Circular Measure 1

CIRCULAR MEASURE

1. RADIAN1.1 Converting Measurements in Radians to Degrees and Vice Versa

Definition:r

Srad

where S is the length of the arc AB,r is the radius of the circle, andθ is the angle subtended at the centre O by the arc AB.

For the full circle, the circumference is 2πr and the angle is 360o.

rad2

2Therefore,

r

r.

Hence, 2π rad = 360o => π rad = 180o

For each of the following measurement in degree, change it to radian.Example: 25.8o

Since 180o = π rad

rad4504.0

1808.258.25

π

(a) 30o =

[0.5237 rad]

(b) 45o =

[0.7855 rad]

(c) 81.5o =

[1.423 rad](d) 122.8o =

[2.144 rad]

(e) 154o 20’ =

[2.694 rad](f) 225o 36’ =

[3.938 rad]

(g) -50o =

[-0.02773 rad]

25.8o

O

A

B

O

A

B

r

r

S

Circular Measure 2

For each of the following measurement in radian, change it to degree.Example: 0.75 radSince π rad = 180o

97.42

18075.0rad75.0

π

Always use π = 3.142

(a) 0.6 rad =

[34.37o]

(b) 1.54 rad =

[88.22o]

(c) rad5

18

[206.24o]

(d) rad7

32

[139.13o]

(e) 1.2π rad =

[216o]

(f) rad3

π

[60o]

(g) -0.25 rad =

[-14.32o]

(h) - rad3

12

[-133.67o]

(i) (1 + π) rad =

[237.29o]

0.75 rad

O

PQ

Circular Measure 3

2. ARC LENGTH OF A CIRCLE

2.1 Arc Length, Radius and Angler

S => S = rθ

For each of the following, find the length of the arc S given the values of r and θ.Example:r = 10 cm and θ = 1.5 radSince S = rθ

= 10 × 1.5= 15 cm

(a) r = 20 cm and θ = 0.6 rad

[12 cm](b) r = 15 cm and θ = 1.4 rad

[21 cm]

(c) r = 30 cm and θ =6

rad

[15.71 cm](d) r = 10 cm and θ = 60o

[10.47 cm]

(e) r = 1.5 m and θ = 82.8o

[2.168 m]

For each of the following, calculate the length of the radius r given the values of S and θ.Example:S = 4.5 cm and θ = 1.5 radSince S = rθ

4.5 = 1.5r

5.1

5.4r

= 3 cm

(a) S = 9 cm and θ = 1.5 rad

[6 cm]

(b) S = 4.5 cm and θ = 1.8 rad

[2.5 cm]

(c) S = 9 m and θ = 30o

[17.19 m](d) S = 15 cm and θ = 32.8o

[26.20 cm]

(e) S = 10 cm and θ =8

rad

[25.46 cm]

O

P

Q

1.5 radS

10 cm

O

A

B

1.5 rad

r 4. 5

cm

Circular Measure 4

For each of the following, find the angle θ in radian given the values of r and S.Example:r = 5 cm and S = 10 cmSince S = rθ

10 = 5θ

θ =5

10

= 2 rad

(a) r = 9 cm and S = 36 cm

[4 rad]

(b) r = 7 cm and S = 5.6 cm

[0.8 rad]

(c) r = 20 cm and S = 0.7 m

[3.5 rad](d) r = 1.2 cm and S = 20 cm

cm]3

216[

(e) r = 3.5 cm and S = 112 cm

[32 cm]

For each of the following, find the angle θ in degree given the values of r and S.Example:r = 8 cm and S = 20 cmSince S = rθ

20 = 8θ

θ =8

20rad

=

1805.2

= 143.22o

(a) r = 10 cm and S = 6 cm

[34.37o]

(b) r = 5 cm and S = 5.4 cm

[57.29o]

(c) r = 5.6 cm and S = 12 cm

[122.76o](d) r = 7.15 cm and S = 0.28 m

[224.34o]

(e) r = 5.6 cm and S = 3π cm

[96.43o]

O

A

B

10

cm

5cm

O

A

B

20cm

8cm

Circular Measure 5

3. AREA OF SECTOR3.1 Area of Sector, Radius and Angle

Area of a circle, A = πr2

Area for a sector OPQ of angle θo, 2

360rA

Converting to radian, 2

rad2

radr

A =2

1r2θ, θ must be in radian.

