NCERT Solutions Class 9 Maths Chapter 5 End-of-Chapter Exercises — 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, End-of-Chapter Exercises. All 26 questions solved.

Twenty-six questions covering the whole chapter. Almost all of them reduce to one of four facts:

$$\ell = 2\sqrt{r^2 – d^2} \qquad \text{(chord, radius, distance)}$$

• the perpendicular from the centre bisects the chord, and equal chords are equidistant from the centre;
• the central angle is twice the inscribed angle, so angles in the same segment are equal and the angle in a semicircle is $90°$;
• opposite angles of a cyclic quadrilateral sum to $180°$.

Fifteen questions carry an asterisk, marking them as enrichment that is not assessed. All are solved in full.

Key insight. When a question gives you a chord, draw the perpendicular from the centre; when it gives you an angle, look for the central angle that is twice it. Nearly every question here opens with one of those two moves, and the rest is arithmetic.

Question 1

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?

Solution. The perpendicular from the centre bisects the chord, giving a right-angled triangle with hypotenuse $r = 13$ and one leg $d = 5$:

$$\left( \frac{\ell}{2} \right)^2 = 13^2 – 5^2 = 169 – 25 = 144 \quad\Longrightarrow\quad \frac{\ell}{2} = 12$$

$$\ell = 24 \text{ cm}$$

Question 2

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?

Solution. The angle at the centre is twice the angle at any point on the remaining part of the circle, so the inscribed angle is half the central one:

$$\frac{70°}{2} = 35°$$

Question 3

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.

Solution. The radius is half the diameter: $r = 13$ cm. Half the chord is $12$ cm.

$$d = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25} = 5 \text{ cm}$$

Question 4

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?

Solution.

$$\ell = 2\sqrt{15^2 – 9^2} = 2\sqrt{225 – 81} = 2\sqrt{144} = 24 \text{ cm}$$

Question 5

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

Solution. Let $AB$ be a chord of a circle with centre $O$.

Since $A$ and $B$ both lie on the circle, $OA = OB$ — both are radii. So $O$ is equidistant from $A$ and $B$.

The set of all points equidistant from two fixed points is exactly the perpendicular bisector of the segment joining them. Hence $O$ lies on the perpendicular bisector of $AB$.

Therefore the perpendicular bisector of any chord passes through the centre. $\blacksquare$

This is the fact behind the circumcircle construction: the centre must lie on the perpendicular bisector of every chord, so two of them are enough to locate it.

Question 6

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

Solution. $\angle ACB$ is the angle subtended by $AB$ at a point on the circle, and $AB$ is a diameter, so the central angle $\angle AOB$ is a straight angle, $180°$.

The inscribed angle is half the central angle:

$$\angle ACB = \frac{180°}{2} = 90°$$

A second argument, without the doubling theorem. Join $OC$. Then $OA = OB = OC$ (radii), so both $\triangle OAC$ and $\triangle OBC$ are isosceles. Writing $\angle OAC = \angle OCA = a$ and $\angle OBC = \angle OCB = b$, the angles of $\triangle ABC$ are $a$, $b$ and $a + b$:

$$a + b + (a+b) = 180° \quad\Longrightarrow\quad 2(a+b) = 180° \quad\Longrightarrow\quad a + b = 90°$$

so the angle at $C$ is $90°$. This is the argument Question 24 asks for from a figure.

Question 7

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

Solution. In a cyclic quadrilateral $ABCD$, opposite angles are supplementary. The opposite pairs are $A$ with $C$, and $B$ with $D$.

$$\angle C = 180° – 75° = 105°$$ $$\angle D = 180° – 110° = 70°$$

Check: all four angles are $75° + 110° + 105° + 70° = 360°$, as a quadrilateral’s must be ✓

Question 8

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

Solution. In $PQRS$, the vertices $P$ and $R$ are opposite, so their angles are supplementary:

$$(2x + 10) + (3x – 20) = 180$$ $$5x – 10 = 180$$ $$5x = 190$$ $$x = 38$$

Substituting back:

$$\angle P = 2(38) + 10 = 86° \qquad \angle R = 3(38) – 20 = 94°$$

Check: $86° + 94° = 180°$ ✓

Question 9

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

Solution. Half the chord is $8$ cm, and this time the radius is the unknown — it is the hypotenuse, so the two legs are added:

$$r = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm}$$

Question 10

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

Solution. The equal sides are adjacent, so the quadrilateral is a kite: two sides of $5$, then two of $12$.

