Question 1

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

Accepted Answer

Construction. Draw $AB=5$ cm. At $A$ draw a ray making $70^\circ$ with $AB$; at $B$ draw a ray making $60^\circ$ with $BA$. The two rays meet at $C$.To draw the circumcircle, construct the perpendicular bisectors of any two sides. They meet at the circumcentre $O$; with radius $OA$ draw the circle — it passes through all three vertices.Where is the centre? First find the third angle. $\angle C = 180^\circ - 70^\circ - 60^\circ = 50^\circ$ All three angles — $70^\circ$, $60^\circ$, $50^\circ$ — are less than $90^\circ$, so the triangle is acute-angled. Hence the circumcentre lies inside the triangle.Useful rule of thumb: the circumcentre is inside for an acute triangle, on the hypotenuse (at its midpoint) for a right triangle, and outside for an obtuse triangle.

Question 2

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

Accepted Answer

Construction. Draw $AB=5$ cm. At $A$ draw a ray making $100^\circ$ with $AB$ and cut off $AC=4$ cm on it. Join $BC$. Construct the perpendicular bisectors of two sides; their meeting point $O$ is the circumcentre.Where is the centre? The given angle $\angle A = 100^\circ$ is more than $90^\circ$, so the triangle is obtuse-angled. $\angle A = 100^\circ > 90^\circ$ Therefore the circumcentre lies outside the triangle — on the far side of $BC$ from $A$.The reason: the central angle standing on $BC$ would have to be $2\times 100^\circ = 200^\circ$, which is more than a straight angle. That is only possible if $O$ and $A$ lie on opposite sides of $BC$.

Question 3

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

Accepted Answer

Construction. Draw $AB=6$ cm. With centre $A$ and radius $7$ cm draw an arc; with centre $B$ and radius $7$ cm draw another arc. They cross at $C$. Join $AC$ and $BC$.Construct the perpendicular bisectors of $AB$ and $BC$. They meet at $O$, the circumcentre.Measurement. On measuring you will find that all three distances are the same: $OA = OB = OC$ That is exactly what makes $O$ the centre of a circle through $A$, $B$ and $C$. The common value is the circumradius $R$, which can be checked by calculation. Using $R = \dfrac{abc}{4\times\text{area}}$ with $a=7$, $b=7$, $c=6$: $\text{area} = \frac{6}{4}\sqrt{4(7)^2-6^2}$ $= 6\sqrt{10}\ \text{cm}^2$ $R = \frac{7\times 7\times 6}{4\times 6\sqrt{10}} = \frac{49\sqrt{10}}{40}$ $R \approx 3.87\ \text{cm}$ So each of $OA$, $OB$, $OC$ measures about $3.9$ cm. Since $\triangle ABC$ is isosceles and acute, $O$ lies inside it, on the perpendicular bisector of $AB$.

Question 4

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

Accepted Answer

Any circle through $A$ and $B$ has $AB$ as a chord, so its centre $O$ must be equidistant from $A$ and $B$ — that is, $O$ lies on the perpendicular bisector of $AB$.Let $M$ be the midpoint of $AB$ and let $OM = d$. By the Baudhāyana–Pythagoras theorem in $\triangle OMA$: $r^2 = \left(\frac{AB}{2}\right)^2 + d^2$ The radius $r$ is smallest when $d$ is smallest, and the smallest $d$ can be is $0$ — when $O$ is at $M$ itself. $r_{\min} = \frac{AB}{2}$ So the least possible radius is half the distance $AB$, and it is achieved by the circle having $AB$ as a diameter. There is exactly one such circle.Two points do not determine a circle — infinitely many circles pass through them, one for each position of the centre on the perpendicular bisector. Among all of them, the one with $AB$ as diameter is the smallest.

Question 5

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

Accepted Answer

Let the circle have centre $O$ and let $AB$ be any chord. Joining $OA$ and $OB$ forms $\triangle OAB$.Why it is isosceles. $A$ and $B$ both lie on the circle, so both are at the same distance from the centre — the radius $r$. $OA = OB = r$ $\therefore\ OA = OB$ A triangle with two equal sides is isosceles, so $\triangle OAB$ is isosceles with $AB$ as its base.As a consequence, the base angles are equal: $\angle OAB = \angle OBA$ This tiny fact is the engine behind most of this chapter. Every chord gives you a free isosceles triangle, and isosceles triangles give you equal angles to work with.

Question 6

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

Accepted Answer

Let $AB$ and $DE$ be two chords of the same circle with centre $O$, and suppose the bases are equal, $AB = DE$. Compare $\triangle OAB$ and $\triangle ODE$. $OA = OD = r$ $OB = OE = r$ $AB = DE$ All three pairs of corresponding sides are equal, so by the SSS congruence rule: $\triangle OAB \cong \triangle ODE$ Two immediate consequences follow from CPCT (corresponding parts of congruent triangles): $\angle AOB = \angle DOE$ that is, equal chords subtend equal angles at the centre — which is Theorem 2 of this chapter — and the two triangles also have equal heights, so equal chords are equidistant from the centre.The argument needs the two circles to be the same size. Two equal chords in circles of different radii give triangles that are not congruent, because the radii differ.

Question 7

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

Accepted Answer

Let $C$ be the centre, $AB$ a chord, and let $CM$ be perpendicular to $AB$ with $M$ on $AB$. We must show $M$ is the midpoint of $AB$.Compare the two right triangles $\triangle CMA$ and $\triangle CMB$. $\angle CMA = \angle CMB = 90^\circ$ $CA = CB = r$ $CM = CM$ By the RHS congruence rule: $\triangle CMA \cong \triangle CMB$ $\therefore\ AM = BM$ So $M$ is the midpoint of $AB$; the perpendicular from the centre bisects the chord. This is Theorem 5.You can also see it with the Baudhāyana–Pythagoras theorem: $AM^2 = CA^2 - CM^2 = CB^2 - CM^2 = BM^2$, and since lengths are positive, $AM = BM$.

