NCERT Solutions Class 9 Maths Chapter 5 In-text Questions — 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 7 in-text Activity and Exercise prompts solved.

Also here. Solutions to the end-of-chapter questions for this chapter are on their own page.

This chapter builds its theorems out of paper folding and measurement before proving any of them, and these seven prompts are where that happens. Four are Activities — things to do with a compass, a ruler and a folded circle — and each is designed so that the theorem on the next page is the thing you have just discovered.

Key insight. Every one of these prompts turns on the same fact: all points of a circle are the same distance from the centre. The paper-folding activity works because a fold that brings the two ends of a chord together must pass through the centre; equal chords are equidistant because you can rotate one onto the other; and no circle passes through three collinear points because no point is equidistant from all three.

Question 1

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

Solution. The point of this prompt is that the circle is not a human invention — it keeps appearing wherever the same physical rule acts equally in all directions.

Things that are circular in nature:

• The Sun and the full Moon as we see them, and the disc of a planet.
• Ripples spreading on still water when a stone is dropped in — the wave travels at the same speed in every direction, so every point of the ripple is the same distance from where the stone fell. That is the definition of a circle, acted out.
• The cross-section of a tree trunk, and its growth rings.
• The iris and pupil of an eye.
• A raindrop or a bubble seen head-on; the opening of a flower such as a sunflower’s disc.
• The rainbow, which is an arc of a circle.

Why nature keeps producing them. In each case there is a centre and something spreading out equally in all directions — a ripple, a growth ring, light bending through raindrops. When nothing favours one direction over another, the shape you get is the set of points at a fixed distance from a centre, which is exactly a circle.

Question 2

Think, Draw and Infer.

Question 2 (i)

$A$, $B$ 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 bisectors 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?

Solution. No such point $P$ exists, and the reason is the whole answer.

Where $P$ would have to lie. A point equidistant from two points lies on their perpendicular bisector — that is what a perpendicular bisector is. So

• $PA = PB$ forces $P$ onto the perpendicular bisector of $AB$;
• $PB = PC$ forces $P$ onto the perpendicular bisector of $BC$.

$P$ would have to be on both, so it would have to be where they cross.

Why they never cross. $A$, $B$ and $C$ all lie on one line $\ell$. The perpendicular bisector of $AB$ is perpendicular to $\ell$, and so is the perpendicular bisector of $BC$ — both are perpendicular to the same line, so they are parallel.

They are also distinct: the first passes through the midpoint of $AB$, the second through the midpoint of $BC$, and those midpoints differ because $A \ne C$. Two distinct parallel lines never meet, so there is no point on both, and no $P$ with $PA = PB = PC$.

Can a circle pass through three collinear points? No. A circle through all three would have a centre equidistant from all three — and that is precisely the point just shown not to exist.

Can a line cut a circle in three distinct points? No — at most two. Put the circle as $x^2 + y^2 = r^2$ and the line as $y = mx + c$. Substituting gives

$$x^2 + (mx + c)^2 = r^2$$

which is a quadratic in $x$, and a quadratic has at most two roots. So a line meets a circle in $0$, $1$ or $2$ points, never three. (This is the same fact as the previous one seen from the other side: three points of a circle can never be collinear.)

Question 2 (ii)

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

Solution. Yes — infinitely many.

By rotation. Rotate $\triangle ABC$ about the centre $O$ of its circumcircle through any angle $\theta$. A rotation preserves all lengths and angles, so the image $\triangle A’B’C’$ is congruent to $\triangle ABC$. And rotation about $O$ maps the circumcircle onto itself, so $A’$, $B’$, $C’$ still lie on that same circle — which is therefore their circumcircle too.

Since $\theta$ can be any of infinitely many angles, there are infinitely many such triangles.

By reflection. Reflecting $\triangle ABC$ in any diameter of the circle does the same job: the reflection is congruent to the original and the circle is unchanged.

What this says. A circumcircle does not determine its triangle. It fixes the size of the triangle’s parts relative to the circle — through the relation between a chord and the angle it subtends — but leaves the triangle free to sit anywhere around the circle, and free to be flipped.

