NCERT Solutions Class 9 Maths Chapter 5 Think and Reflect — I’m Up and Down, and Round and Round

NCERT Class 9 Mathematics — Ganita Manjari Part I, I’m Up and Down, and Round and Round. All 9 Think and Reflect prompts answered.

Also here. Every exercise in this chapter is solved, one page per exercise.

Chapter 5 sets three Think and Reflect boxes into its opening pages, holding nine prompts between them. They come early, before the theorems, and they are where the chapter’s central ideas are set up: the symmetry of a circle, the idea of a locus, and what it takes to pin a circle down.

The five prompts in the last box form a single investigation, working towards the fact that three points determine a circle — so they are best attempted in order.

Key insight. A locus is the set of all points satisfying some condition, and two of these prompts turn on the same one: the points equidistant from $A$ and $B$ form the perpendicular bisector of $AB$. That single fact explains how to find a circle’s centre by folding, why infinitely many circles pass through two points, and where their centres live.

Prompt 1 comes from the box on page 93, at the very start of the chapter.

Question 1

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?

Solution. Fold the paper in half, twice — the centre is where the two creases cross.

The instructions in full.

1. Fold the circular paper so that one half lies exactly on top of the other, and crease it.
2. Open it out and fold again in a different direction, so the halves match exactly once more. Crease.
3. The two creases meet at a single point. That point is the centre.

Why it works. A fold that makes the two halves match exactly is a reflection that maps the circle onto itself — and the only such lines are the diameters. So each crease is a diameter.

Every diameter passes through the centre. Two distinct diameters therefore both pass through the centre, and two distinct lines meet at only one point. So their intersection must be the centre.

A second method, using the chapter’s own theorems: draw any two chords, construct their perpendicular bisectors, and find where those meet. That works for a circle drawn on paper too, not just a paper circle — and it is the construction behind the circumcircle.

Prompts 2 to 4 come from the box on page 94, alongside the discussion of a circle’s symmetry.

Question 2

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?

Solution. A shape has rotational symmetry of order $n$ if turning it through $\frac{360°}{n}$ about its centre leaves it looking the same. A line of reflection symmetry is a mirror line across which the shape maps onto itself.

Square. Rotations of $90°$, $180°$, $270°$ and $360°$ all leave it unchanged — order $4$. It has $4$ lines of symmetry: two through the midpoints of opposite sides, two along the diagonals.

Regular pentagon. Rotations through multiples of $\frac{360°}{5} = 72°$ — order $5$. It has $5$ lines of symmetry, each through a vertex and the midpoint of the opposite side.

Regular hexagon. Rotations through multiples of $\frac{360°}{6} = 60°$ — order $6$. It has $6$ lines of symmetry: three through opposite vertices, three through the midpoints of opposite sides.

The pattern. A regular $n$-sided polygon has rotational symmetry of order $n$ and exactly $n$ lines of reflection symmetry.

And the circle. As the number of sides grows, the polygon looks more and more like a circle — and a circle is the limiting case: it has infinitely many lines of symmetry (every diameter) and rotational symmetry through every angle, not just special ones. That is what makes it the most symmetric shape in the plane, and it is why the fold trick of Question 1 works.

Question 3

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

Solution. The longest chord. A chord at distance $d$ from the centre has length $2\sqrt{r^2 – d^2}$, which is greatest when $d = 0$ — that is, when the chord passes through the centre. So the longest chord is the diameter:

$$2r = 2 \times 5 = 10 \text{ units}$$

A smallest chord? No. Push a chord away from the centre and it gets shorter: at $d = 4$ it is $2\sqrt{25 – 16} = 6$; at $d = 4.9$ it is about $1.99$; at $d = 4.99$ about $0.63$. The length can be made as small as you like by taking $d$ close enough to $5$.

But it never reaches a smallest value. At $d = 5$ the formula gives $0$, and a “chord” of length zero is a single point where the line touches the circle — a tangent, not a chord, since a chord needs two distinct points of the circle.

So the chord lengths fill the range from just above $0$ up to $10$, with the maximum attained and the minimum not. There is a longest chord but no shortest one.

Question 4

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?

Solution. Let the two given points be $A$ and $B$. We want all points $P$ with $PA = PB$.

The answer: the perpendicular bisector of $AB$.

Why every such point lies on it. Suppose $PA = PB$. Let $M$ be the midpoint of $AB$ and join $PM$. Then in $\triangle PAM$ and $\triangle PBM$:

$$PA = PB \; \text{(given)}, \qquad AM = BM \; \text{(midpoint)}, \qquad PM = PM \; \text{(common)}$$

By SSS the triangles are congruent, so $\angle PMA = \angle PMB$. These two angles are also a linear pair, adding to $180°$, so each is $90°$. Hence $PM \perp AB$ and $P$ lies on the perpendicular bisector.

Why every point of the line qualifies. Conversely, if $P$ is on the perpendicular bisector, the same two triangles are congruent by SAS ($AM = BM$, right angles, $PM$ common), so $PA = PB$.

The two directions together say the locus is exactly the perpendicular bisector — no more and no less.

Where this is used. It is the reason the centre of a circle through $A$ and $B$ must lie on the perpendicular bisector of $AB$ — the centre is equidistant from both, being at a radius from each. That fact drives the whole of the next box, and the circumcircle construction.

Prompts 5 to 9 come from the box on page 95, which works up to the fact that three points determine a circle.

Question 5

How many circles pass through two points on a plane?