Question 8

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.

Accepted Answer

Let $O$ be the centre of the circle and let $AD$ be the altitude from $A$ to $BC$.Step 1: the altitude is the perpendicular bisector of $BC$. In $\triangle ABD$ and $\triangle ACD$: $AB = AC$ $\angle ADB = \angle ADC = 90^\circ$ $AD = AD$ By RHS congruence $\triangle ABD \cong \triangle ACD$, so $BD = DC$. Thus the line $AD$ is perpendicular to $BC$ and passes through its midpoint — it is the perpendicular bisector of $BC$.Step 2: the centre lies on that perpendicular bisector. $O$ is equidistant from $B$ and $C$ because both are on the circle: $OB = OC = r$ Every point equidistant from $B$ and $C$ lies on the perpendicular bisector of $BC$. Hence $O$ lies on line $AD$.Therefore the altitude from $A$ passes through the centre $O$.The same argument shows that in an isosceles inscribed triangle, the altitude, the median from $A$, the angle bisector of $\angle A$ and the line $AO$ are all the same line.

Question 9

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.

Accepted Answer

Let $O$ be the centre and $r = 5$ cm. Let $AB = 6$ cm and $CD = 8$ cm be the two chords, with midpoints $M$ and $N$.The perpendicular from the centre bisects a chord, so $OM \perp AB$ with $AM = 3$ cm, and $ON \perp CD$ with $CN = 4$ cm.Distance of the 6 cm chord. In right $\triangle OMA$: $OM^2 = 5^2 - 3^2 = 25 - 9 = 16$ $OM = 4\ \text{cm}$ Distance of the 8 cm chord. In right $\triangle ONC$: $ON^2 = 5^2 - 4^2 = 25 - 16 = 9$ $ON = 3\ \text{cm}$ Both chords are parallel, so $OM$ and $ON$ lie along one and the same perpendicular line. The chords are on opposite sides of the centre, so the two distances add: $MN = OM + ON = 4 + 3 = 7\ \text{cm}$ The distance between the midpoints is 7 cm.If the two chords had been on the same side of the centre, you would subtract instead: $4 - 3 = 1$ cm. Always read which side the question puts them on.

Question 10

Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true. (Theorem 6: chords of a circle that are of equal length are equidistant from the centre.)

Accepted Answer

Let the circle have centre $O$ and radius $r$, and let $AB$ and $DE$ be two chords with $AB = DE$. Drop perpendiculars $OM$ and $ON$ from $O$ to $AB$ and $DE$.The perpendicular from the centre bisects the chord (Theorem 5), so: $AM = \frac{AB}{2}$ Since $AB = DE$, these halves are equal: $AM = DN$.Now apply the Baudhāyana–Pythagoras theorem in the right triangles $\triangle OMA$ and $\triangle OND$: $OM^2 = OA^2 - AM^2 = r^2 - AM^2$ $ON^2 = OD^2 - DN^2 = r^2 - DN^2$ The right-hand sides are equal because $AM = DN$. Hence: $OM^2 = ON^2$ $OM = ON$ Distances are positive, so $OM = ON$: the two equal chords are the same distance from the centre. That is Theorem 6.

Question 11

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

Accepted Answer

Reading the figure: $C$ is the centre. $AB$ and $GF$ are two chords. $CE \perp AB$ with $E$ on $AB$, and $CH \perp GF$ with $H$ on $GF$. We are told $CE = CH$, i.e. the two chords are equidistant from the centre.Step 1. Compare the right triangles $\triangle CEA$ and $\triangle CHF$. $\angle CEA = \angle CHF = 90^\circ$ $CA = CF = r$ $CE = CH$ By the RHS congruence rule, $\triangle CEA \cong \triangle CHF$, so $AE = FH$ by CPCT.Step 2. The perpendicular from the centre bisects the chord, so $E$ is the midpoint of $AB$ and $H$ is the midpoint of $GF$. $AB = 2\,AE$ $\therefore\ AB = GF$ So chords that are equidistant from the centre have equal length. This is Theorem 7, the converse of Theorem 6.

Question 12

Solve the previous question using the Baudhāyana–Pythagoras theorem.

Accepted Answer

Same setting: centre $C$, radius $r$, $CE \perp AB$, $CH \perp GF$ and $CE = CH$.Because the perpendicular from the centre bisects the chord, $AE = \frac{1}{2}AB$ and $FH = \frac{1}{2}GF$. Apply the theorem in each right triangle: $AE^2 = CA^2 - CE^2 = r^2 - CE^2$ $FH^2 = CF^2 - CH^2 = r^2 - CH^2$ Since $CE = CH$, the two right-hand sides are equal, so: $AE^2 = FH^2$ $AE = FH$ Doubling both sides: $AB = 2\,AE = 2\,FH = GF$ Hence $AB = GF$, proved without using congruence at all — only lengths.Both routes are valid. Congruence is quicker to write; the Baudhāyana–Pythagoras route is easier to turn into a numerical formula, which is exactly what Exercise Set 5.5 asks for.

Question 13

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

Accepted Answer

Let $O$ be the centre, $AB$ the chord, and $M$ the foot of the perpendicular from $O$ to $AB$. Then $OA = r = 7$ cm and $OM = d = 6$ cm, and $M$ is the midpoint of $AB$.In right $\triangle OMA$: $AM^2 = OA^2 - OM^2 = 7^2 - 6^2$ $AM^2 = 49 - 36 = 13$ $AM = \sqrt{13}\ \text{cm}$ The chord is twice this half-length: $AB = 2\sqrt{13}\ \text{cm}$ $AB \approx 7.21\ \text{cm}$ So the chord is $2\sqrt{13}$ cm, roughly 7.2 cm.