Question 3

Activity: Take a paper circle. Fold the circle from the boundary, inwards. Open the fold. The crease is now a chord (Fig. 5.13 B). Now fold the paper again, so that the end points of the chord meet. Open the fold (Fig. 5.13 C). Measure the lengths of the parts into which the chord is divided. Measure the angle between the creases. Measure the distance from the centre to the midpoint of the chord.

Solution. What to do. Fold once from the edge inwards and open it — the crease is a chord $AB$. Fold again so that $A$ lands exactly on $B$, and open it: the second crease crosses the first.

What you will find, whatever circle you start with:

Why it must come out that way. Folding $A$ onto $B$ creates a crease every point of which is equidistant from $A$ and $B$ — that is what the fold does. So the second crease is the perpendicular bisector of $AB$. The centre $O$ satisfies $OA = OB$ (both are radii), so $O$ lies on that crease too.

Reading the two facts together gives the theorem the chapter states next:

The perpendicular from the centre of a circle to a chord bisects the chord.

And the distance from the centre to the midpoint of the chord, measured along that second crease, is what is meant by “the distance from the centre to the chord” — the perpendicular distance, which is the shortest one.

The follow-up in the text. Draw another chord of the same length and measure its distance from the centre: it comes out the same. Rotating the circle about its centre carries any chord to another chord of equal length and carries its perpendicular distance along with it — so equal chords are equidistant from the centre.

Question 4

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 centre in a table.

Solution. The method. Draw one circle and several chords of clearly different lengths, including one through the centre. For each, drop a perpendicular from the centre and measure both the chord and that perpendicular distance.

A representative table, for a circle of radius $5$ cm:

What the table shows. As the chord gets longer, its distance from the centre gets smaller. The longest chord of all is the diameter, at distance $0$; a chord shrinks towards nothing as it slides out towards the edge.

Why, exactly. The perpendicular from $O$ bisects the chord, so it makes a right-angled triangle with legs $d$ and $\frac{\ell}{2}$ and hypotenuse the radius $r$. By the Baudhāyana–Pythagoras theorem,

$$d^2 + \left(\frac{\ell}{2}\right)^2 = r^2 \qquad\Longrightarrow\qquad \ell = 2\sqrt{r^2 – d^2}$$

Check the table: $r = 5$, $d = 3$ gives $\ell = 2\sqrt{25 – 9} = 2 \times 4 = 8$ ✓, and $d = 4$ gives $\ell = 2\sqrt{25-16} = 6$ ✓

Since $r$ is fixed, increasing $d$ decreases $r^2 – d^2$ and so decreases $\ell$. That is the result the activity is designed to reveal, and it answers the question the section asks:

Of two unequal chords, the longer one is nearer the centre.

Question 5

Exercise: A circle with centre $O$ is drawn, and $A$, $B$, $C$, $D$ are points on the circle (see 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.

Solution. The naming matters here: arc $AKB$ means the arc from $A$ to $B$ that passes through $K$, and arc $CLD$ the arc from $C$ to $D$ passing through $L$. Naming a middle point is the only way to say which of the two arcs between the endpoints is meant.

Arc $AKB$. In Fig. 5.19, $A$ is low on the left, $B$ is high on the left and $K$ sits between them on the left of the circle. Measuring $\angle AOB$ on the side that contains $K$:

$$\angle AOB \approx 95° \quad (\text{measured from the figure})$$

This is less than $180°$, so arc $AKB$ is a minor arc.

Arc $CLD$. $C$ is at the top, $D$ at the bottom, and $L$ is out on the right — so this arc sweeps the whole right-hand side of the circle:

$$\angle COD \text{ (through } L) \approx 210°$$

This is more than $180°$, so arc $CLD$ is a major arc.

A check that needs no protractor. Look at how much of the circle each arc covers. Arc $AKB$ takes up a little over a quarter of the circumference; arc $CLD$ takes up well over half. An arc that is more than half the circle must subtend more than $180°$, because the whole circle is $360°$. Neither of these is anywhere near the boundary, so the classification does not depend on measuring accurately.

The general rule. For any two distinct points on a circle, the two arcs between them subtend angles adding to $360°$. So unless the points are diametrically opposite (giving two semicircles of $180°$ each), one arc is always minor and the other always major.