Solution. Infinitely many.

The centre of any circle through $A$ and $B$ must be equidistant from them — each distance being a radius. By Question 4, such centres are exactly the points of the perpendicular bisector of $AB$.

That line contains infinitely many points, and each one gives a circle: take the centre there and the radius as its distance to $A$. Every one of those circles passes through both $A$ and $B$.

So there are as many circles through two points as there are points on a line — infinitely many.

(By contrast, through three non-collinear points there is exactly one circle: the perpendicular bisectors of two of the segments meet at a single point, leaving no freedom.)

Question 6

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 $B$?

Solution. Not all radii. $AB$ is a chord of every such circle, and no chord exceeds the diameter, so

$$AB \leq 2r \qquad \text{that is} \qquad r \geq \frac{AB}{2}$$

No circle of radius less than $\frac{AB}{2}$ can contain both points. Every radius from $\frac{AB}{2}$ upwards is achievable, by placing the centre at the right distance along the perpendicular bisector.

The smallest circle. Radius exactly $\frac{AB}{2}$, with the centre at the midpoint of $AB$. Then $AB$ is a diameter of it. This is the case mentioned in Exercise Set 5.1, Question 4.

The largest circle. There is none. Moving the centre further and further along the perpendicular bisector makes the radius grow without limit — every point of that infinite line is a valid centre, and there is no furthest one. The radii take every value from $\frac{AB}{2}$ upwards, with no largest.

Question 7

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

Solution. They increase.

Let $M$ be the midpoint of $AB$ and let the centre $O$ sit on the perpendicular bisector at distance $h = OM$ from $M$. Then $\triangle OMA$ is right-angled at $M$, with legs $h$ and $\frac{AB}{2}$, and hypotenuse the radius:

$$r = \sqrt{h^2 + \left( \frac{AB}{2} \right)^2}$$

As $h$ grows, so does $h^2$, and so does $r$. The increase never stops, which is the “no largest circle” of Question 6.

Checking the extreme: at $h = 0$ the centre is at $M$ itself and $r = \frac{AB}{2}$ — the smallest circle, as expected.

Question 8

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

Solution. Less curved — the arc through $A$ and $B$ flattens out.

Curvature is the rate at which a curve bends, and for a circle it depends only on the radius: a small circle turns sharply, a large one gently. Since Question 7 shows the radius increases as the centre moves away, the circles become less curved.

Picture the extreme: a bicycle wheel curves noticeably over a short arc, while the Earth’s surface — an arc of a circle thousands of kilometres in radius — looks flat. The arc through $A$ and $B$ behaves the same way: as the centre recedes, the arc joining $A$ to $B$ straightens towards the segment $AB$ itself, though it never quite becomes it.

Question 9

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?

Solution. Two quite different counts, and the contrast is the point.

With $A$ and $B$ merely on the boundary: infinitely many. The two points may sit anywhere along the sides — on the same side, on adjacent sides, on opposite sides — and the square may be of any size large enough and at any angle. Each choice gives a different square, and there are infinitely many choices.

With $A$ and $B$ as corners: exactly three. Now the segment $AB$ must join two vertices of the square, and there are only two ways that can happen:

• $AB$ is a side. Then the square lies on one side of $AB$ or the other — two squares.
• $AB$ is a diagonal. Then the square is completely determined: its other two vertices are the points making $AB$ a diagonal, and there is exactly one such square.

$$2 + 1 = 3$$

The contrast is the same one the box has been building all along: a weak condition (be somewhere on the boundary) leaves infinite freedom, while a strong condition (be a named vertex) pins the figure down to finitely many — exactly as two points leave infinitely many circles but three points leave only one.

Common mistakes

• Question 1, folding along a chord that is not a diameter. Only a fold along a diameter makes the two halves coincide exactly. If the edges do not match, the crease is not a diameter and the method fails.
• Question 1, using only one fold. One diameter contains the centre but does not locate it. A second, in a different direction, is what fixes the point.
• Question 2, counting only the diagonals of the square. A square has four lines of symmetry: the two diagonals and the two lines through the midpoints of opposite sides.
• Question 2, giving a regular pentagon rotational symmetry of order $10$. It has $5$ rotations and $5$ mirror lines; the total number of symmetries is $10$, but the order of rotational symmetry is $5$.
• Question 3, answering “the smallest chord is $0$”. A chord of length zero is not a chord — it is a single point, where the line touches the circle. Lengths get arbitrarily small without any smallest one existing.
• Question 4, answering “a circle”. That is the locus of points at a fixed distance from one point. Equidistance from two points gives a straight line.
• Question 5, answering “one”. Exactly one circle passes through three non-collinear points; through two there are infinitely many.
• Question 6, saying the largest circle has $AB$ as a diameter. That is the smallest one. There is no largest.
• Question 8, confusing “less curved” with “shorter”. The arc between $A$ and $B$ gets flatter, not shorter — and the whole circle gets very much longer.
• Question 9, answering “two” for the corners case. The diagonal configuration is easily missed, and it gives a third square.

Practise next

• Exercise Set 5.1 — the constructions these prompts set up, including Question 4 there, which is Question 6 above written as an exercise.
• The End-of-Chapter Exercises, Question 5 — proves formally what Question 4 above establishes informally: that the perpendicular bisector of a chord passes through the centre.
• The End-of-Chapter Exercises, Question 22 — the longest-chord question of Question 3 above, asked as a statement to be justified.