The quick route. Draw the diagonal joining the two vertices where a $5$ meets a $12$. It splits the kite into two triangles, each with sides $5$, $12$ and that diagonal. Now notice

$$5^2 + 12^2 = 25 + 144 = 169 = 13^2$$

so if the diagonal is $13$, both triangles are right-angled — and the angle opposite a diameter is exactly $90°$, so the diagonal being a diameter of length $13$ is consistent. Each triangle then has area $\frac{1}{2}(5)(12) = 30$, and the kite has area

$$2 \times 30 = 60 \text{ square units}$$

Confirmation by Brahmagupta’s formula for a cyclic quadrilateral with sides $a$, $b$, $c$, $d$ and semi-perimeter $s$:

$$s = \frac{5 + 5 + 12 + 12}{2} = 17$$ $$\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = 60$$

The two agree.

Question 11

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

Solution. The four vertices divide the circle into four arcs, and the position of the centre depends entirely on those arcs.

The criterion. The centre lies inside the quadrilateral exactly when none of the four arcs is greater than a semicircle. If one arc exceeds $180°$, all four vertices are crowded into less than half the circle, the centre falls on the far side of that arc’s chord, and it is outside.

How to test it without drawing the circle. An arc is greater than a semicircle exactly when the chord cutting it off subtends an angle of more than $90°$ at a point on the other arc. So for each side, look at the angle it subtends at either of the two non-adjacent vertices:

• side $AB$ subtends $\angle ACB$ (equivalently $\angle ADB$);
• side $BC$ subtends $\angle BDC$; and so on.

If any of these four angles exceeds $90°$, the centre is outside; if all are less than $90°$, it is inside. If one equals $90°$, that side is a diameter and the centre lies on the quadrilateral’s boundary.

Example. In a rectangle, each side subtends $90°$ at the far vertices — the diagonals are diameters — and indeed the centre sits exactly at the intersection of the diagonals, inside. In a very “flat” cyclic quadrilateral with all four vertices close together on one part of the circle, one arc is nearly the whole circle, and the centre is well outside.

Question 12

*When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.

Solution. Let chords $AB$ and $CD$ of a circle with centre $O$ satisfy $AB = CD$, and let them meet at $P$.

Step 1 — equal chords are equidistant. Drop perpendiculars $OM \perp AB$ and $ON \perp CD$. By Theorem 6, $OM = ON$. Also, $M$ and $N$ are the midpoints of their chords.

Step 2 — compare two right-angled triangles. In $\triangle OMP$ and $\triangle ONP$:

$$\angle OMP = \angle ONP = 90°, \qquad OP = OP \text{ (common)}, \qquad OM = ON$$

By RHS, $\triangle OMP \cong \triangle ONP$, so

$$MP = NP$$

Step 3 — build up the segments. Since $M$ and $N$ are the midpoints, $AM = \frac{1}{2}AB$ and $CN = \frac{1}{2}CD$, and these are equal. Adding or subtracting the equal pieces $MP$ and $NP$:

$$AP = AM + MP = CN + NP = CP$$ $$BP = BM – MP = DN – NP = DP$$

So the segments of one chord equal the corresponding segments of the other. $\blacksquare$

Question 13

*Draw a circle in which a chord of $6$ cm length stands at a distance of $3$ cm from the centre.

Solution. First find the radius, then construct.

$$r = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \approx 4.24 \text{ cm}$$

(half the chord is $3$ cm, the distance is $3$ cm).

Construction.

