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Circles introduction
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© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
Primary 6 Challenging Mathematics
Lesson 14:
CIRCLES AND COMPOSITE FIGURES (PART I)
© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
Circles And Composite Figures
Overview
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What are 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.
© 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
© 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
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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
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Diameter Versus Radius
Can you see the relationship between radius and diameter?
diameter = 2 x radius
diameter
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What is a Circumference?
The circumference of a circle is its perimeter.
© 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).
© 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
227
© 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.
© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
Area of a Circle
Thus,
area of circle = ππππ x radius x radius
© 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
© 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
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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
© 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
© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
Circles And Composite Figures Challenging Word Problems
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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
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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:
= -
© 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.
© 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
© 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
© 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
© 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
© 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
© 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
© 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.
= -
227
© 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.
= -
© 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
© 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.
© 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
© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
Challenging Word Problem: Example 4
Solution:
308 – 154 = 154
Area of shaded region is 154 cm2.
© 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?
© 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.
© 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.
© 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.
© 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
© Blue Orange Pte Ltd Primary 6 Challenging Mathematics
End