Question 14

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 $2\sqrt{r^2-d^2}$.

Accepted Answer

Let $O$ be the centre and $AB$ the chord. Drop the perpendicular $OM$ from $O$ to $AB$, so $OM = d$.Step 1. By Theorem 5, the perpendicular from the centre bisects the chord, so $M$ is the midpoint of $AB$ and $AM = \frac{AB}{2}$ Step 2. $\triangle OMA$ is right-angled at $M$, with hypotenuse $OA = r$. By the Baudhāyana–Pythagoras theorem: $OM^2 + AM^2 = OA^2$ $d^2 + \left(\frac{AB}{2}\right)^2 = r^2$ $\left(\frac{AB}{2}\right)^2 = r^2 - d^2$ $\frac{AB}{2} = \sqrt{r^2-d^2}$ $AB = 2\sqrt{r^2-d^2}$ This is the general chord-length formula. It also confirms two facts you already know: when $d = 0$ the chord is the diameter $2r$ (the longest chord), and as $d$ grows towards $r$ the chord shrinks towards $0$.Read backwards, the same formula gives the distance from the centre: $d = \sqrt{r^2 - (AB/2)^2}$, and the radius: $r = \sqrt{d^2 + (AB/2)^2}$. One formula, three questions.

Question 15

*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.

Accepted Answer

No, we cannot. The relationship between distance and chord length is not proportional — it involves a square root.Let the distances be $d_{AB} = 2d$ and $d_{CD} = d$. By the chord formula: $AB = 2\sqrt{r^2-4d^2}$ $CD = 2\sqrt{r^2-d^2}$ For $CD = 2\,AB$ we would need $\sqrt{r^2-d^2} = 2\sqrt{r^2-4d^2}$, i.e. $r^2-d^2 = 4r^2-16d^2$, i.e. $15d^2 = 3r^2$. That happens only for the one special case $d = r/\sqrt{5}$ — not in general.A concrete counterexample. Take $r = 10$ cm, $d_{CD} = 3$ cm and $d_{AB} = 6$ cm. $AB = 2\sqrt{100-36} = 2\times 8 = 16\ \text{cm}$ $CD$ $= 2\sqrt{100-9}$ $= 2\sqrt{91} \approx 19.08\ \text{cm}$ Here $2\,AB = 32$ cm, but $CD \approx 19.1$ cm. So $CD \neq 2\,AB$.What we can say is the qualitative statement of Theorem 8: since $AB$ is farther from the centre than $CD$, the chord $AB$ is the shorter of the two. $CD > AB$, but not by a factor of $2$.A good habit whenever a question says “can we conclude”: try to build one counterexample. A single counterexample settles it — you do not have to argue about every case.

Question 16

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

Accepted Answer

In $\triangle OAB$, the two sides $OA$ and $OB$ are radii, so $OA = OB = 12$ cm and the triangle is isosceles.The base angles are equal, and the three angles add to $180^\circ$: $\angle OAB$ $= \angle OBA$ $= \frac{180^\circ - 60^\circ}{2}$ $= 60^\circ$ All three angles are $60^\circ$, so $\triangle OAB$ is equilateral. Hence the chord equals the radius: $AB = OA = OB = 12\ \text{cm}$ The chord $AB$ is 12 cm long.Checking with the chord formula: the perpendicular from $O$ to $AB$ makes a $30^\circ\text{-} 60^\circ\text{-} 90^\circ$ triangle, so $d = 12\cos 30^\circ = 6\sqrt{3}$ and $AB = 2\sqrt{144-108} = 2\times 6 = 12$ cm. Same answer.

Question 17

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 $\angle AXB$ is different from $\angle AYB$? (ii) Is it true that if $\angle AXB = \angle AYB$, then $X$ and $Y$ lie on the same side of t

Accepted Answer

(i) No. All points on the same side of $AB$ lie on the same arc, and every angle subtended by $AB$ at a point of that arc is half the central angle $\angle AOB$: $\angle AXB = \angle AYB = \frac{1}{2}\angle AOB$ This is the “angles in the same segment are equal” result. So no two such points can give different angles.(ii) Not necessarily. If $AB$ happens to be a diameter, then every point of the circle — on either side — gives $\angle AXB = 90^\circ$. So equal angles do not force the points onto the same side. $\text{If } AB \text{ is a diameter: } \angle AXB$ $= \angle AYB$ $= 90^\circ$ For a chord that is not a diameter the two sides give supplementary angles $\theta$ and $180^\circ-\theta$, which are equal only when $\theta = 90^\circ$ — the diameter case again.(iii) Yes, provided $X$ and $Y$ are on the same side of $AB$. This is exactly Theorem 10 (concyclicity). If a segment $AB$ subtends equal angles at two points $X$ and $Y$ on the same side of $AB$, then $A$, $B$, $X$, $Y$ lie on one circle. $\angle AXB$ $= \angle AYB \implies A, B, X, Y \text{ are concyclic}$ So the circle drawn through $A$, $B$, $X$ must also pass th

Question 18

Find $x$ in Fig. 5.26. (A cyclic quadrilateral $ADCB$ is drawn; $\angle ADC = 100^\circ$ and $x = \angle ABC$.)

