, with AB = CD, meeting at a point P inside the
circle. To show: the two pieces of AB match the two pieces of CD.
Step 1 — equal chords are equidistant from the centre. Let M and N be the midpoints of AB
and CD. By Theorem 6, OM = ON, and OM ⊥ AB, ON ⊥ CD.
Step 2 — P is equidistant from the two midpoints.

In ΔOMP and ΔONP:

∠OMP = ∠ONP = 90°

OP = OP (common hypotenuse)

OM = ON (Step 1)

By RHS congruence, ΔOMP ≅ ΔONP ⇒ PM = PN

Step 3 — put the pieces together. Since M and N are midpoints, AM = MB = ½AB and CN = ND
= ½CD, and AB = CD gives AM = CN.

AP = AM − PM and CP = CN − PN (P between A and M, say)

AM = CN and PM = PN ⇒ AP = CP

PB = MB + PM and PD = ND + PN ⇒ PB = PD

So {AP, PB} and {CP, PD} are the same pair of lengths:
each segment of one chord equals the corresponding segment of the other.

Page 50 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Why the midpoints are the key: the two chords are placed differently in the circle,
so there is no direct congruence between them. What they do share is the centre.
Going through O — equal chords ⇒ equal distances ⇒ P equally far from the two
midpoints — is what links the two chords to each other.

Tip: depending on which side of the midpoint P falls, AP may pair with CP or with PD.
The statement to remember is that the multiset of pieces is the same: {AP, PB} = {CP,
PD}.

Q13 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?)

The circle is forced: its radius must be 3√2 ≈ 4.24 cm.

r² = d² + (half-chord)² = 3² + 3² = 9 + 9 = 18

r = √18 = 3√2 ≈ 4.24 cm

Construction (direct).

1. Draw the chord AB = 6 cm.
2. Construct the perpendicular bisector of AB; let it cut AB at M.
3. Mark O on that bisector with OM = 3 cm.
4. With centre O and radius OA (≈ 4.2 cm), draw the circle. AB is then a chord of length 6 cm at
distance 3 cm from O.

Answering the hint. Yes — it is the circumcircle of a triangle, and a very recognisable one.

In right triangle OMA: OM = MA = 3 cm, so ∠MOA = 45°

∠AOB = 2 × 45° = 90°

So ΔAOB is a right isosceles triangle with legs 3√2 and hypotenuse AB = 6.

Since the central angle is 90°, any point C on the major arc gives

∠ACB = ½ × 90° = 45°

Page 51 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

So an equivalent construction is: draw ΔABC with AB = 6 cm and ∠ACB = 45°, and draw its
circumcircle. That circle automatically has the chord AB = 6 cm standing 3 cm from the centre.

O
3√2
3

A 6 cm
M B

Half-chord 3 and distance 3 make ΔOMA a 45° right triangle, so ∠AOB = 90° and r = 3√2.

Q14 Show that rectangle is the only parallelogram that can be inscribed in a circle.

Given: a parallelogram ABCD inscribed in a circle. To show: ABCD is a rectangle.

Page 52 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

ABCD is a parallelogram ⇒ opposite angles are equal:

∠A = ∠C and ∠B = ∠D

ABCD is cyclic ⇒ opposite angles are supplementary (Theorem 11):

∠A + ∠C = 180°

Substituting ∠C = ∠A:

2∠A = 180° ⇒ ∠A = 90°, and hence ∠C = 90°

Similarly 2∠B = 180° ⇒ ∠B = ∠D = 90°

All four angles are right angles ⇒ ABCD is a rectangle.

The converse holds too, so nothing is lost: every rectangle is cyclic. Its diagonals are equal and
bisect each other, so their common midpoint is the same distance from all four vertices — that
point is the centre of a circle through A, B, C, D.

Why this rules out the other parallelograms: a rhombus or a general
parallelogram has one pair of acute angles and one pair of obtuse angles. Two equal
angles can only add to 180° if each is exactly 90°, so as soon as a parallelogram is
inscribed in a circle its slant is forced away and it straightens into a rectangle. A
square is a special case — it is a rectangle too.

Q15 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.

Given: rectangle ABCD inscribed in a circle. To show: the diagonals meet at the centre.
Proof 1 — each diagonal is a diameter.

Page 53 of 75

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a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
e m.
∠ABC = 90° (angle of a rectangle)
m l as
.co
B lies on the circle and ∠ABC is subtended by the chord AC.
m a g
l a se
g
An inscribed angle is 90° only when it stands on a diameter (converse of the corollary to
aTheorem 9).

co m
. ag
∴ AC is a diameter. In the same way, ∠BAD = 90° ⇒ BD is a diameter.
em
g l as
a
Two diameters both pass through the centre, and two distinct lines meet in only one point.

co m
m.
∴ their point of intersection is the centre O.

o m l a se
g
.cusing the properties of a rectangle. The diagonals of a arectangle
m
Proof 2 — are equal and

l a
bisectseeach other. Let them meet at P. Then
ag
m a s
agl
PA = PC = ½ AC and PB = PD = ½ BD, with AC = BD
.co
⇒ PA = PB = PC = PD
a s em
aglvertices — P is the centre of the circle through them, i.e. P =
So P is equidistant from all four

m
O.
. co
se m
o m l a
g inscribe a rectangle
m .cthis gives a quick way to find the centre of a circular disc:
Tip:
a
ase(two chords with a common perpendicular pair) and the diagonals cross at the
agl centre. It also explains Q10: the diagonal 13 there was a diameter, so the
circumradius was 6.5.
se m
com g l a
m . a
ase
Q16
agl
Consider all chords of a circle of a fixed length. What is the shape formed by the

m
midpoints of all these chords?

. co
em
m l as
.co g

emradius of the given circle and ℓ the fixed chord length. a
s
Another circle, with the same centre — a concentric circle of radius √(r² − ℓ²/4), where r is the
l a
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Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Let O be the centre and let AB be any chord of length ℓ, with midpoint M.

By Theorem 4, OM ⊥ AB, so OM is the distance from O to the chord.

OM² = OA² − AM² = r² − (ℓ/2)²

OM = √(r² − ℓ²/4), the same value for every such chord

So every midpoint is at a fixed distance from O ⇒ the midpoints lie on a circle with centre O

and that radius.

And every point of that circle is reached. Take any point M with OM = √(r² − ℓ²/4). Draw the
chord through M perpendicular to OM. By Theorem 5 it is bisected at M, and its length is 2√(r² −
OM²) = 2 × (ℓ/2) = ℓ. So M really is the midpoint of a chord of length ℓ. The locus is the whole
concentric circle, not just part of it.

Page 55 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

O

Chords of one fixed length all sit the same distance from O, so their midpoints trace a concentric
circle.

Two special cases: if ℓ = 2r the chords are all diameters and the "circle" of midpoints
shrinks to the single point O. If ℓ is close to 0 the midpoints crowd towards the given
circle itself.

Q17 In a circle with centre O, chords AB and AC are congruent. Explain why this
statement is true: “The centre of the circle lies on the angle bisector of ∠BAC”.

Given: circle with centre O; chords AB and AC with AB = AC. To show: AO bisects ∠BAC.

Page 56 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

In ΔOAB and ΔOAC:

AB = AC (given)

OB = OC = r (radii)

OA = OA (common)

By SSS congruence, ΔOAB ≅ ΔOAC

∴ ∠OAB = ∠OAC (corresponding angles)

That is, the ray AO divides ∠BAC into two equal parts — AO is the bisector of ∠BAC, so

O lies on it.

A second way to see it. AB = AC are equal chords, so they are equidistant from O (Theorem 6).
A point inside an angle that is equidistant from both arms lies on the bisector of that angle.
Hence O lies on the bisector of ∠BAC.

A

O

B C

Page 57 of 75

Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round AglaSem · NCERT Solutions

Equal chords from A make the whole figure symmetric about the line AO — so AO bisects ∠BAC.

The idea behind both proofs: AB = AC makes the picture symmetric about the line
AO. A reflection in AO swaps B and C and maps the circle to itself, so it fixes the
centre — which means O must lie on the mirror line, and that mirror line bisects
∠BAC.

Q18 Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre
of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Radius = 13 cm.

Let d be the distance from the centre O to the longer chord (24 cm).

Both chords are on the same side, and the longer chord is the nearer one (Theorem 8),

so the distance to the 10 cm chord is d + 7.

For the 24 cm chord: r² = d² + 12² = d² + 144

For the 10 cm chord: r² = (d + 7)² + 5² = d² + 14d + 49 + 25

Equating: d² + 144 = d² + 14d + 74

144 − 74 = 14d

70 = 14d ⇒ d = 5 cm

r² = 5² + 12² = 25 + 144 = 169 ⇒ r = 13 cm

Page 58 of 75

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a g
Class 9 Maths Chapter 5 I'm Up and Down, and Round and Roundl AglaSem · NCERT Solutions

co m
e m.
m l as
.co 10 cm a g
a s em
ag l
7
.com ag
s e m
a 24 cm
agl
5 m
co
em.
m l as
.co O a g
a sem
agl
m a s
m .co agl
l a se
a g

co m
m .
m as e
.co a g l
se m
g l a
a Same side of the centre: the two distances differ by 7, they do not add to 7.

se m
com g l a
m . a
ase
Check it yourself: distance to the 10 cm chord = 5 + 7 = 12 cm, and 12² + 5² = 144 +

agl
25 = 169 = 13² ✓. Both chords fit the same circle of radius 13 cm.

. com
Why "same side" changes the equation: with both chords on one side, the

m a s emthem is the
perpendiculars point the same way from O, so the gap between
o
.cdifference
m Q3) the gap would be the sum. agl sides (as in Exercise 5.3
of the two distances. Had they been on opposite

l a se
ag
.c
s e m
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a

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m .
m ase
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