© T Madas Composite Shapes with Circular Parts.

Composite Shapeswith

Circular Parts

A semicircle has a radius of 9 cm.

• Calculate its area

• Calculate its perimeter

A = π x r

A = π x 92

A ≈ 254

127 cm2

A semicircle has a radius of 9 cm.

C = π x r

C = π x 9

C ≈ 56.5

2 x2 x

28.25 cm

P = + 18

= 46.25 cm

A quarter-circle has a radius of 16 cm.

A = π x r

A = π x 182

A ≈ 804 cm2

201 cm2

A quarter-circle has a radius of 16 cm.

C = π x r

C = π x 16

C ≈ 100.53

2 x2 x

25.1 cm

P = + 16

= 57.1 cm+ 16

mCalculate the perimeter and area of the following composite shape.

4 cm10+ 8+ 10P = + x π x 41

28P = x π+ 4

40.6P ≈ cm

C = x π x r2

mCalculate the perimeter and area of the following composite shape.

10+ 8+ 10P = + x π x 412

40.6P ≈ cm

C = x π x r2

10x 8A = + π x 412

80A = x π+ 8

105A ≈ cm2

A =π x r 2

x π+ 4

Calculate the perimeter & area of the grey region below.

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

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

Calculate the perimeter & area of the following shape

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

Calculate the perimeter & area of the following shape

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

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

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

P = 4 x 4 + 2 x 6.28≈28.57m

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

P = 2 x 16 + 2 x12.57≈57.14m2

[ ]x 4

Calculate the perimeter & area of the following shape:

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

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

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

28 cm7 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 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

Harder Problems

Find the area of the heart in terms of a

Area of square:

2SA a=

2 semi-circles = circle

CA = ( )2

2ap= 21

TA = 1SA 2a

Note that the circle’s radius is:

CA+ = 214 ap+ = 1

4p+ ( )2

41a p= +

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.

solution

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

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

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

Area of the square:

sA = 24a=

4 quarter-circles = circle

GA = SA 24a 2( )a

CA- = 2ap- = 4 p-

Find the exact area of the orange “petal”

Billy wants a hint ...

one of the blue regions:

area of the squareless the area of the quarter circle

both blue regions

2( )a=

Find the exact area of the orange “petal”

2a 214 ap- 1 1

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

Vase equals Square

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

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

1st hint 2nd hint

Vase equals Square

1st hint 2nd hint

Vase equals Square

1st hint 2nd hint

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

No complex calculations needed !

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

working with the green section:

A r e a

These semicircles

both have a radius

A r e a

These semicircles

both have a radius

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

2( )A= 12 p´

A= 12 p´

A r e a

8aA p=

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

8aA p=

P =p 2a´ p´ a´ Û

P e r i m e t e r

8aA p=

P = 2ap Û

P = ap

Spiral Galaxies

Comets

Yin Yang

Marbles ?

Look at the shape below, consisting of three sections.Calculate the area and perimeter of these sections.

The blue sections are congruentArea of the blue sections

This semicircle

has a radius of 6a

2( )A= 12 p´

A= 12 p´

Area of the blue sections

This semicircle has a radius of

2( )A= 12 p´

This semicircle has a radius of

72ap 2( )1

2- p´

3a´ Û

4a´ 1

2- p´

9a´ Û

8 18a ap p-

Area of the blue sections

72ap Û2

8 18a ap p-

A= Û2( )ap 18

A= Û2( )ap 972

A= Û26

12aA p=

Area of the orange section

12aA p=

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

2( )A=p2a´

12ap-A=

A= Û2( )ap 14

A= Û2( )ap 312

perimeter of a blue section

P = p´3a´ Û1

P = 6ap Û3

p´ 23a´1

2+ p´ a´1

P = Û( )ap 16

P = ( )ap 16

P = ap

2( )´2( )´

perimeter of a blue section

P = p´3a´ Û1

P = 6ap Û3

P = Û2 ( )ap 16

P = 2 ( )ap 16

P = ap

perimeter of the orange section

P = ap

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

Find the grey area enclosed by the 3 circles in terms of 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:1

2= 2a´ 2a´ sin60´ o

22a= sin60´ o

22a= 32´

2( )a= 12p-

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

Area of semicircle: 212 ap

The grey area:23a 21

2 ap- 3

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

Solution

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

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

2 ap13´

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

3 ( )´

2 ap13´

Solution

2a 23 ap+

6a= 2 ap+

2 ( )a= 3 p+

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

What percentage of the largest circle is shaded?

2A rp=

3A =Annulus:

So: 725pp 28%=

p 23´ = 9p4A =p 24´ = 16p

5A =p 25´ = 25p

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

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 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

2 p´ 1.5´ =3p

2 p´ 2´ = 4p

3p 11´ = 33p

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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