Accepted Answer

Reading the figure: the four points lie on the circle in the order $A$, $D$, $C$, $B$. The marked angle $100^\circ$ is at $D$, and $x$ is the angle at $B$. $\angle ADC$ and $\angle ABC$ are therefore opposite angles of the cyclic quadrilateral.Method 1 — opposite angles of a cyclic quadrilateral. They add up to a straight angle: $\angle ADC + \angle ABC = 180^\circ$ $100^\circ + x = 180^\circ$ $x = 80^\circ$ Method 2 — through the centre. $\angle ADC$ stands on the arc $ABC$, so the central angle over that arc is: $2\times 100^\circ = 200^\circ$ The remaining central angle, over arc $ADC$, is: $360^\circ - 200^\circ = 160^\circ$ and $x$ is half of it: $x = \frac{160^\circ}{2} = 80^\circ$ Both routes give $x = \mathbf{80^\circ}$.

Question 19

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?

Accepted Answer

Use the chord formula with $r = 13$ cm and $d = 5$ cm. $\text{chord} = 2\sqrt{r^2-d^2}$ $= 2\sqrt{13^2-5^2}$ $= 2\sqrt{169-25} = 2\sqrt{144}$ $= 2\times 12 = 24\ \text{cm}$ The chord is 24 cm long.

Question 20

An arc of a circle subtends an angle of $70^\circ$ at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

Accepted Answer

The angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle (Theorem 9).

$\angle \text{at the centre}$

$= 2 \times \angle \text{at the circle}$

$70^\circ = 2 \times \angle \text{at the circle}$

$\angle \text{at the circle} = 35^\circ$

So the arc subtends $35^\circ$ at a point on the major arc.

If the point is taken on the minor arc instead, the angle is $\frac{1}{2}(360^\circ-70^\circ) = 145^\circ$. The two answers are supplementary, which is why opposite angles of a cyclic quadrilateral add to $180^\circ$.

Question 21

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.

Accepted Answer

First the radius: $r = \frac{26}{2} = 13\ \text{cm}$ The perpendicular from the centre bisects the chord, so half the chord is $12$ cm. In the right triangle formed by the radius, half the chord and the distance $d$: $d^2 = r^2 - \left(\frac{\text{chord}}{2}\right)^2$ $d^2 = 13^2 - 12^2 = 169 - 144 = 25$ $d = 5\ \text{cm}$ The chord is 5 cm from the centre.

Question 22

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?

Accepted Answer

Apply the chord formula with $r = 15$ cm and $d = 9$ cm. $\text{chord} = 2\sqrt{15^2-9^2}$ $= 2\sqrt{225-81} = 2\sqrt{144}$ $= 2\times 12 = 24\ \text{cm}$ The chord is 24 cm long.The right triangle here has sides $9$, $12$, $15$ — a scaled $3\text{-} 4\text{-} 5$ triple. Spotting these saves time in an exam.

Question 23

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

Accepted Answer

Let $O$ be the centre and $AB$ a chord of the circle. We must show that $O$ lies on the perpendicular bisector of $AB$.Key fact. The perpendicular bisector of a segment is precisely the set of points equidistant from the two end points.Now $A$ and $B$ both lie on the circle, so: $OA = OB = r$ Therefore $O$ is equidistant from $A$ and $B$, and so $O$ lies on the perpendicular bisector of $AB$.Equivalently, in the language of Theorem 4: let $M$ be the midpoint of $AB$. Then $\triangle OMA \cong \triangle OMB$ by SSS, so $\angle OMA = \angle OMB$; these two angles are on a straight line, so each is $90^\circ$. $\angle OMA + \angle OMB = 180^\circ$ $\angle OMA = \angle OMB = 90^\circ$ So $OM$ is perpendicular to $AB$ at its midpoint — that is, the line $OM$ is the perpendicular bisector of $AB$, and it contains $O$.This is why the circumcentre of a triangle is found by intersecting perpendicular bisectors: the centre must lie on the perpendicular bisector of every chord, in particular of all three sides.

Question 24

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

Accepted Answer

The answer is $90^\circ$ — the angle in a semicircle is a right angle.Reasoning. Let $O$ be the centre. The arc $AB$ not containing $C$ subtends the angle $\angle AOB$ at the centre. Since $AB$ is a diameter, $A$, $O$, $B$ are collinear, so that central angle is a straight angle: $\angle AOB = 180^\circ$ By Theorem 9, the angle at a point on the circle is half the central angle: $\angle ACB = \frac{1}{2}\times 180^\circ = 90^\circ$ An alternative proof using isosceles triangles. Join $OC$. Then $OA = OB = OC = r$, so $\triangle OAC$ and $\triangle OBC$ are both isosceles. Let $\angle OAC = \angle OCA = a$ and $\angle OBC = \angle OCB = b$. The angles of $\triangle ABC$ add to $180^\circ$: $a + b + (a+b) = 180^\circ$ $2(a+b) = 180^\circ$ $\angle ACB = a+b = 90^\circ$ This is the corollary stated in the chapter, and it holds for every position of $C$ on the circle.

Question 25

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

Accepted Answer

In a cyclic quadrilateral, opposite angles are supplementary — they add up to $180^\circ$.

Finding $\angle C$ (opposite to $\angle A$):

$\angle A + \angle C = 180^\circ$

$75^\circ + \angle C = 180^\circ$

$\angle C = 105^\circ$

Finding $\angle D$ (opposite to $\angle B$):

$\angle B + \angle D = 180^\circ$

$110^\circ + \angle D = 180^\circ$

$\angle D = 70^\circ$

Check. All four angles of any quadrilateral must total $360^\circ$:

$75^\circ + 110^\circ + 105^\circ + 70^\circ$

$= 360^\circ$

Angle	A	B	C	D
Measure	75°	110°	105°	70°

Question 26

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

Accepted Answer

In the quadrilateral $PQRS$, the vertices $P$ and $R$ are opposite each other, so $\angle P$ and $\angle R$ are supplementary. $\angle P + \angle R = 180^\circ$ $(2x+10) + (3x-20) = 180$ $5x - 10 = 180$ $5x = 190$ $x = 38$ Now substitute $x = 38$: $\angle P = 2(38)+10 = 86^\circ$ $\angle R = 3(38)-20 = 94^\circ$ Check: $86^\circ + 94^\circ = 180^\circ$. ✓So $x = \mathbf{38}$, $\angle P = \mathbf{86^\circ}$ and $\angle R = \mathbf{94^\circ}$.