1. Draw segment $AB = 6$ cm.
2. Construct the perpendicular bisector of $AB$, meeting it at its midpoint $M$.
3. On that bisector, mark $O$ with $OM = 3$ cm.
4. With centre $O$ and radius $OA$ (which measures $3\sqrt2 \approx 4.2$ cm), draw the circle. It passes through $B$ as well.

The hint. “Is it a circumcircle of a suitable triangle?” — yes. Take any third point $C$ on the circle; then this is the circumcircle of $\triangle ABC$. More directly, mark $C$ on the bisector at distance $r$ from $O$ on the far side; then $\triangle ABC$ is isosceles with base $6$ cm and the circle is its circumcircle. Constructing the circumcircle of that triangle is an equivalent route to the same figure.

Question 14

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

Solution. Suppose a parallelogram $ABCD$ is inscribed in a circle. Two facts now apply at once.

From the parallelogram: opposite angles are equal.

$$\angle A = \angle C$$

From the circle: opposite angles of a cyclic quadrilateral are supplementary.

$$\angle A + \angle C = 180°$$

Substituting the first into the second:

$$\angle A + \angle A = 180° \quad\Longrightarrow\quad 2\angle A = 180° \quad\Longrightarrow\quad \angle A = 90°$$

So $\angle A = \angle C = 90°$, and by the same argument applied to $B$ and $D$, all four angles are right angles. A parallelogram with four right angles is a rectangle. $\blacksquare$

Question 15

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

Solution. Let rectangle $ABCD$ be inscribed in a circle, with diagonals meeting at $P$.

Step 1 — each diagonal is a diameter. The angle at $B$ is $90°$, and it is the angle subtended by $AC$ at a point on the circle. An inscribed angle of $90°$ corresponds to a central angle of $180°$, which means $AC$ is a diameter. The same argument at $A$ shows $BD$ is a diameter.

Step 2 — two diameters meet at the centre. Every diameter passes through the centre $O$, so $O$ lies on both $AC$ and $BD$. Two distinct lines meet in at most one point, and $AC$ and $BD$ meet at $P$. Hence

$$P = O$$

So the diagonals intersect at the centre. $\blacksquare$

As a corollary, the diagonals of the rectangle are both equal to the diameter, and each is bisected at the centre — consistent with what is known about rectangles from elsewhere.

Question 16

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

Solution. Let the circle have centre $O$ and radius $r$, and let every chord have the same fixed length $\ell$.

For each such chord, the perpendicular from $O$ meets it at its midpoint $M$, and the distance $OM$ satisfies

$$OM = \sqrt{r^2 – \left( \frac{\ell}{2} \right)^2}$$

Since $r$ and $\ell$ are both fixed, this distance is the same number for every chord — call it $d$.

So every midpoint is at the fixed distance $d$ from the fixed point $O$. That is precisely the definition of a circle: the midpoints form a circle of radius $d$, concentric with the original.

Every point of that inner circle is attained: given any point at distance $d$ from $O$, the chord perpendicular to that radius at that point has length $2\sqrt{r^2 – d^2} = \ell$.

Question 17

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

Solution. The two chords meet at $A$, so $\angle BAC$ is the angle between them.

Step 1 — the centre is equidistant from the two chords. $AB = AC$ are equal chords of the same circle, and equal chords are equidistant from the centre (Theorem 6). So the perpendicular distance from $O$ to line $AB$ equals the perpendicular distance from $O$ to line $AC$.

Step 2 — use the locus. The set of points inside an angle that are equidistant from its two arms is exactly the bisector of that angle.

Since $O$ is equidistant from the arms $AB$ and $AC$ of $\angle BAC$, it lies on the bisector of $\angle BAC$. $\blacksquare$

A second route. $\triangle OAB$ and $\triangle OAC$ have $OA$ common, $OB = OC$ (radii) and $AB = AC$ (given), so they are congruent by SSS. Hence $\angle OAB = \angle OAC$, which says exactly that $AO$ bisects $\angle BAC$.

Question 18

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.

