© t madas. the area of a circle the circumference of a circle r we can use these formulae to find...

© T Madas

The area of a circle The circumference of a circle

=A r 2 =C r2

r

We can use these formulae to find areas of sectors and lengths of arcs

Sector Arc

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=A r 2 =C r2

6 cm


50°

Area of sector = ´ 26 ´50

360

= ´ 36´50

36010

= 5 .» 215 7cm

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=A r 2 =C r2

6 cm


50°

Length of arc = ´ ´ 62 ´50

360

= ´ 12´50

36030

= 53

.» 5 24cm

© T Madas

13cm

24 c

m

Find the area of the circular segment, from the information given.

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13cm

24 c

m 12 c

m

x

x 2+ 212 = 213 Ûx 2 - 212= 213 Ûx 2 = 25 Ûx = 5 cm

5 cm


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13cm

12 c

m

x 2

5 cm

+ 212 = 213 Ûx 2 - 212= 213 Ûx 2 = 25 Ûx = 5 cm

12 ´ 5 ´ 24 = 260 cm

Area of the triangle


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13cm

12 c

m

12

5 cm

13= sin Û

( )- 1sin 1213= Û

» °67.38

134.76°

Area of sector = ´ 213 .´

134 76

360.» 2198 74 cm

138.74 cm2

12 ´ 5 ´ 24 = 260 cm

Area of the triangle


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6 cm

Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively.

Calculate the area and the perimeter of each sector, correct to 1 decimal place.

16

18

5 cm

A = πr 2

area of a circle

A =π x 5 2

area of the sector

A = 16

π x 5 2x ≈ 13.1 cm2

A = πr 2

area of a circle

A =π x 6 2

area of the sector

A = 18

π x 6 2x ≈ 14.1 cm2

© T Madas

6 cm

Two circular sectors have radii of 5 cm and 6 cm and represent one sixth and one eighth of a circle respectively.

Calculate the area and the perimeter of each sector, correct to 1 decimal place.

16

18

5 cm

C = r

circumference of a circle

C = x πx 5

length of the arc

16

x ≈ 5.2 cm

2 π2

L = x πx 5 2

C = r

circumference of a circle

C = x πx 6

length of the arc

18

x ≈ 4.7 cm

2 π2

L = x πx 6 2

© T Madas

The diagram below shows a composite shape consisting of:

a semicircle, centre at O and a radius of 6 cm a circular sector, centre at A corresponding to an angle

of 60°.1. Calculate the area of the composite shape in terms of

π2. Round your answer to part (1), to 3 significant

figures.O

A B

C

60°

6 cm

area of the semicircle

A = 12 x π r 2 ⇔

A = 12 x πx 62 ⇔

A = π18

area of the sector

A = 16 x π r 2 ⇔

A = 16 x π x 122 ⇔

A = π2460°= 16of a circle

42π

© T Madas

The diagram below shows a composite shape consisting of:

a semicircle, centre at O and a radius of 6 cm a circular sector, centre at A corresponding to an angle

of 60°.1. Calculate the area of the composite shape in terms of

π2. Round your answer to part (1), to 3 significant

figures.O

A B

C

60°

6 cm

A = π42 = x π42

42π≈ [3 s.f.]cm2132

© T Madas

The diagram below shows a pond consisting of 2 concentric sectors corresponding to the same angle θ, and two semicircles at each end.The radii of the two sectors are r and R as shown in the diagram.1. Find a formula for the area A of this pond in terms

of r, R and θ.2. Find the area of a pond with this shape if r = 7 m, R

= 9 m and θ = 45°

θ

r

R

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= 9 m and θ = 45°

θ

r

R

πR 2 x θ

360

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= 9 m and θ = 45°

θ

r

R

πr 2 x θ

360–πR

2 x θ360

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= 9 m and θ = 45°

θ

r

R

R – r πr 2 x θ

360–πR

2 x θ360

+ π R – r2

2A =

© T Madas



= 9 m and θ = 45°

πr 2 x θ

360–πR

2 x θ360

+ π R – r2

2A =πr

2 x θ360

–πR 2 x θ

360+ π R – r

2

2A =

© T Madas



= 9 m and θ = 45°

πr 2 x θ

360–πR

2 x θ360

+ π R – r2

2A =

π4

+R 2– r 2

πθ360

R – r2

A =

π4

+92 – 7245π360

9 – 72

A =

π4

+x 32π8

x 4A =

+ π4πA =

A = 5π≈ 15.7 m2

© T Madas

8 cm

3 cm45°

The prism shown has a cross section of a circular sector of radius 8 cm corresponding to a central angle of 45°.The thickness of the prism is 3 cm.1. Calculate the volume of the prism, correct to 3 s.f.2. Calculate the surface area of the prism, correct to 3 s.f.

8 cm45°

A = πr 2

area of a circle

A =π x 8 2

area of the given sector

volume of the prism

A = 45360

π x 8 2 x

V = 45360

π x 8 2 x x 3

V = 24πV ≈75.4 cm3 [3 s.f.]

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8 cm

3 cm45°


8 cm45°


A = 45360

π x 8 2 x


A = 45360

π x 8 2 x

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8 cm

3 cm45°


8 cm45°


A = 45360

π x 8 2 x

x 2

x 2

area of the side rectanglex 2

A =3x 8 x 2

area of the curved surface

A = 45360

π2 x xx 8

2πr= 16π

= 48

x 3= 6π

total surface area

A = 16π + 48 + 6π≈ 117 cm2 [3 s.f.]

© T Madas

All measurements in cm

72

141

The diagram below models the action of a single windscreen wiper rotating through an angle of 162° on a rectangular windscreen measuring 141 cm by 72 cm. The wiper AB is 45 cm long and OA = 15 cm.Calculate what percentage of the screen gets wiped.

O 15

B45A

A = πr 2

area of a circle

A =π x 60 2

area of the “larger” sector

A = 162360

π x 60 2 x = 1620π162°

© T Madas


72

141


O 15

B45A162°

A = πr 2

area of a circle

A =π x 60 2


A = 162360

π x 60 2 x = 1620π

area of the “inner” sector

A = 162360

π x 15 2 x = 101.25π

© T Madas


72

141


O 15

B45A162°

A = πr 2

area of a circle

A =π x 60 2


A = 162360

π x 60 2 x = 1620π

area of the “inner” sector

A = 162360

π x 15 2 x = 101.25π

area swept by the wiper is1620π – 101.25π= 1518.75π cm2

≈ 4771.3 cm2

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area wipedtotal area


72

141


O 15

B45A162°

area swept by the wiper is1620π – 101.25π= 1518.75π cm2

≈ 4771.3 cm2

x 100%age =

1518.75π141 x 72

x 100%age =

47%%age ≈

© T Madas

-1tan ( )

Two circles of respective radii of 3 cm and 4 cm are overlapping each other in such a way so that their centres are 5 cm apart.Calculate the perimeter of the compound shape, correct to 2 significant figures.

5 cm

4 cm

A B

3,4,5

= Alarm Bells !

C

3 cm ∆ABC is right angled

By trig on ∆ABC :θ

tanq = 43 Û

q = 43 Û

53.13q » °

© T Madas


5 cm

4 cm

A B

C

3 cm

θ

Now some angle calculations:If θ ≈ 53.13°

CBA ≈ 36.87°CAD ≈ 106.26°CBD ≈ 73.74°

D

106.26°

36.87°

73.74° reflexCAD ≈ 253.74°reflexCBD ≈ 286.26°

253.74°

286.26°

Simplify the diagram by showing only the relevant information

© T Madas


4 cm

A B

C

3 cm

D

253.74°

286.26°

This arc corresponds to an angle of 253.74° on a circle with radius of 3 cm

This arc corresponds to an angle of 286.26° on a circle with radius of 4 cm

© T Madas


4 cm

A B

C

3 cm

D

253.74°

286.26°

The smaller of the 2 arcs2 p´ 3´ ´ 253.74

360 13.29 cm»

The larger of the 2 arcs2 p´ 4´ ´ 286.26

360 19.98 cm»

The required perimeter

33.3 cm [2 s.f.]

© T Madas

In a circle of radius 5 m the tangents at the endpoints of a chord AB of length 8 m meet at point C.

Calculate the perimeter and the area of the finite region bounded by the two tangents and the circle

O

A

B

C

8 m

5 m

5 m

4 m

3 m

OBD is right angledOD = 3 mTrig to find BOD

D-1tan ( )

tanq = 43 Û

q = 43 Û

53.13q » °

θ

© T Madas

O

A

B

C

5 m

5 m

Dθ

The area of the sector

p 25´ ´ 53.13360

211.59 m»

The length of the arc2 p´ 5´ ´ 53.13

360 4.64 m»

area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 m



© T Madas

O

A

B

C

5 m

5 m

Dθ

area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 m

OBC is right angledOB = 5 mBOC = 53.13°Trig to find BC

BCOB = tanq Û

5BC 53.13tan °= 4

3= Û

3BC 20= Û203 mBC =

4 m

3 m

BC = 20/3 m



© T Madas

O

A

B

C

5 m

5 m

Dθ

area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 mBC = 20/3 m

Area of OBC12 5´ 20

3´ 1006= 50

3= 216.67m»

area of the sector ≈ 11.59 m2

half the shaded area ≈ 5.08 m2

shaded area ≈ 10.16 m2



© T Madas

O

A

B

C

5 m

5 m

Dθ

area of the sector ≈ 11.59 m2length of the arc ≈ 11.59 mBC = 20/3 m

E

Length BC = 20/3 m ≈ 6.67 m

length of the arc ≈ 11.59 m

Total perimeter ≈ 36.5 m

11.59

6.67

x 2x 2

≈ 23.18

≈ 13.34



© T Madas

A circular sector has radius r and an area A .The sector corresponds to an arc of length L .Write an expression for A in terms of r and L .

r

A L

© T Madas


rθ

A L

=A ´360

2r

=L ´360

2 r

© T Madas

2 r2 r

2r 2r


=A ´360

2r

=L ´360

2 r

Û =A2

360

r

Û =L360

2 r

Û =360A 2r

Û =360L 2 r

Û

Û

=360A 2r

=360L 2 r

Û

Û

=2

360A

r

= 180L

r

Now what?

© T Madas


=2

360A

r

= 180L

r

2

360A

rÛ =180L

rÛ =360 rA 2180 Lr

360 r 360 r

=A

2

rL2

© T Madas

Harder Examples

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A B

CD

The following pattern is produced inside a square whose side length is 8 cmCalculate the shaded area

The shaded region can be split into 8 congruent shapes

take one of these shapes

add the right angled triangle next to it

they make up a 45° sector

of a circle whose radius is half the square’s diagonal

8 cm

45°P l a n

© T Madas

A B

CD


8 cm

45°

4

4Pythagoras:

x

2x = 24 24+ Û2x = 32 Û

x = 3232

Area of sector:

18´ p ( )232´ = 4p

Area of triangle:

cm8

© T Madas

8 cm

In a square with side length of 8 cm, 2 quarter circular arcs are drawn, as shown below.Find the area of the shaded region

A B

CD

60°

The bottom part of the shaded area consists of:

•an equilateral triangle with side length of 8 cm

•plus 2 circular segments

Look at this shaded section:

It is a circular sector, one sixth of a circle of radius 8 cm.The area of the circular segment equals the area of this sector less the area of the equilateral triangle.

© T Madas

8 cm


A B

CD

60°

16´


p 28´ 12- 8´ 8´ sin60´ °

323p 32- 3

2´

3sin602

°=

323 16 3p= -

32π

3 –

16 3

© T Madas

8 cm


A B

CD

60°

16´


p 28´ 12- 8´ 8´ sin60´ °

323p 32- 3

2´ 323 16 3p= -

Now look at one of the non - shaded regions

32π

3 –

16 3

30°

The area of one of the non shaded regions equals:

One twelfth of a circle of radius 8 cm

Less the area of the segment112´ p 28´ 32

3 16 3( )p- -

© T Madas

8 cm


A B

CD

60°

32π

3 –

16 3

30°

112´ p 28´ 32

3 16 3( )p- -

112´ p 28´ 32

3 16 3( )p- -

6412p= 32

3p- 16 3+

163p= 32

3p- 16 3+

-- 163p= 16 3+

163p-16 3=

© T Madas

8 cm


A B

CD

60°

32π

3 –

16 3

30°

112´ p 28´ 32

3 16 3( )p- -

6412p= 32

3p- 16 3+

163p= 32

3p- 16 3+

-- 163p= 16 3+

163p-16 3=

16π

3

16 3

16 316π3

© T Madas

8 cm


A B

CD

16 316π3

The shaded region is 64 cm2 less the two non shaded regions we just found 16

32 16 3( )p- -64 242 cm»

16π

3

16 3

© T Madas

´r360

r 2 +h2 = L2

L2´

360=A

=L

r360

Û L2´360

=A L

r360

Û

L2 ´360

1

=A L

r360

Û

L2

L360=A Û

=LrA

© t madas. the area of a circle the circumference of a circle r we can use these formulae to find...

Documents

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