Question 27

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

Accepted Answer

Half the chord is $8$ cm, and the perpendicular from the centre meets the chord at its midpoint. That gives a right triangle with legs $8$ cm and $6$ cm and hypotenuse $r$. $r^2 = \left(\frac{16}{2}\right)^2 + 6^2$ $r^2 = 8^2 + 6^2 = 64 + 36 = 100$ $r = 10\ \text{cm}$ The radius is 10 cm. (Another $3\text{-} 4\text{-} 5$ triangle, scaled by $2$.)

Question 28

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

Accepted Answer

The sides come in equal adjacent pairs, so the quadrilateral is a kite — call it $ABCD$ with $AB = BC = 5$ and $CD = DA = 12$.

Method 1 — Brahmagupta's formula for a cyclic quadrilateral with sides $a,b,c,d$ and semi-perimeter $s$:

$s = \frac{5+5+12+12}{2} = 17$

$\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$

$= \sqrt{(17-5)(17-5)(17-12)(17-12)}$

$= \sqrt{12\times 12\times 5\times 5}$

$= 12\times 5 = 60\ \text{sq units}$

Method 2 — split it into two right triangles. In a cyclic kite the two angles between the unequal sides are right angles, because the diagonal $AC$ turns out to be a diameter. So the quadrilateral is two copies of a $5\text{-} 12\text{-} 13$ right triangle glued along the hypotenuse:

$\text{Area} = 2\times\frac{1}{2}\times 5\times 12$

$= 60\ \text{sq units}$

Both methods give 60 square units. The diagonal joining the two “mixed” vertices is $\sqrt{5^2+12^2} = 13$ units, so the circumradius is $6.5$ units.

Question 29

*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?

Accepted Answer

The idea. You do not need to draw the circle — the angles of the quadrilateral already tell you which side of each edge the centre falls on.Side-by-side test. The centre $O$ lies on the same side of a chord as the arc that the chord's inscribed angle stands on. For a chord $PQ$ of the quadrilateral with opposite vertex $R$: $\angle PRQ < 90^\circ \implies O \text{ and } R \text{ are on the same side of } PQ$ $\angle PRQ > 90^\circ \implies O \text{ and } R \text{ are on opposite sides}$ $\angle PRQ = 90^\circ \implies O \text{ lies on } PQ$ Apply this to each of the four sides. If for every side the centre falls on the inner side, the centre is inside; if it falls outside for even one side, the centre is outside.The quick version. Split the quadrilateral by a diagonal into two triangles. The centre of the circumcircle of the quadrilateral is the circumcentre of both triangles. So: $\text{all angles of both triangles acute} \implies \text{centre inside}$ If either triangle has an obtuse angle, its circumcentre falls outside that triangle, and you then check whether it landed inside the other one.Best practical method: check whether any side of the quadrilateral is a di

Question 30

*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.

Accepted Answer

Let chords $AB$ and $CD$ of a circle with centre $O$ meet at a point $P$ inside the circle, and suppose $AB = CD$.Step 1: they are equidistant from the centre. Equal chords are equidistant from the centre (Theorem 6). Let $M$ and $N$ be the midpoints of $AB$ and $CD$, so $OM \perp AB$, $ON \perp CD$ and: $OM = ON$ Step 2: $P$ is equidistant from the two midpoints. Compare the right triangles $\triangle OMP$ and $\triangle ONP$: $\angle OMP = \angle ONP = 90^\circ$ $OP = OP$ $OM = ON$ By RHS congruence, $\triangle OMP \cong \triangle ONP$, so by CPCT: $PM = PN$ Step 3: read off the segments. $M$ and $N$ are midpoints, so $AM = MB = \frac{1}{2}AB$ and $CN = ND = \frac{1}{2}CD$, and these halves are equal. Now: $AP = AM - PM = CN - PN = CP$ $PB = MB + PM = ND + PN = PD$ So $AP = CP$ and $PB = PD$: the segments of one chord match the corresponding segments of the other.The signs in Step 3 flip depending on which side of the midpoint $P$ falls, but because $PM = PN$ the same cancellation works in every configuration — the two chords are mirror images of each other in the line $OP$.

Question 31

*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?)

Accepted Answer

Step 1: find the radius first. Half the chord is $3$ cm and the distance from the centre is $3$ cm, so: $r^2 = 3^2 + 3^2 = 9 + 9 = 18$ $r = 3\sqrt{2} \approx 4.24\ \text{cm}$ Step 2: construction. Draw a line segment $AB = 6$ cm. Construct its perpendicular bisector and mark $M$, the midpoint. On the perpendicular bisector, mark a point $O$ with $OM = 3$ cm. With centre $O$ and radius $OA$ (which will measure about $4.24$ cm), draw the circle. It passes through $A$ and $B$, and $AB$ is exactly $3$ cm from $O$.Step 3: the hint's route. Yes — this circle is the circumcircle of a suitable triangle. Since $OM = AM = MB = 3$ cm, the angle $\angle AOB$ is $90^\circ$. Take any third point $C$ on the major arc; then: $\angle ACB = \frac{1}{2}\angle AOB = 45^\circ$ So the circle is the circumcircle of any triangle $ABC$ with $AB = 6$ cm and $\angle C = 45^\circ$ — for instance the isosceles right triangle $AOB$ extended, or simply $\triangle ABC$ with $AB = 6$ cm, $\angle A = \angle B = 67.5^\circ$.Constructing the triangle first and then its circumcircle is the alternative route: draw $AB = 6$ cm, make $\angle C = 45^\circ$ at any

