Arcs and chords of a circle

The classroom clock has a minute hand $15$ cm long. From $7{:}00$ to $7{:}20$ , its tip moves along an arc from $A$ to $B$ . What is the measure of that arc? How long is the segment $AB$ ?

Chords of a circle

Explore: which chord is the longest?

On the circle $(O, 5\text{ cm})$ , joining two points gives a chord. A chord $4$ cm from the centre is $6$ cm long, one $3$ cm from the centre is $8$ cm long, and one through the centre is $10$ cm long. The closer to the centre, the longer the chord.

Why? For a chord $AB$ not through the centre, the triangle $OAB$ gives $AB<OA+OB=2R$ . If $AB$ passes through the centre, then $AB=2R$ .

The triangle $OAB$ is isosceles ( $OA=OB=R$ ). Hence the foot $H$ of the perpendicular from $O$ to $AB$ is the midpoint of $AB$ , and by Pythagoras' theorem:

$AB=2HA=2\sqrt{R^2-OH^2}.$

The chord

AB

and its distance

OH

from the centre

Angles at the centre and arcs

Two points $A$ , $B$ on a circle split it into two arcs, both written $\overset{\frown}{AB}$ ; to tell them apart, we name a point $M$ on one and write $\overset{\frown}{AMB}$ .

The angle at the centre

\angle AOB

subtending the minor arc

AB

The measure of an arc

If a point $C$ lies on the arc $AB$ , then $\overset{\frown}{AB}=\overset{\frown}{AC}+\overset{\frown}{CB}$ (the angle $\angle AOB$ splits into two angles).

Now we answer the question about the clock from the start.

Example The minute hand of a clock

The minute hand $OA$ is $15$ cm long. From $7{:}00$ to $7{:}20$ , its tip moves from $A$ to $B$ . Find the measure of the arc $AB$ travelled by the tip and the length $AB$ (give your answer to 2 decimal places).

Solution. Each minute the minute hand turns $\dfrac{360^\circ}{60}=6^\circ$ , so in $20$ minutes it turns $20\cdot6^\circ=120^\circ$ . Hence $\angle AOB=120^\circ$ and $\overset{\frown}{AB}=120^\circ$ .

Draw $OH\perp AB$ . The triangle $OAB$ is isosceles, so $OH$ is also an angle bisector, hence $\angle AOH=60^\circ$ . In the triangle $OHA$ , right-angled at $H$ , we have

$HA=OA\cdot\sin60^\circ=15\cdot\frac{\sqrt3}{2},\qquad AB=2HA=15\sqrt3\approx25.98\ \text{(cm)}.$

So the arc measures $120^\circ$ and $AB\approx25.98$ cm.

When three points split a circle into three arcs, we add the measures of arcs.

Key terms

English	Tiếng Việt
chord	dây cung
angle at the centre	góc ở tâm
arc	cung
minor arc / major arc	cung nhỏ / cung lớn
subtended arc	cung bị chắn
semicircle	nửa đường tròn
measure of an arc	số đo cung

Summary

A chord joins two points of a circle; a diameter is the longest chord.
If $OH\perp AB$ , then $H$ is the midpoint of $AB$ and $AB=2\sqrt{R^2-OH^2}$ .
A minor arc equals the angle at the centre subtending it; a major arc is $360^\circ$ minus the minor arc; a semicircle is $180^\circ$ .
If $C$ lies on the arc $AB$ , then $\overset{\frown}{AB}=\overset{\frown}{AC}+\overset{\frown}{CB}$ .

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

An angle at the centre has $\angle AOB=65^\circ$ . Find the measures of the minor arc $AB$ and the major arc $AB$ .

What you need and how to start

What you need: the minor arc equals the angle at the centre; the major arc is $360^\circ$ minus the minor arc.

Steps:

Read off the minor arc, then subtract it from $360^\circ$ .

Show solution

We have $\overset{\frown}{AB}_{\text{minor}}=65^\circ$ , so the major arc measures $360^\circ-65^\circ=295^\circ$ .

So the arcs measure $65^\circ$ and $295^\circ$ .

Answer. Minor arc $65^\circ$ , major arc $295^\circ$ .

Exercise 2

Consider the circle $(O, 5\text{ cm})$ and a chord $AB$ at distance $3$ cm from the centre. Find $AB$ .

What you need and how to start

What you need: $AB=2\sqrt{R^2-OH^2}$ .

Steps:

Find half the chord, $HA$ , then double it.

Show solution

Draw $OH\perp AB$ ; then $HA=\sqrt{5^2-3^2}=4$ (cm), so $AB=2HA=8$ cm.

So $AB=8$ cm.

Answer. $AB=8$ cm.

Exercise 3

Consider the circle $(O, 13\text{ cm})$ and a chord $AB=24$ cm. Find the distance from $O$ to $AB$ .

What you need and how to start

What you need: the foot of the perpendicular, $H$ , is the midpoint of $AB$ .

Steps:

Find $HA=12$ cm.
Use Pythagoras' theorem in the triangle $OHA$ .

Show solution

Draw $OH\perp AB$ ; then $HA=\dfrac{24}{2}=12$ (cm). In the triangle $OHA$ , right-angled at $H$ , we have $OH=\sqrt{13^2-12^2}=5$ (cm).

So the distance from $O$ to $AB$ is $5$ cm.