Solution. Let $d$ be the distance from the centre to the longer chord ($24$ cm). Since the longer chord is nearer the centre and the chords are on the same side, the shorter chord is at distance $d + 7$.

Write $r^2$ twice, once from each chord:

$$r^2 = 12^2 + d^2 = 144 + d^2$$ $$r^2 = 5^2 + (d+7)^2 = 25 + d^2 + 14d + 49$$

Setting them equal:

$$144 + d^2 = 74 + d^2 + 14d$$ $$144 = 74 + 14d$$ $$14d = 70 \quad\Longrightarrow\quad d = 5$$

So the $24$ cm chord is $5$ cm from the centre and the $10$ cm chord is $12$ cm from it. Then

$$r = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm}$$

Check the other chord: $\sqrt{25 + 144} = 13$ ✓ And $12 – 5 = 7$ cm apart ✓

Question 19

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

Solution. Join the centre $O$ to each of the six vertices. This cuts the hexagon into six triangles, and by symmetry the six central angles are equal:

$$\text{each central angle} = \frac{360°}{6} = 60°$$

Side length. Each triangle has two sides equal to $r$ with $60°$ between them, so its base angles are $\frac{180° – 60°}{2} = 60°$ as well — the triangle is equilateral. Hence each side of the hexagon equals the radius:

$$\text{side} = r$$

Distance from the centre. Each side is a chord of length $r$, so

$$d = \sqrt{r^2 – \left( \frac{r}{2} \right)^2} = \sqrt{r^2 – \frac{r^2}{4}} = \sqrt{\frac{3r^2}{4}} = \frac{\sqrt3}{2}\,r$$

(This distance is the hexagon’s apothem, about $0.87\,r$.)

Question 20

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

Solution. Note that here $O$ is a vertex of the quadrilateral, not the centre. The vertices run $M$, $N$, $O$, $P$ around the circle.

The two angles are equal. Both $\angle MOP$ and $\angle MNP$ are subtended by the same chord $MP$ — the first from the vertex $O$, the second from the vertex $N$. Going round the circle in order $M \to N \to O \to P$, both $N$ and $O$ lie on the same arc relative to the chord $MP$. So they are angles in the same segment, and therefore equal:

$$\angle MOP = \angle MNP$$

What the diameter gives. Since $MN$ is a diameter, it subtends a right angle at every other point of the circle:

$$\angle MON = 90° \qquad \text{and} \qquad \angle MPN = 90°$$

So the quadrilateral has right angles in both triangles cut off by the diameter, and the two angles named in the question are equal to each other.

Question 21

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

Solution. Two facts, one from the straight line and one from the circle.

From the straight line. $E$ lies on the extension of $CD$ beyond $D$, so $\angle CDE$ and the interior angle $\angle CDA$ form a linear pair:

$$\angle CDE + \angle CDA = 180°$$

From the circle. $ABCD$ is cyclic, and $\angle CDA$ and $\angle ABC$ are opposite angles, so

$$\angle CDA + \angle ABC = 180°$$

Compare. Both $\angle CDE$ and $\angle ABC$ are equal to $180° – \angle CDA$. Two quantities equal to the same thing are equal to each other:

$$\angle CDE = \angle ABC \qquad \blacksquare$$

The same argument works at every vertex, because both facts used are available at each one.

Question 22

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

Solution. Two justifications, both short.

From the formula. A chord at distance $d$ from the centre has length

$$\ell = 2\sqrt{r^2 – d^2}$$

Since $d^2 \geq 0$, we have $r^2 – d^2 \leq r^2$, so $\ell \leq 2r$. Equality holds exactly when $d = 0$ — that is, when the chord passes through the centre and is a diameter.

From the triangle inequality. Let $AB$ be any chord and $O$ the centre. If $O$ does not lie on $AB$, then $OAB$ is a genuine triangle and

$$AB < OA + OB = r + r = 2r$$

If $O$ does lie on $AB$, then $AB$ is a diameter and $AB = 2r$ exactly. Either way $AB \leq 2r$. $\blacksquare$

Question 23

*Let $A$ be any point within a given circle with centre $O$. Show that the shortest chord of the circle that passes through point $A$ is the one that is perpendicular to $OA$.