Question 6

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 $P$, $Q$, $R$. What do you notice? Repeat for a different arc $AKB$.

Solution. What to do. Draw the circle, mark $A$ and $B$, and pick $K$ on one of the two arcs. Measure $\angle AOB$ on the $K$ side. Then pick three points $P$, $Q$, $R$ anywhere on the other arc and measure $\angle APB$, $\angle AQB$ and $\angle ARB$.

What you will notice — two things, and the second is the surprise.

1. The three angles at $P$, $Q$ and $R$ are all equal. It does not matter where on the outer arc the point is taken.

2. Each of them is exactly half the angle at the centre.

A representative set of measurements:

Repeating with a different arc $AKB$ — a longer or shorter one — changes the centre angle and changes the three equal angles with it, but the halving holds every time. That is what makes it a theorem rather than a coincidence of one drawing:

The angle subtended by an arc at the centre is twice the angle it subtends at any point of the circle outside the arc.

Why the three agree. Each is half of the same central angle, so each equals the same number — which is why the position of $P$ on the outer arc is irrelevant. This is the fact the chapter uses next to prove that opposite angles of a cyclic quadrilateral add to $180°$, and it is what Question 7 below rests on.

Question 7

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

Solution. Yes, it can. Three checks, all of which pass.

1. The angles sum to $360°$, as any quadrilateral’s must:

$$80° + 110° + 100° + 70° = 360° \ \checkmark$$

2. The opposite angles add to $180°$. In quadrilateral $ABCD$ the opposite pairs are $(\angle A, \angle C)$ and $(\angle B, \angle D)$:

$$\angle A + \angle C = 80° + 100° = 180° \ \checkmark$$ $$\angle B + \angle D = 110° + 70° = 180° \ \checkmark$$

3. That condition is not just necessary but sufficient. The chapter’s Theorem 11 says a cyclic quadrilateral must have opposite angles summing to $180°$ — which alone would only tell us nothing is wrong. Theorem 12 is the converse:

If two opposite angles of a quadrilateral add up to $180°$, then its vertices lie on a circle.

So these angles do not merely fail to rule a cyclic quadrilateral out; they guarantee one exists. Note also that once one pair sums to $180°$, the other must too, since all four sum to $360°$ — so the two checks in step 2 are not independent, and either one is enough.

What a “no” would look like. Change $\angle C$ to $90°$ and $\angle D$ to $80°$: the total is still $360°$, so it is still a perfectly good quadrilateral — but now $\angle A + \angle C = 170° \ne 180°$, and no circle passes through all four vertices.

Common mistakes

• Question 2(i), answering “the circumcentre”. Three points have a circumcentre only when they form a triangle. Collinear points do not, and that is exactly what the question is probing.
• Question 2(i), saying the bisectors “meet at infinity”. At this level the honest statement is that they are parallel and distinct, so they do not meet at all, so no such $P$ exists.
• Question 3, folding anywhere for the second fold. The second fold must bring $A$ exactly onto $B$. Any other fold gives a crease that is not the perpendicular bisector and misses the point.
• Question 4, measuring the distance to the end of the chord. The distance from the centre to a chord means the perpendicular distance, to its midpoint — the shortest one.
• Question 4, expecting longer chords to be further out. It is the other way round: the longest chord, the diameter, passes through the centre at distance $0$.
• Question 5, ignoring the middle letter. “Arc $AKB$” names the arc through $K$. Reading it as just “the arc from $A$ to $B$” leaves it undecided which of the two arcs is meant, and they have different answers — one minor, one major.
• Question 6, taking $P$, $Q$, $R$ on the wrong arc. They must lie on the arc outside $AKB$. A point on arc $AKB$ itself gives the supplementary angle, not half the centre angle.
• Question 7, checking only that the angles total $360°$. Every quadrilateral does that. What makes it cyclic is the opposite angles summing to $180°$.

Practise next

• Exercise Set 5.4 and Exercise Set 5.5 — chords and their distances from the centre, which is where Questions 3 and 4 lead.
• Exercise Set 5.6 — angles subtended by arcs, the exercise counterpart of Question 6.
• The Think and Reflect prompts for this chapter — including the one asking how Jamuna found the centre of a paper circle, which is Question 3 turned into a puzzle.