Question 32

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

Accepted Answer

Let a parallelogram $ABCD$ be inscribed in a circle. We show it must be a rectangle.Fact 1 — parallelogram. Opposite angles of a parallelogram are equal: $\angle A = \angle C$ Fact 2 — cyclic quadrilateral. Opposite angles of a cyclic quadrilateral are supplementary: $\angle A + \angle C = 180^\circ$ Substituting Fact 1 into Fact 2: $\angle A + \angle A = 180^\circ$ $2\angle A = 180^\circ$ $\angle A = 90^\circ$ Then $\angle C = 90^\circ$ as well, and the same argument applied to the pair $\angle B$, $\angle D$ gives $\angle B = \angle D = 90^\circ$.All four angles are right angles, so $ABCD$ is a rectangle. Conversely, every rectangle can be inscribed — its diagonals are equal and bisect each other, so their common midpoint is equidistant from all four vertices.A square is a special rectangle, so squares are inscribable too. A non-rectangular rhombus is not — its opposite angles are equal but not $90^\circ$, so they cannot sum to $180^\circ$.

Question 33

*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.

Accepted Answer

Let rectangle $ABCD$ be inscribed in a circle with centre $O$, and let the diagonals $AC$ and $BD$ meet at $P$.Step 1: each diagonal is a diameter. Every angle of a rectangle is $90^\circ$. In particular $\angle ABC = 90^\circ$, and $\angle ABC$ is the angle subtended by the chord $AC$ at the point $B$ on the circle. By the corollary to Theorem 9, an inscribed angle is $90^\circ$ only when it stands on a diameter: $\angle ABC = 90^\circ \implies AC \text{ is a diameter}$ The same argument with $\angle BCD = 90^\circ$ shows $BD$ is a diameter too.Step 2: the diameters meet at the centre. Every diameter passes through the centre $O$. Two distinct diameters meet at exactly one point, and that point is $O$. Since $AC$ and $BD$ meet at $P$: $P = O$ Alternative route. The diagonals of a rectangle bisect each other, so $P$ is the midpoint of both. Hence: $PA = PB = PC = PD$ A point equidistant from all four vertices of a cyclic quadrilateral is the centre of its circumcircle, so again $P = O$.

Question 34

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

Accepted Answer

Let the circle have centre $O$ and radius $r$, and let every chord have the same fixed length $\ell$. Let $M$ be the midpoint of one such chord.The line $OM$ is perpendicular to the chord (Theorem 4), so $OM$ is the distance of the chord from the centre. By the chord formula: $\ell = 2\sqrt{r^2 - OM^2}$ $OM^2 = r^2 - \frac{\ell^2}{4}$ $OM = \sqrt{r^2-\frac{\ell^2}{4}}$ The right-hand side depends only on $r$ and $\ell$, which are both fixed. So every such midpoint is the same distance from $O$.A set of points all at a fixed distance from a fixed point is a circle. Therefore the midpoints trace out a circle concentric with the given circle, of radius: $\rho = \sqrt{r^2-\frac{\ell^2}{4}}$ Two edge cases are worth noting. If $\ell = 2r$ (all chords are diameters) then $\rho = 0$ and the “circle” shrinks to the single point $O$. If $\ell$ is very small, $\rho$ is close to $r$ and the midpoints hug the original circle.Every point of that inner circle really is achieved: rotate one chord about the centre and its midpoint sweeps the whole inner circle. So the answer is the complete circle, not just part of it.

Question 35

*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 $\angle BAC$”.

Accepted Answer

We are told $AB = AC$, and both are chords through the common point $A$.Step 1: equal chords are equidistant from the centre. Drop perpendiculars $OM$ to $AB$ and $ON$ to $AC$. By Theorem 6: $OM = ON$ Step 2: a point equidistant from the two arms lies on the bisector. The distance from $O$ to the line $AB$ is $OM$ and to the line $AC$ is $ON$. These are equal, and the standard angle-bisector characterisation says a point inside an angle that is equidistant from both arms lies on the bisector of that angle. Hence $AO$ bisects $\angle BAC$.Step 3 (the same thing by congruence). Compare $\triangle OMA$ and $\triangle ONA$: $\angle OMA = \angle ONA = 90^\circ$ $OA = OA$ $OM = ON$ By RHS congruence $\triangle OMA \cong \triangle ONA$, so by CPCT: $\angle OAM = \angle OAN$ that is, $\angle OAB = \angle OAC$. So the ray $AO$ is the bisector of $\angle BAC$, and the centre $O$ lies on it.Notice $\triangle ABC$ here is isosceles with $AB=AC$, so this is the same result as Exercise Set 5.3 Q2 seen from a different angle: in an isosceles inscribed triangle the bisector from the apex, the altitude, the median and the line to the centre all

Question 36

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.

Accepted Answer

Let the radius be $r$, and let $d_1$ and $d_2$ be the distances from the centre to the $10$ cm chord and the $24$ cm chord respectively.The perpendicular from the centre bisects each chord, so the half-lengths are $5$ cm and $12$ cm: $d_1 = \sqrt{r^2-5^2}$ $d_2 = \sqrt{r^2-12^2}$ The longer chord is nearer the centre, so $d_2 < d_1$. Both chords are on the same side, so the distance between them is the difference: $d_1 - d_2 = 7$ $\sqrt{r^2-25} - \sqrt{r^2-144} = 7$ Move one root across and square: $\sqrt{r^2-25} = 7 + \sqrt{r^2-144}$ $r^2-25$ $= 49 + 14\sqrt{r^2-144} + r^2-144$ $70 = 14\sqrt{r^2-144}$ $\sqrt{r^2-144} = 5$ $r^2 = 144 + 25 = 169$ $r = 13\ \text{cm}$ Check. $d_1 = \sqrt{169-25} = 12$ cm and $d_2 = \sqrt{169-144} = 5$ cm, and $12 - 5 = 7$ cm. ✓The radius is 13 cm.Had the chords been on opposite sides, the equation would be $d_1 + d_2 = 7$, which has no solution here — $d_1$ alone already exceeds $7$ for any valid radius. The words “same side” are doing real work.

Question 37

*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.

Accepted Answer

Join the centre $O$ to all six vertices. This cuts the hexagon into six congruent triangles, and the full turn at the centre is shared equally: $\angle \text{at centre per triangle}$ $= \frac{360^\circ}{6}$ $= 60^\circ$ Side length. Each of these triangles has two sides equal to $r$ (radii), so it is isosceles; with the apex angle $60^\circ$ the base angles are also $60^\circ$. The triangle is equilateral, so the side equals the radius: $\text{side} = r$ Distance from the centre (the apothem). Drop the perpendicular from $O$ to a side; it bisects that side, giving a right triangle with hypotenuse $r$ and one leg $\frac{r}{2}$: $d^2 = r^2-\left(\frac{r}{2}\right)^2 = \frac{3r^2}{4}$ $d = \frac{\sqrt{3}}{2}r \approx 0.866\,r$ So each side is $r$ long and lies $\frac{\sqrt{3}}{2}r$ from the centre.This is why a hexagon is the easiest regular polygon to construct: set your compass to the radius and step it six times around the circle. The chord you cut off each time is exactly the radius.

Question 38

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.

Accepted Answer

The vertices lie on the circle in the order $M$, $N$, $O$, $P$. Both $\angle MOP$ and $\angle MNP$ stand on the same chord $MP$, and the vertices $O$ and $N$ are on the same side of $MP$ — they lie on the same arc.

Conclusion: the two angles are equal. Angles in the same segment are equal, because each is half the central angle standing on $MP$:

$\angle MOP$

$= \frac{1}{2}\angle \text{(central angle on } MP)$

$\angle MNP$

$= \frac{1}{2}\angle \text{(central angle on } MP)$

$\therefore\ \angle MOP = \angle MNP$

What the diameter adds. Because $MN$ is a diameter, the angles it subtends at the other two vertices are right angles:

$\angle MON = \angle MPN = 90^\circ$

So $MNOP$ has right angles at $O$ and $P$ when measured across the diameter, and the pair $\angle MOP$, $\angle MNP$ remain equal to each other whatever the exact positions of $O$ and $P$.

This is the same idea as Theorem 10 read forwards: if $M$, $P$ subtend equal angles at $N$ and $O$ on the same side, those four points are concyclic — which they are, by construction.

Question 39

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 a point on the extension of side $CD$).

Accepted Answer

Careful reading: $E$ lies on the extension of $CD$ beyond $D$, so $\angle ADE$ is the exterior angle at $D$ and $\angle ADC$ is the interior angle at $D$.Step 1: the exterior and interior angles at $D$ are supplementary — they sit on the straight line $CDE$: $\angle ADE + \angle ADC = 180^\circ$ Step 2: opposite angles of a cyclic quadrilateral are supplementary: $\angle ABC + \angle ADC = 180^\circ$ Step 3: compare. Both $\angle ADE$ and $\angle ABC$ are the supplement of the same angle $\angle ADC$, so they must be equal: $\angle ADE = 180^\circ - \angle ADC = \angle ABC$ Hence the exterior angle at a vertex equals the interior angle at the opposite vertex. The same argument works at every vertex of the quadrilateral.This gives a fast test for concyclicity: if in a quadrilateral the exterior angle at one vertex equals the interior opposite angle, the four vertices lie on a circle — because the opposite angles must then sum to $180^\circ$.

Question 40

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

Accepted Answer

Let the circle have centre $O$ and radius $r$, and let $AB$ be any chord at distance $d$ from the centre, where $d \ge 0$.Argument 1 — the chord formula. From Exercise Set 5.5 Q2: $AB = 2\sqrt{r^2-d^2}$ Since $d^2 \ge 0$, we always have $r^2 - d^2 \le r^2$, so: $AB \le 2\sqrt{r^2} = 2r$ and $2r$ is exactly the diameter. Equality holds only when $d = 0$, i.e. when the chord passes through the centre — when it is a diameter.Argument 2 — the triangle inequality. Join $OA$ and $OB$. If $A$, $O$, $B$ are not collinear, $\triangle OAB$ exists and the third side is shorter than the sum of the other two: $AB < OA + OB = r + r = 2r$ If $A$, $O$, $B$ are collinear the chord passes through the centre and $AB = 2r$. Either way $AB \le 2r$.So the diameter is the longest chord, and it is the only chord of that length.The chapter's Comment makes the same point in words: the chord nearest the centre is the longest, and the chord containing the centre is at distance zero.

Question 41

*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 $OA$.

Accepted Answer

Let a chord through $A$ have length $\ell$ and let its distance from the centre be $d$. The chord formula says: $\ell = 2\sqrt{r^2-d^2}$ Because $r$ is fixed, $\ell$ is smallest exactly when $d$ is largest. So the question becomes: among all chords through $A$, which is farthest from the centre?Bounding $d$. Let $M$ be the foot of the perpendicular from $O$ to the chord, so $OM = d$. The point $M$ lies on the chord, and $A$ is also on the chord, so $\triangle OMA$ is right-angled at $M$ (or degenerate). The hypotenuse is $OA$: $OM^2 + MA^2 = OA^2$ $d^2 = OA^2 - MA^2 \le OA^2$ $d \le OA$ So the distance can never exceed $OA$, and it reaches $OA$ precisely when $MA = 0$, that is when $M = A$ — which means $OA$ itself is perpendicular to the chord at $A$.Conclusion. The maximum distance is $d = OA$, attained by the chord through $A$ perpendicular to $OA$, and that chord therefore has the minimum length: $\ell_{\min} = 2\sqrt{r^2-OA^2}$ Every other chord through $A$ is closer to the centre and hence longer. The longest chord through $A$ is the diameter along the line $OA$, of length $2r$.Two nice checks: if $A$ is at

Question 42

How would you use the following figure (Fig. 5.30) to justify the statement that the angle in a semicircle is $90^\circ$?