Solution. Every chord through $A$ has length $2\sqrt{r^2 – d^2}$, where $d$ is the perpendicular distance from $O$ to that chord. The radius $r$ is fixed, so

$$\text{the chord is shortest exactly when } d \text{ is largest.}$$

How large can $d$ be? Let $M$ be the foot of the perpendicular from $O$ to the chord, so $OM = d$. The point $A$ lies on the chord, and $OM$ is the shortest distance from $O$ to any point of that line. Hence

$$d = OM \leq OA$$

So $d$ can never exceed the fixed length $OA$.

When is the maximum reached? Exactly when $M$ coincides with $A$ — that is, when $OA$ is itself perpendicular to the chord.

Therefore the chord through $A$ perpendicular to $OA$ has the greatest possible distance from the centre, and so is the shortest chord through $A$. Its length is

$$2\sqrt{r^2 – OA^2} \qquad \blacksquare$$

(The longest chord through $A$ is the one along $OA$ — the diameter, with $d = 0$.)

Question 24

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

Solution. Fig. 5.30 shows a semicircle on diameter $BC$ with centre $O$, a point $A$ on the arc, and the segment $OA$ drawn dashed. The tick marks show that $OB$, $OC$ and $OA$ are all equal — they are radii. The angles at the two ends of the diameter are marked $a$ and $b$.

The argument the figure sets up.

$\triangle OAB$ is isosceles because $OA = OB$, so its base angles are equal:

$$\angle OAB = \angle OBA = a$$

$\triangle OAC$ is isosceles because $OA = OC$, so likewise:

$$\angle OAC = \angle OCA = b$$

The angle at $A$ in the large triangle is the sum of these two:

$$\angle BAC = \angle OAB + \angle OAC = a + b$$

Now use the angle sum of $\triangle ABC$:

$$a + b + (a + b) = 180°$$ $$2(a + b) = 180°$$ $$a + b = 90°$$

So $\angle BAC = 90°$, whatever point $A$ is chosen on the semicircle. $\blacksquare$

The figure’s dashed segment $OA$ is the whole trick: drawing that one radius turns a single triangle into two isosceles ones, and the rest is the angle sum.

Question 25

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

Solution. The cleanest route is by reflection in the diameter.

Step 1 — $C$ and $C’$ are mirror images in $AB$. The chord $CC’$ is perpendicular to the diameter $AB$, and a perpendicular from the centre to a chord bisects it — so $AB$ meets $CC’$ at its midpoint and at right angles. That is exactly the statement that $AB$ is the perpendicular bisector of $CC’$, so reflecting in the line $AB$ swaps $C$ and $C’$.

Step 2 — likewise for $D$ and $D’$. The same argument applies to $DD’$, so reflection in $AB$ swaps $D$ and $D’$.

Step 3 — the segments are mirror images. Reflection in $AB$ therefore maps the segment $CD$ onto the segment $C’D’$. A reflection maps a midpoint to a midpoint, so it maps $M$, the midpoint of $CD$, onto $M’$, the midpoint of $C’D’$.

Step 4 — conclude. A segment joining a point to its mirror image is always perpendicular to the mirror line (the mirror line is its perpendicular bisector). Since $M’$ is the image of $M$ under reflection in $AB$,

$$MM’ \perp AB \qquad \blacksquare$$

(If $M$ happens to lie on $AB$ then $M’ = M$ and the segment degenerates to a point — which occurs only when $CD$ is itself symmetric about the diameter.)

Question 26

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

Solution. Fig. 5.31 shows a cyclic quadrilateral $ABCD$ with centre $O$, and the four radii $OA$, $OB$, $OC$, $OD$ drawn — the tick marks confirming they are equal. Four base angles are labelled: $p$ at $A$, $q$ at $B$, $u$ at $C$, $v$ at $D$.

The argument. The four radii split the quadrilateral into four triangles $OAB$, $OBC$, $OCD$, $ODA$, and each is isosceles, since two of its sides are radii. So in each triangle the two base angles are equal. Write them as:

$$\triangle OAB: \; \angle OAB = \angle OBA = p$$ $$\triangle OBC: \; \angle OBC = \angle OCB = q’$$ $$\triangle OCD: \; \angle OCD = \angle ODC = u$$ $$\triangle ODA: \; \angle ODA = \angle OAD = v$$

Each vertex angle of the quadrilateral is made of two of these:

$$\angle A = p + v, \quad \angle B = p + q’, \quad \angle C = q’ + u, \quad \angle D = u + v$$

Add all four. Every one of $p$, $q’$, $u$, $v$ appears exactly twice, and the angles of a quadrilateral sum to $360°$:

$$2(p + q’ + u + v) = 360° \quad\Longrightarrow\quad p + q’ + u + v = 180°$$

Now take an opposite pair.

$$\angle A + \angle C = (p + v) + (q’ + u) = p + q’ + u + v = 180°$$

and likewise

$$\angle B + \angle D = (p + q’) + (u + v) = 180°$$

So each pair of opposite angles sums to $180°$. $\blacksquare$

What makes the figure work is that every base angle gets counted once in each of the two opposite-angle sums — which is why both pairs give the same total.

Common mistakes

• Questions 1, 3, 4 and 9, forgetting to halve the chord. The right-angled triangle uses half the chord. Question 9 is the one going the other way: there the radius is the hypotenuse, so the two legs are added, not subtracted.
• Question 2, doubling instead of halving. The central angle is twice the inscribed one, so going from centre to circumference divides by $2$.
• Questions 7 and 8, adding adjacent angles. Only opposite angles are supplementary. Read the vertex order to identify the pairs: in $PQRS$ the opposite pairs are $P$–$R$ and $Q$–$S$.
• Question 10, using Heron’s formula on the whole quadrilateral. Heron’s formula is for triangles. Brahmagupta’s is its cyclic-quadrilateral counterpart — or split into two triangles along a diagonal.
• Question 11, assuming the centre is always inside. It is not: a cyclic quadrilateral whose vertices are crowded into one part of the circle leaves the centre outside.
• Question 13, taking the radius as $6$ or $3$ cm. Neither: the radius is the hypotenuse of a triangle with legs $3$ and $3$, so $3\sqrt2 \approx 4.24$ cm.
• Question 14, assuming a rhombus works. A rhombus has equal sides but its angles are not right angles unless it is a square — and the argument shows only rectangles can be inscribed.
• Question 16, answering “a circle” without the radius. The radius $\sqrt{r^2 – (\ell/2)^2}$ is the content of the answer; “a circle” alone does not say which.
• Question 18, adding the distances. The chords are on the same side here, so the distances subtract — unlike Exercise Set 5.3 Question 3, where they were on opposite sides and added. The word “same” or “opposite” is what decides.
• Question 20, treating $O$ as the centre. In $MNOP$, $O$ is a vertex. Reading it as the centre makes $\angle MOP$ a central angle and gives a different, wrong answer.
• Question 22, checking a few chords. Drawing several chords and seeing none longer is not a proof. The formula or the triangle inequality covers every chord at once.
• Question 23, maximising the chord instead of minimising it. Shortest chord means largest distance from the centre. The longest chord through $A$ is the diameter along $OA$, which is the opposite extreme.
• Questions 24 and 26, arguing from the figure’s appearance. The figures supply a construction — the radius $OA$, or the four radii — and the argument is the isosceles triangles that construction creates.

Practise next

• Chapter 6, Measuring Space: Perimeter and Area — the next chapter, which returns to measurement, including the circumference and area of the circle studied here.
• Exercise Set 5.6 — worth revisiting alongside Questions 6, 20, 24 and 26, all of which turn on the central-angle-is-twice-the-inscribed-angle rule.
• Exercise Set 5.3, Question 3 — the parallel-chords problem with the chords on opposite sides, to set beside Question 18 here.