Answer. $5$ cm.

Exercise 4

Consider the circle $(O, R)$ and a chord $AB=R$ . Find the measures of the minor arc $AB$ and the major arc $AB$ .

What you need and how to start

What you need: a triangle with three equal sides is equilateral.

Steps:

Show that the triangle $OAB$ is equilateral and deduce $\angle AOB$ .

Show solution

We have $OA=OB=AB=R$ , so the triangle $OAB$ is equilateral, hence $\angle AOB=60^\circ$ . Therefore the minor arc measures $60^\circ$ and the major arc measures $300^\circ$ .

So the arcs measure $60^\circ$ and $300^\circ$ .

Answer. Minor arc $60^\circ$ , major arc $300^\circ$ .

Exercise 5

Ha starts her homework at $3{:}00$ and finishes at $3{:}45$ . Through how many degrees do the minute hand and the hour hand of the clock turn in that time?

What you need and how to start

What you need: the minute hand turns $6^\circ$ per minute; the hour hand turns $30^\circ$ per hour.

Steps:

Multiply the number of minutes by $6^\circ$ .
The hour hand turns for $\dfrac{45}{60}$ of an hour.

Show solution

The minute hand turns $45\cdot6^\circ=270^\circ$ . The hour hand turns $\dfrac{45}{60}\cdot30^\circ=22.5^\circ$ .

So the minute hand turns $270^\circ$ and the hour hand $22.5^\circ$ .

Answer. The minute hand turns $270^\circ$ and the hour hand $22.5^\circ$ .

Exercise 6

Consider the circle $(O, 8\text{ cm})$ and a chord $AB$ with $\angle AOB=120^\circ$ . Find $AB$ (give your answer to 2 decimal places). (Revise trigonometric ratios of acute angles.)

What you need and how to start

What you need: the altitude $OH$ of the isosceles triangle $OAB$ is also an angle bisector.

Steps:

Deduce that $\angle AOH=60^\circ$ .
Find $HA=OA\cdot\sin60^\circ$ , then double it.

Show solution

Draw $OH\perp AB$ . The triangle $OAB$ is isosceles, so $\angle AOH=60^\circ$ . In the triangle $OHA$ , right-angled at $H$ , we have $HA=8\sin60^\circ=4\sqrt3$ (cm), hence $AB=8\sqrt3\approx13.86$ cm.

So $AB\approx13.86$ cm.

Answer. $AB=8\sqrt3\approx13.86$ cm.

Exercise 7

(1.5 marks) Consider the circle $(O, 10\text{ cm})$ and a chord $AB=16$ cm. Let $H$ be the midpoint of $AB$ .

a) Prove that $OH\perp AB$ and find $OH$ .

b) Find the angle $\angle AOB$ (give your answer to the nearest degree).

c) Find the measure of the major arc $AB$ .

What you need and how to start

What you need: in an isosceles triangle, the median from the apex is also an altitude and an angle bisector.

Steps:

Use the isosceles triangle $OAB$ , then Pythagoras' theorem.
Find $\tan\angle AOH$ , then $\angle AOB=2\angle AOH$ .
Subtract the minor arc from $360^\circ$ .

Show solution

a) The triangle $OAB$ is isosceles and $OH$ is a median, so it is also an altitude: $OH\perp AB$ . We have $HA=8$ cm, so $OH=\sqrt{10^2-8^2}=6$ (cm).

b) In the triangle $OHA$ , right-angled at $H$ , we have $\tan\angle AOH=\dfrac{8}{6}$ , so $\angle AOH\approx53.13^\circ$ . Since $OH$ is the angle bisector, $\angle AOB\approx106.26^\circ\approx106^\circ$ .

c) The major arc $AB$ measures $360^\circ-106.26^\circ\approx254^\circ$ .

So $OH=6$ cm, $\angle AOB\approx106^\circ$ and the major arc $AB$ is about $254^\circ$ .

Answer. a) $OH=6$ cm; b) $\angle AOB\approx106^\circ$ ; c) about $254^\circ$ .

Exercise 8 Challenge

Consider the circle $(O, 10\text{ cm})$ and two parallel chords $AB=12$ cm and $CD=16$ cm. Find the distance between the two chords.

What you need and how to start

What you need: the distance from the centre to a chord is $\sqrt{R^2-\left(\frac{\text{chord}}{2}\right)^2}$ .

Steps:

Find the distance from $O$ to each chord.
Look at two cases: the chords on the same side of $O$ or on opposite sides.

Show solution

The distance from $O$ to $AB$ is $\sqrt{10^2-6^2}=8$ (cm), and to $CD$ it is $\sqrt{10^2-8^2}=6$ (cm). The chords are parallel, so both perpendiculars from $O$ lie on one line.

If the chords are on the same side of $O$ , the distance is $8-6=2$ (cm); if they are on opposite sides, it is $8+6=14$ (cm).

So the distance between the chords is $2$ cm or $14$ cm.

Answer. $2$ cm or $14$ cm.

a) $OH\perp AB$ .	True	False
b) $OH=3$ cm.	True	False
c) The minor arc $AB$ measures $90^\circ$ .	True	False
d) The circle has a chord of length $11$ cm.	True	False

Chords of a circle

Explore: which chord is the longest?

Angles at the centre and arcs

The measure of an arc

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer