Lesson 2, Topic 1
In Progress

# The Length of an Arc

Lesson Progress
0% Complete

Recall, the circumference of a circle is 2πr. In the diagram below, the arc length l or XY subtends an angle θ° at the centre O with radius r.

In general, the length of an arc of a circle is proportional to the angle at which the arc subtends at the centre.

Therefore, the arc length “l” or XY is given as

l = $$\frac {θ}{360}\scriptsize \: \times \: 2 \pi r$$

Questions 1

1. In terms of π, what is the length of an arc which subtends an angle of 30o at the centre of a circle of radius $$\scriptsize 3\normalsize \frac {1}{2} \: cm$$?

Solution:

Using XY = $$\frac {θ}{360}\scriptsize \: \times \: 2 \pi r$$

θ = 30o

radius = $$\scriptsize 3\frac {1}{2} = \normalsize \frac {7}{2}$$

|XY| = $$\frac {30}{360}\scriptsize \: \times \: 2 \pi \: \times \: \normalsize \frac{7}{2}$$

|XY| = $$\scriptsize 7 \normalsize \frac{\pi}{12}$$

|XY| = $$\frac{7 \pi}{12}$$

Question 2

What angle does an arc 5.5cm in length subtend at the centre of a circle of diameter 7cm? (WAEC)

Solution:

Using AB = $$\frac {θ}{360}\scriptsize \: \times \: 2 \pi r$$

AB = 5.5

θ = ?

$$\scriptsize \pi = \frac{22}{7}$$

5.5 = $$\frac {θ}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7} \: \times \: \scriptsize 7$$

$$\frac {11}{2} = \frac {44\theta}{360}$$

i.e. $$\frac {11}{2} = \frac {11\theta^{\circ}}{90}$$

i.e. θ = $$\frac {11 \: \times \: 90}{2 \: \times \: 11}$$

i.e. θ = 45°

Question 3

3. An arc length of 28cm subtends an angle of 24o at the centre of a circle. In the same circle, what angle does an arc of length 35cm subtend?

Solution

Recall the length of an arc is proportional to the angle it subtends at the centre.

i.e. 28 ≡ 24°

35 ≡ θ°

$$\frac{28}{24} = \frac{35}{\theta}$$

θ° = $$\frac {35 \: \times \: 24}{28}$$

θ° = $$\frac {840}{28}$$

θ° = 30°

Question 4

In the diagram below, the radius of the circle is 10cm. Calculate the length of the minor arc XY.

Solution:

The angle subtends at the centre by arc XY

XY = 2 × 63 = 126o (Angle at the centre is twice the angle at Circumference)

Using XY = $$\frac {θ}{360}\scriptsize \: \times \: 2 \pi r$$

θ = 126o

∴ length of XY = $$\frac {126}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7} \: \times \: \scriptsize 10$$

= $$\frac {42 \: \times \: 11}{3 \: \times \: 7} = \frac {14 \: \times \: 11}{7}$$

Length XY = 22cm

Question 5

In the diagram below four pencils are held together in a “square” by an elastic band.

If the pencils are of diameter 7mm, what is the length of the band in this position?

Solution:

Length of quadrant arc |AB|= $$\frac {90}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: \normalsize \frac{7}{2}$$

|AB| = $$\frac{11}{2}$$

Distance between the four separate quadrant arc length = r + r = 2r

= 2 × $$\frac{7}{2}$$

= 7

Total length = 4(quadrant Arc AB + 2r)

i.e. Total length of band = 4 $$\left [ \frac{11}{2} \scriptsize \: + \: 7 \right ]$$

= 22 + 28 = 50mm

Question 6

The elastic band in question 5 is used to hold seven of the same pencils as shown in the diagram below. What is the length of the band in the position?

Solution

Length of quadrant arc |AB|= $$\frac {60}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: \normalsize \frac{7}{2}$$

i.e Length of quadrant arc |AB| = $$\frac {11}{3}\scriptsize = 3.6667$$

|AB| = 3.7cm

Distance between arcs = r + r = 2r

= 2 × $$\frac{7}{2}$$ = 7

Total length of band = 6 [Arc length + 2r]

= $$\scriptsize 6 \left [ \normalsize \frac {11}{3}\scriptsize \: + \: 7 \right]$$

= $$\scriptsize 6 \left [ \normalsize \frac {11 \: + \: 21}{3}\right]$$

= $$\scriptsize 6 \: \times \: \left [ \normalsize \frac {32}{3}\right] \\ \scriptsize = 2 \: \times \: 32 \\ \scriptsize = 64 mm$$

Question 7

In the diagram above AB is a chord of a circle of radius 10cm, M is the mid-point of AB and If MN = 2cm, calculate

(a) AB

(b) The angle arc ANB subtends at the centre of the circle

(c) The difference in length between AB and arc ANB to the nearest mm.