Accepted Answer

Reading the figure: $O$ is the centre, the horizontal segment is a diameter with end points marked, $A$ is a point on the semicircle, the dashed segment is the radius $OA$, and the base angles at the two ends are labelled $a$ and $b$. The tick marks show that the three radii are equal.Step 1: two isosceles triangles. Call the diameter $PQ$, so $OP = OQ = OA = r$. The radius $OA$ splits $\triangle PAQ$ into $\triangle OPA$ and $\triangle OQA$, and both are isosceles. $OP = OA \implies \angle OPA = \angle OAP = a$ $OQ = OA \implies \angle OQA = \angle OAQ = b$ Step 2: add up the angles of $\triangle PAQ$. Its three angles are $a$ at $P$, $b$ at $Q$, and $\angle PAQ = a+b$ at $A$: $a + b + (a+b) = 180^\circ$ $2(a+b) = 180^\circ$ $a+b = 90^\circ$ Step 3: read off the answer. The angle at $A$ is $a+b$, so: $\angle PAQ = 90^\circ$ The figure does the work by turning one unknown angle into two equal pairs, which then have to split a straight angle in half. Nothing about the position of $A$ was used, so the result holds for every point on the semicircle.

Question 43

*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 $AB$.

Accepted Answer

Set the picture up on the diameter. Let $O$ be the centre and take $AB$ as a horizontal line. The chords $CC'$ and $DD'$ are perpendicular to $AB$, so each is vertical, and $AB$ — being a diameter perpendicular to them — bisects each of them.

Step 1: the diameter is a line of symmetry. Because $AB$ is perpendicular to the chord $CC'$ and passes through the centre, it bisects $CC'$. So $C$ and $C'$ are mirror images of each other in the line $AB$. The same holds for $D$ and $D'$.

$C \xleftrightarrow{\ \text{reflection in } AB\ } C'$

$D \xleftrightarrow{\ \text{reflection in } AB\ } D'$

Step 2: midpoints inherit the symmetry. Reflection is a rigid motion, so it sends the segment $CD$ to the segment $C'D'$, and therefore sends the midpoint $M$ of $CD$ to the midpoint $M'$ of $C'D'$.

Step 3: the segment joining a point to its mirror image is perpendicular to the mirror. That is exactly what reflection means: $AB$ is the perpendicular bisector of $MM'$. Hence:

$MM' \perp AB$

The same argument in coordinates. Put $O$ at the origin with $AB$ along the $x$-axis. Then $C=(p,\,h)$ forces $C'=(p,\,-h)$, and $D=(q,\,k)$ forces $D'=(q,\,-k)$. The midpoints are:

$M = \left(\frac{p+q}{2},\ \frac{h+k}{2}\right)$

$M' = \left(\frac{p+q}{2},\ -\frac{h+k}{2}\right)$

The two points share the same $x$-coordinate, so $MM'$ is a vertical segment — perpendicular to the $x$-axis, which is $AB$.

If $h+k = 0$ the two midpoints coincide on $AB$ itself and the segment $MM'$ degenerates to a point; the statement is then vacuously true.

Question 44

*How would you use the following figure (Fig. 5.31) to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^\circ$?

Accepted Answer

Reading the figure: $ABCD$ is a cyclic quadrilateral with centre $O$ joined to all four vertices. The tick marks show that $OA = OB = OC = OD = r$, and four base angles are labelled $p$ (at $A$), $q$ (at $B$), $u$ (at $C$) and $v$ (at $D$).Step 1: four isosceles triangles. Joining the centre to the vertices splits the quadrilateral into $\triangle OAB$, $\triangle OBC$, $\triangle OCD$ and $\triangle ODA$. Every one of them has two sides equal to the radius, so in each triangle the two base angles are equal. Writing the base angles as $p, q, u, v$ going round: $\angle OAB = \angle OBA = q$ $\angle OBC = \angle OCB = u$ $\angle OCD = \angle ODC = v$ $\angle ODA = \angle OAD = p$ Step 2: add up the angles of the quadrilateral. The interior angle at each vertex is the sum of the two base angles meeting there: $\angle A = p+q,\ \ \angle B = q+u$ $\angle C = u+v,\ \ \angle D = v+p$ The four interior angles of a quadrilateral total $360^\circ$: $(p+q)+(q+u)+(u+v)+(v+p)$ $= 360^\circ$ $2(p+q+u+v) = 360^\circ$ $p+q+u+v = 180^\circ$ Step 3: read off a pair of opposite angles. $\angle A + \angle C = (p+q)+(u+v) = 180^\circ$ and likewise

Quantity	Value
Central angle per side	60°
Side length	\(r\)
Distance from centre	\(\frac{\sqrt{3}}{2}r\)
Perimeter	\(6r\)

I’m Up and Down, and Round and Round

Exercise Set 5.1

Exercise Set 5.2

Exercise Set 5.3

Exercise Set 5.4

Exercise Set 5.5

Exercise Set 5.6

End-of-Chapter Exercises

Chord	Distance	Length
CD	3 cm	≈ 19.08 cm
AB	6 cm	16 cm
Is \(CD = 2AB\)?		No