For each of the following, find the area of the sector A given the values of r and θ.Example:r = 3 cm and θ = 1.8 rad

Since 22

1rA

8.132

1 2

= 8.1 cm2

(a) r = 5 cm and θ = 0.8 rad

[10 cm2]

(b) r = 2.11 cm and θ = 1.78 rad

[3.962 cm2]

(c) r = 2.7 cm and θ =3

21 rad

[6.075 cm2]

(d) r = 12 mm and θ =5

4rad

[180.98 mm2]

(e) r = 1.5 m and θ = 122.8o

[2.411 m2]

(f) r =5

3m and θ = 85o

[0.2671 m2]

(g) r = 13.2 cm and θ = 67.2o

[102.19 cm2]

O

P

Q

1.8 rad

3cm

3cm

O

r

r

P

Q

Circular Measure 6

For each of the following, find the length of the radius r given the values of θ and A.Example: θ = 1.2 rad and A = 15 cm2

Since 22

1rA

15 2.12

1 2 r

2.1

2152 r

= 25r = 5 cm

(a) θ = 3 rad and A = 6 cm2

[2 cm]

(b) θ = 1.4 rad and A = 118.3 cm2

[13 cm]

(c) θ =6

rad and A = 3π m2

[6 m]

(d) θ =7

31 rad and A = 78.25 cm2

[109.55 cm]

(e) θ = 102o and A = 215 cm2

[15.54 cm]

(f) θ = 65o and A = 78 cm2

[11.73 cm]

(g) θ = 102o and A = 1215 cm2

[36.94 cm]

O

P Q

r r1.2 rad

Circular Measure 7

For each of the following, find the angle θ in radian given the values of r and A.

Example: r = 5 cm and A =8

39 cm2

Since 22

1rA

252

1

8

39

258

275

4

3 rad

(a) r = 2.4 cm and A = 1.44 cm2

[0.5 rad](b) r = 8 cm and A = 64 cm2

[2 rad]

(c) r = 0.6 m and A = 3.24 m2

[18 rad]

For each of the following, find the angle θ in degree given the values of r and A.Example: r = 4 cm and A = 16 cm2

Since 22

1rA

242

116

16

216

= 2 rad

=142.3

1802

= 114.58o

(a) r = 15 cm and A = 50π cm

[80o]

(b) r = 8 cm and A = 48 cm2

[85.93o]

(c) r = 3.2 m and A = 5 m2

[55.95o]

O

A

B

16 cm2

4 cm

4cm

5 cm

A

BO

cm2938_5

cm

Circular Measure 8

2.2 Perimeter of Segments of a Circle

Example:For the shaded segment in the given diagram beside, find its perimeter.

The perimeter of the shaded segment is the total length of the arc ABand the chord AB.

SAB = rθ= 10 × 1.2= 12 cm

To find the length of the chord AB, considerthe triangle OAB in the diagram beside.

)142.3

1806.0sin(10

2

1

rad6.0sin10

2

1

AB

AB

AB = 2 × 5.646= 11.292 cm

perimeter of shaded segment = 12 + 11.292= 23.292 cm

3.2 Area of Segments of a Circle

Example:With reference to the sector in the same diagram beside,find the area of the shaded segment.

The area of the shaded segment is the difference betweenthe area of the sector OAB and the triangle OAB.

Area of sector OAB =2

1r2θ

2.1102

1 2

= 60 cm2

To find the area of the triangle OAB,please refer to the diagram beside.AC is perpendicular to the radius OB.

142.3

1802.1sin

rad2.1sin

AOAC

AO

AC

2cm60.46

9320.010102

1

75.68sin2

1

2

1ofArea

OAOB

ACOBOAB

Area of shaded segment = 60 – 46.60= 13.40 cm2

Note:The length of the chord

AB =2

sin2

r

Area of segment

sin2

1

sin2

1

2

1

2

22

r

rr

O

A

B

10 cm

10 cm

=

=

0.6 rad

0.6 rad

O

A

B

10 cm

10 cm

1.2 rad

C

Note:Area of triangle OAB

= sin2

1 2r

O

A

B

10 cm

10 cm

1.2 rad

O

A

B

10 cm

10 cm

1.2 rad

Circular Measure 9

For each of the following diagrams, find(i) the perimeter and(ii) the area of the shaded segment.

(a)

[Perimeter = 4.974 cm, Area = 0.2579 cm2]

(b)

[Perimeter = 9.472 cm, Area = 1.489 cm2]

(c)

[Perimeter = 10.19 cm, Area = 1.145 cm2]

(d)

[Perimeter = 8.599 cm, Area = 4.827 cm2]

O

P

Q

6cm6

cm

0.8rad

O

A

B

5 cm

5 cm

0.5 rad

O

P

Q

3.8

cm

3.8 cm

1 rad

O A

B

4.2 cm

4.2cm

60o