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© Blue Orange Pte Ltd Primary 6 Challenging Mathematics Primary 6 Challenging Mathematics Lesson 14: CIRCLES AND COMPOSITE FIGURES (PART I)

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Page 1: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Primary 6 Challenging Mathematics

Lesson 14:

CIRCLES AND COMPOSITE FIGURES (PART I)

Page 2: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Circles And Composite Figures

Overview

Page 3: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What are Circles?

Page 4: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What are Circles?

A circle is a simple shape consisting of points in a plane which are the same distance from a given point called the centre.

Page 5: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What are Circles?

A circle is a simple shape consisting of points in a plane which are the same distance from a given point called the centre.

centre

Page 6: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What is a Radius?

Draw a line from the centre to any point on the edge of the circle. This line is the radius of the circle.

centre

radius

radius

Page 7: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What is a Diameter?

“Fold” the circle into half and you get a line as shown. This line is known as the diameter of the circle.

diameter

Page 8: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Diameter Versus Radius

Can you see the relationship between radius and diameter?

diameter = 2 x radius

diameter

Page 9: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What is a Circumference?

The circumference of a circle is its perimeter.

Page 10: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What is a Circumference?

The circumference of any circle is about 3.14 times the diameter. This approximate value , 3.14, is represented by ππππ (read as pi).

Page 11: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

What is a Circumference?

ππππ is or approximately 3.14.

Circumference of a circle = π x diameter or 2 x π x radius

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Page 12: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Area of a Circle

If we cut a circle according to the diagram below and rearrange them, we get something that looks like a rectangle.

Page 13: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Area of a Circle

Thus,

area of circle = ππππ x radius x radius

Page 14: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Overview: Example 1

The figure is made up of a rectangle and two semicircles. Find the area of the figure. (Take π = ) 22

7

20 m

14 m

Page 15: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Overview: Example 1

20 x 14 = 280 m2

Area of rectangle is 280 m2.

x 7 x 7 = 154 m2

Area of the two semicircles is 154 m2.

280 + 154 = 434

Area of the figure is 434 m2. 20 m

14 m

227

Page 16: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Overview: Example 2

The figure is made up of four identical quadrants. Find the area and perimeter of the figure. (Take ππππ = ) 22

7

7 cm

Page 17: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

For one quadrant,

area = ¼ x x 7 x 7 = 38.5 cm2

perimeter = 7 + 7 + ¼ x 2 x x 7

= 7 + 7 + 11 = 25 cm

Since there are 4 quadrants,

area = 38.5 x 4 = 154 cm2

perimeter = 25 x 4 = 100 cm

Overview: Example 2

7 cm

227

227

or alternatively, area of 4 quadrants =area of 1 circle

Page 18: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Circles And Composite Figures Challenging Word Problems

Page 19: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 1

The figure is made up of a square and two quadrants. Find the area and perimeter of the shaded part. (Take ππππ = 3.14)

12 cm

Page 20: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 1

Solution:

Let’s divide the shaded area into half as shown in the diagram below.

The area of that shaded region:

= -

Page 21: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 1

Solution:

¼ x 3.14 x 12 x 12 = 113.04

Area of quadrant is 113.04 cm2.

½ x 12 x 12 = 72

Area of triangle is 72 cm2.

Page 22: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 1

Solution:

113.04 – 72 = 41.04.

41.04 x 2 = 82.08

Area of shaded region is 82.08 cm2.

12 cm

Page 23: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 1

Solution:

For the perimeter, it is formed by 2 times the“edge” of a quadrant.

2 x ¼ x 2 x 3.14 x 12 = 37.68

Perimeter of shaded region is 37.68 cm.

12 cm

Page 24: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 2

ABCD is a square of side 14 cm. The unshaded parts of the figure are 4 quadrants. What fraction of the square is shaded? (Take π = ) 22

7

Page 25: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 2

Solution:

Note that the 4 quadrants form a circle.

14 x 14 = 196

The area of the square is 196 cm2.

x 7 x 7 = 154

The area of the 4 quadrants is 154 cm2.

227

Page 26: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 2

Solution:

196 – 154 = 42

The area of the shaded area is 42 cm2.

of the square is shaded.

42196

=314

314

Page 27: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 3

The figure shows two squares of side 8 cm and 14 cm respectively. Find the area of the shaded part. (Take ππππ = ) 22

7

14 cm

8 cm

Page 28: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 3

Solution:

¼ x x 14 x 14 = 154

Area of quadrant is 154 cm2.

8 x 8 = 64

Area of smaller square is 64 cm2.

= -

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Page 29: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 3

Solution:

½ x (14 + 8) x 8 = 88

Area of triangle is 88 cm2.

154 + 64 – 88 = 130

Area of shaded region is 130 cm2.

= -

Page 30: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 4

The figure below shows 4 circles each touching the big circle exactly at one point. Given that each small circle has a radius of 7 cm, find the total area of the shaded region. (Take ππππ = ) 22

7

Page 31: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 4

Solution:

Note that the shaded area of the two figures are identical.

Page 32: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 4

Solution:

½ x x 14 x 14 = 308

Area of large semicircle is 308 cm2.

x 7 x 7 = 154

Area of small circle is 154 cm2.

227

227

Page 33: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 4

Solution:

308 – 154 = 154

Area of shaded region is 154 cm2.

Page 34: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 5

The figure below is made up of circles of 3 different diameters. What fraction of the figure is shaded?

Page 35: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 5

Solution:

Note that the shaded area of the two figures are identical.

Page 36: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 5

Solution:

Let radius of smallest circle be 10 cm (or any number you can conveniently choose).

½ x ππππ x 40 x 40 = 800ππππArea of large semicircleis 800ππππ cm2.

Page 37: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 5

Solution:

4 x ½ x ππππ x 10 x 10 = 200ππππArea of 4 smallest semicircleis 200ππππ cm2.

800ππππ – 200ππππ = 600ππππArea of 4 shaded regionis 600ππππ cm2.

Page 38: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

Challenging Word Problem: Example 5

Solution:

ππππ x 40 x 40 = 1600ππππArea of largest circle is 1600ππππ cm2.

of the area is shaded.

1600ππππ600ππππ

83=

83

Page 39: Math - P6 (Circles)

© Blue Orange Pte Ltd Primary 6 Challenging Mathematics

End