Solution

(a) Calculate AB

Consider the sector ANBO

AO, ON and OB are radii

θ1

OB = OA = ON = 10cm

OM = 10 – 2 = 8cm

Using Pythagoras on right angled AMO

AO2 = OM2 + AM2

AM2 = AO2 – OM2

AM = $$\scriptsize \sqrt{AO^2 \: – \: OM^2}$$

AM = $$\scriptsize\sqrt{10^2 \: – \: 8^2}$$

AM = $$\scriptsize \sqrt{36}$$

AM = $$\scriptsize 6cm$$

AB = 2 x 6 = 12cm

(b) Calculate The angle that are ANB subtends at the centre of the circle

Consider Isosceles $$\scriptsize \Delta ABO \; \; \theta_1 = \theta_2$$

Cos θ1 = $$\frac{adj}{hyp}\\ = \frac {8}{10} \\ = \frac {4}{5} \\ = \scriptsize 0.8000$$

θ1 = cos – 1 (0.8000)

θ1 = 36.9o

Angle ANB

Subtends 2(θ1) at the centre

i.e. Angle ANB = 2 x 36.9o

= 73.8

= 74o

(c) The difference in length between AB and arc ANB to the nearest mm.

Arc length ANB

= $$\frac {73.8}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 10 \\ = \frac {73.8 \: \times \: 44 }{36 \: \times \: 7} \\ = \frac{73.8 \: \times \: 11}{63}$$

Arc length ANB = 12.8857143

i.e. ANB = 12.9cm

Difference = Arc length ANB – AB

= 12.8857143 – 12

= 0.8857143 cm

= 8.857143 mm

$$\scriptsize \equiv 9 mm$$

Question 8

Two friction wheels are of diameter 20mm and 200mm respectively. They touch at P and rotate without slipping _ see the diagram below.

Calculate the number of turns made by the small wheel when the large wheel rotates through 60o.

Solution:

Arc length for Big wheel rotating through 60o

= $$\frac {60}{360}\scriptsize \: \times \: 2 \: \times \: \normalsize \frac{22}{7} \scriptsize \: \times \: 100 \\ = \frac {22 \: \times \: 100 }{3 \: \times \: 7} \\ = \frac{2200 }{21}$$

Arc length for Big wheel = 104.761905mm

Circumference of small circle = 2πr = πD

= $$\frac{22}{7} \: \times \: \scriptsize 20 \\ = \frac{440}{7} \\ \scriptsize = 62.8571429$$

Numbers of turns by small wheel = $$\frac{104.761905}{62.8571429} \\ \scriptsize = 1.666667$$

= $$\scriptsize1 \frac{2}{3} \; turns$$

Question 9

Water is taken from a well 11m deep in a bucket, the rope winding onto a drum 35cm in diameter.

a. Through what angle does the handle turn in winding up 1 metre of rope?

b. How many revolutions of the handle does it take to bring the bucket up from the bottom?

c. If the arm of the handle is 42cm long, how far does the hand of the winder travel when bringing the bucket up from the bottom?

Solution

(a)

r = $$\frac{35}{2} \\ \scriptsize = 17.5cm \\ \scriptsize 0.175m$$

Let the angle be θ  be proportional to 1m.

i.e 100cm = $$\frac {\theta}{360} \;\scriptsize \times \: 2 \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: 17.5$$

i.e θ = $$\frac {360 \: \times \: 7 \: \times \: 100}{2 \: \times \: 22\: \times \: 17.5} = \frac{25200}{770}$$

i.e θ = $$\frac{25200}{770} \\ = \scriptsize 327.272727$$

i.e θ = 327.27o or $$\scriptsize 327 \frac{3}{11}^o$$

(b) Circumference of drum = πD or 2πr

= $$\frac{22}{7} \scriptsize \: \times \: 35 \\ \scriptsize = 10cm$$

∴ Number of revolutions = $$\frac{11.0}{1.10} \: or \: \frac{1100}{110}$$

= 10 revolutions

(c)

Circumference of winder’s hand

= $$\scriptsize 2 \: \times \: \frac{22}{7} \: \times \: \scriptsize 42$$

2 x 22 x 6 = 264 cm

Total distance moved to bring up water = 264 x 10 revolutions = 2,640cm

= 26.4m

Evaluation Questions

1. Complete the table below. Make a rough sketch in each case.

2. The minute – hand of a clock is 6cm long. How far does the end of the hand travel in 35 minutes?

3. In a circle of radius 6cm a chord is drawn 3cm from the centre.   (a) Calculate the angle subtended by the chord at the centre of the circle.

(c) Hence, find the length of the minor arc cut off by the chord.

4. The diagram below shows a circular wire clip of a radius of 7mm with a gap of 7mm between the ends.

Calculate the total length of wire in the clip to the nearest mm, given that about 4mm are used altogether in making the turnovers at the ends. error: