© t madas composite shapes with circular parts

© T Madas

Composite Shapeswith

Circular Parts

© T Madas

A semicircle has a radius of 9 cm.

• Calculate its area

• Calculate its perimeter

9 cm

A = π x r

2

A = π x 92

A ≈ 254

cm2

127 cm2

© T Madas

A semicircle has a radius of 9 cm.



9 cm

C = π x r

C = π x 9

C ≈ 56.5

cm

2 x2 x

28.25 cm

P = + 18

28.25

= 46.25 cm

© T Madas

A quarter-circle has a radius of 16 cm.



16 cm

A = π x r

2

A = π x 182

A ≈ 804 cm2

201 cm2

© T Madas

A quarter-circle has a radius of 16 cm.



16 cm

C = π x r

C = π x 16

C ≈ 100.53

cm

2 x2 x

25.1 cm

P = + 16

25.1

= 57.1 cm+ 16

© T Madas

8 cm

10 c

mCalculate the perimeter and area of the following composite shape.

4 cm10+ 8+ 10P = + x π x 41

22x

28P = x π+ 4

40.6P ≈ cm

C = x π x r2

© T Madas

8 cm

10 c

mCalculate the perimeter and area of the following composite shape.

4 cm

A2

A1

10+ 8+ 10P = + x π x 412

2x

28P =

40.6P ≈ cm

C = x π x r2

10x 8A = + π x 412

2x

80A = x π+ 8

105A ≈ cm2

A =π x r 2

x π+ 4

© T Madas

6 cm

Calculate the perimeter & area of the grey region below.

© T Madas

6 cm


C = x π x r2

A =π x r 2

The perimeter of the grey area is equal to …… the circumference of a circle of radius …… 3 cm

P = x π x 32

18.8P ≈ cm

© T Madas

6 cm


C = x π x r2

A =π x r 2

The area of the grey area is equal to …… the area of a square with side 6 cm …… less …… the area of a circle of radius 3 cm

6x 6A = – π x 32

36A = x π– 9

7.73A ≈ cm2

© T Madas

[ ] x

10 cm

Calculate the perimeter & area of the following shape

40 cm

C = x π x r2

A =π x r 2

The perimeter is equal to …… the circumference of a semi-circle of radius 20 cm …… plus …… the circumference of a circle of radius 10 cm …P = x π x 202

20P = π

125.7P ≈ cm

12 x π x 10+ 2

+ 20π40P = π

© T Madas

[ ] x

10 cm

Calculate the perimeter & area of the following shape

40 cm

C = x π x r2

A =π x r 2

The total area is equal to …… the area of a semi-circle of radius 20 cm …… plus …… the area of a circle of radius 10 cm …

A = π x 202

200A = π

942A ≈ cm2

12 + πx 102

+ 100π300A = π

© T Madas

The figure below shows a pond made up of two squares and two identical quarter circles with a radius of 4 m. Calculate to 2 decimal places:

1. the perimeter of the pond.

2. the area of the pond

4 m 4 m

© T Madas




4 m 4 m

4

4

4

4 ?

?

C = 2 x xπ r

C = 2 x xπ 4

C ≈ 25.13m

Each curved edge is ¼ of the circumference of a full circle.

25.13÷ 4 ≈6.28m

6.28

6.28

P = 4 x 4 + 2 x 6.28≈28.57m

© T Madas




4 m

16 m2

16 m2

4 m

Area of a Quarter Circle

A = xπ r 2

A = xπ 4 2

A = xπ 16

A ≈50.27m2

50.27÷4≈12.57m2

12.57 m2

12.57 m2

P = 2 x 16 + 2 x12.57≈57.14m2

© T Madas

[ ]x 4

Calculate the perimeter & area of the following shape:

16 m

4 m

C = x π x r2

A =π x r 2

The perimeter is equal to …… the circumference of …… 8 semi-circles of radius 4 m …… or …… 4 circles of radius 4 m …P = x π x 42

32P = π100.5P ≈ m

© T Madas


16 m

4 m

C = x π x r2

A =π x r 2

The total area is equal to …… the area of a square with side length of 16 m …… plus …… the area of 4 circles of radius 4 m …

16 x 16A = + π x 42

256A = x π+ 64

457A ≈ m2

[ ]x 4

© T Madas


28 cm7 cm

C = x π x r2

A =π x r 2

The perimeter is equal to …… the circumference of a semi-circle of radius 14 cm …… plus …… the circumference of a circle of radius 7 cm …[ ] xP = x π x 142

14P = π

88.0P ≈ cm

12 x π x 7+ 2

+ 14π28P = π

© T Madas


28 cm7 cm

C = x π x r2

A =π x r 2

[ ] x

The total area is equal to …… the area of a semi-circle of radius 14 cm …… less…… the area of a circle of radius 7 cm …

A = π x 142

98A = π

154A ≈ cm2

12 – π x 72

– 49π49A = π

© T Madas

Harder Problems

© T Madas

2( )a

aa

Find the area of the heart in terms of a

Area of square:

2SA a=

2 semi-circles = circle

CA = ( )2

2ap= 21

4 ap=

TA = 1SA 2a

Note that the circle’s radius is:

2a

2rp

CA+ = 214 ap+ = 1

4p+ ( )2

41a p= +

© T Madas

Calculate in terms of π the area of the composite shape drawn below which consists of three semicircles of unit radius and the area enclosed by them.

© T Madas

1


solution

© T Madas

1


the area of this composite consists of a circle of unit radius plus a 1 by 2 rectangle

Ac = πr 2

Ac = π x 12

Ac = π

the total area is

© T Madas

1


the area of this composite consists of a circle of unit radius plus a 1 by 2 rectangle

Ac = πr 2

Ac = π x 12

Ac = π

the total area isπ + 2

© T Madas

2( )

Find the area enclosed by the 4 circles in terms of a

Area of the square:

sA = 24a=

4 quarter-circles = circle

cA =

GA = SA 24a 2( )a

a

2a

2ap

CA- = 2ap- = 4 p-

© T Madas

a

Find the exact area of the orange “petal”

Billy wants a hint ...

© T Madas

one of the blue regions:

area of the squareless the area of the quarter circle

both blue regions

2( )a=

a

Find the exact area of the orange “petal”

2a 214 ap- 1 1

4p-

2 142 1( )a p-

The area of the “petal” is given by the area of the square less the area of the two blue regions:

2a 2 142 1( )a p- - 2a= 22a- 21

2 ap+ 212 ap= 2a-

2( )a= 1-12p

© T Madas

Vase equals Square

© T Madas

Vase equals SquareLook at this vase shaped object

It consists of 6 identical arcs

Each arc is a quarter circle

If the quarter circles to which these arcs correspond have radius a, find the area of this object

a

a

?

© T Madas

Vase equals SquareLook at this vase shaped object

It consists of 6 identical arcs

Each arc is a quarter circle

If the quarter circles to which these arcs correspond have radius a, find the area of this object

a

a

1st hint 2nd hint

?

© T Madas

Vase equals Square

a

a

1st hint 2nd hint

© T Madas

2a

Vase equals Square

a

a

1st hint 2nd hint

The area of this object is equal to the area of the square on the right.

No complex calculations needed !

4a 2

© T Madas

a

In the following Yin – Yang symbol calculate the area and perimeter of each of its two identical sections

© T Madas

a


working with the green section:

A r e a

These semicircles

both have a radius

of 4a

© T Madas

a



A r e a

These semicircles

both have a radius

of 4a

The area of the green section of the Yin Yang is equal to the area of a semicircle

© T Madas

a


2( )A= 12 p´

2a´

A= 12 p´

2

4a´

A=2

8ap

Û

Û


A r e a

2

8aA p=

© T Madas

a


working with the green section;the required perimeter is given by:

the circumference of a circle

of diameter

plus the circumference of a

semicircle of diameter

P e r i m e t e r

2

8aA p=

2a

a

© T Madas

a


P =p 2a´ p´ a´ Û

P e r i m e t e r

2

8aA p=

12

+

P = 2ap Û

2ap+

P = ap

© T Madas

a


The blue sections are congruentArea of the blue sections

This semicircle

has a radius of 6a

2( )A= 12 p´

6a´

A= 12 p´

2

36a´

A=2

72ap

Û

Û

2

72ap

© T Madas

a


Area of the blue sections

This semicircle has a radius of

2a

2( )A= 12 p´

2a´

This semicircle has a radius of

3a

2

72ap 2( )1

2- p´

3a´ Û

A= 12

p´2

4a´ 1

2- p´

2

9a´ Û

A=2

8ap 2

18ap-

2 2

8 18a ap p-

© T Madas

a


Area of the blue sections

A=2

72ap Û2

8ap

2 2

8 18a ap p-

2

18ap-

2

72ap+

A= Û2( )ap 18

118

- 172

+

A= Û2( )ap 972

472

- 172

+

A= Û26

72ap

2

12aA p=

2

12ap

2

12ap

© T Madas

a


Area of the orange section

Û2

12ap-

2

12ap-

2

12aA p=

2

12ap

2

12ap

This is best found by subtracting the areas of the two blue sections we just found from the whole circle

2( )A=p2a´

Û2

12ap-

2

12ap-A=

2

4ap

A= Û2( )ap 14

112

- 112

-

A= Û2( )ap 312

112

- 112

-

2

12ap

© T Madas

a


perimeter of a blue section

P = p´3a´ Û1

2

P = 6ap Û3

ap+

p´ 23a´1

2+ p´ a´1

2+

2ap+

P = Û( )ap 16

13

+ 12

+

P = ( )ap 16

26

+ 36

+

P = ap

© T Madas

2( )´2( )´

a


perimeter of a blue section

P = p´3a´ Û1

2

P = 6ap Û3

ap+

p´12

+

P = Û2 ( )ap 16

13

+

P = 2 ( )ap 16

26

+

P = ap

perimeter of the orange section

23a´

P = ap

© T Madas

The generalisations of the Yin Yang shape:

The circle in every case :

•is divided by curved lines of equal lengths•the resulting regions have equal perimeters•the resulting regions have equal areas

© T Madas

2a

2a

Find the grey area enclosed by the 3 circles in terms of a

a

The grey area is equal to the area of an equilateral triangle of side 2a less a semicircle of radius a

60°

Area of Triangle:1

2= 2a´ 2a´ sin60´ o

22a= sin60´ o

22a= 32´

23a=

23a

© T Madas

2( )a= 12p-

Find the grey area enclosed by the 3 circles in terms of a

a

The grey area is equal to the area of an equilateral triangle of side 2a less a semicircle of radius a

Area of Triangle:

23a

Area of semicircle: 212 ap

The grey area:23a 21

2 ap- 3

© T Madas

Three cylindrical broomsticks each of radius a are held together by an elastic band.How long is the elastic band in terms of a ?

60°

a

Solution

all the distances between the centres of the circles are 2a , so we have an equilateral triangle at the centre.

© T Madas


a

120°

Solution

Draw radii as shown towards the elastic band.

The radii must be at right angles at the points of contact with the elastic band. (tangent – radius)

We can now work another useful angle

60°

© T Madas

2 ap13´

2a


a

120°

Solution

We can now calculate some lengths.

Each straight piece (not in contact with the circles) has length 2a

Each arc corresponds to one third of a circle

Finally do all the adding

60°

© T Madas

Three concentric circles have radii of 3, 4 and 5 units, as shown opposite.

What percentage of the largest circle is shaded?

34

5

2A rp=

3A =Annulus:

So: 725pp 28%=

p 23´ = 9p4A =p 24´ = 16p

5A =p 25´ = 25p

4A 3A- 16p= 9p- 7p=

725

=

© T Madas

Marcus has two circular railway lines, one with radius of 1.5 metres and the other with radius of 2 metres.

He runs an engine clockwise round each track at the same speed from the start line of the diagram.

Where would the engine on the outer track be, out of A, B, C or D when the engine on the inner track has made 11 complete circuits?

A

B

C

D

1.5

2

© T Madas

A circuit on the inner track:

A circuit on the outer track:

11 complete circuits on the inner track:

Both engines travel at the same speed, so the engine on the outer track must also cover a distance of 33π, with each circuit in the outer track being 4π

A

B

C

D

1.5

2

2 p´ 1.5´ =3p

2 p´ 2´ = 4p

3p 11´ = 33p

334pp

334= 1 circuits48 =

The engine on the outer track will be at point B when the engine on the inner track has completed 11 circuits.

© t madas composite shapes with circular parts

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