What is a Circle?
- Round in shape
- a shape that is curved and without sharp angles
1. Circumference : outline, edge, boundary
2. Centre O : equal length from point O to the circumference
3. Diameter : line passing through O to the circumference
4. Radius : distance from O to the circumference
Diameter = 2 x radius = 2r
Radius = ½ of diameter = ½ d
d = 2r , r = ½ d
CIRCUMFERENCE OF A CIRCLE
Circumference is the Outline of a circle; or its perimeter.
Circumference = π d [π x d]
= π x 2r [ d = 2 x r ]
= 2 π r [number is always placed in front]
~ Circumference is also when a circle makes one complete turn.
Area of a Circle = π x r x r [ r x r = r2, cm x cm = cm2]
= π x r2
= π r2
Example:
The diameter of bicycle wheel is 14cm. Find the circumference. (Take π =22/7)
Method
Step1: Circle/Underline keywords, identify topic of question
Step2: Draw Circle(if need) and write formula for circumference
circumference = π d
Step 3 : Solve : Use given values and formula
Circumference = 22/7 x 14
= 22 x 2 = 44 cm
Example : Find the area of the circle (π = 22/7 )
Step 1: Circle/Underline keywords , identify topic = circle area
Step 2: Draw Circle and write required formula : area=πr2
Step 3: Solve : Use given values and formula
D=28 , r = d/2 = 28/2 = 14cm
area= π r2
= 22/7 x 14 x 14
= 616cm2
SHAPES RELATED/ASSOCIATED WITH A CIRCLE
Circle Semicircle Quarter-Circle(Quadrant)
Circle = 2 semicircles 1 semicircle = ½ circle 1 quadrant = ½ semicircle
= 4 quadrants = 2 quadrants = ¼ circle
SEMICIRCLE
½ of a circle; 2 semicircles of the same radius form a circle
PERIMETER OF A SEMICIRCLE
Perimeter of a semicircle = ½ x 2 π r + d
= π r + d
= π r + 2r
Area of a Circle =1/2 π x r x r [ r x r = r2, cm x cm = cm2]
= ½ π x r2
= ½ π r2
QUADRANT
= ¼ of a circle
PERIMETER OF A QUARTER CIRCLE(QUADRANT)
Perimeter of quadrant = ¼ of circumference of circle + radius + radius
= ¼ x 2 π r + (r + r)
= ½ π r + 2r
= ½ π r + d
AREA OF A QUARTER CRICLE (QUADRANT)
Area of a quadrant = ¼ of area of a circle
= ¼ x π r x r [ r x r = r2, cm x cm = cm2]
= ¼ π x r2
Example
The circumference of a circle of radius 7 cm is 44cm. what is the perimeter of a quadrant?
Method
Step1: Circle/Underline keywords (circumference 44cm, 7cm radius, perimeter of quarter circle)
Step2: Draw Circle and write required formula for circumference
circumference = π d
Step 3: Solve : Use given values and formula
[Perimeter = the line that bounds the shape, trace with finger]
Perimeter of quadrant = r + r + ¼ circumference
= 7 + 7 + ¼ x 2 x 7 x 22/7
= 14 + 11 = 25cm
Example
The figure shows two identical semicircles of radius 14cm. Find the perimeter of the figure (take π = 22/7).
Step 1: Circle/Underline keywords [ to identify topic and recall method to solve]
[ Semicircle, perimeter => perimeter = outline, semicircle formula]
- semicircle = ½ of circle circumference = ½ x 2 π r = π r
Step 2: Trace outline and write parts to calculate
Calculate value for the parts [perimeter of semicircle]
Perimeter of ½ circle circumference = ½ x 2 π r
= ½ x 2 x 22/7 x 14
= ½ x 2 x 22/7 x 142
= 22 x 2
= 44 cm
Step 3: Solve by adding all the parts of the perimeter
Perimeter of figure = 5 + 44 + 5 + 44
= 98 cm
RELATED/ASSOCIATED PORTION OF A CIRCLE
(1) Curved-section
-> part
of a circle
(2) Square / Rectangle
- Quarter Circle if curved part is touching diagonal of square
- Semicircle if curved part is on side of square/rectangle
Example 1:
ABCD
is a square. AB = 7cm. Find the green area. [curved
section -> part of a circle]
Method
Step 1: Circle/Underline keywords
Step 2: Draw Circle and
write required
formula
: area of square = side x side; area of circle =πr2
Step 3: Solve : Use given values and formula
r = 7
area
of square = 7 x 7 = 40cm2
area=π r2
=1/4 x 22/7 x 7
x 7
= 38.5 cm2
Area = 49 – 38.5 cm2 = 10.5 cm2
Example 2
The figure shows a right-angle isosceles triangle in a circle. O is the centre of the circle. The area of the triangle 64cm2. Find the total area of the shaded part.
Step1: Write formulae and method
Area of circle =πr2
Area of triangle = ½ x base x height
Area of shaded area = Area of circle – Area of Triangle
Step2: Find the required data
Given = 64 cm2
Required data = Radius
Area of triangle = ½ x r x 2 r
r2 = 64
r = 8cm
Step3 : Solve
Area of shaded area =πr2 – 64
= 3.14 x 8 x 8 – 64
= 136.96 cm2
Example 3
The figure is made up of three identical circles. Point A, B and C are the centres of the respective circles. The radius of each circle is 14cm. Find the total area of the shaded parts. (Take π = 22/7)
Step1: Write Formula, data and method
Formula : Area of Circle = π r2
Drawn triangle = equilateral (the sides are radii of the circles)
= 60o angle
Step2: Find and ‘deconstruct’ the composite shapes
“Shift” the shaded portion of each part to the unshaded portion
Each = segment of circle with angle = 60o
Step3 : Solve
There are three 60o segment of circles = 60o + 60o + 60o = 180o
Area of shaded parts = Area of semi-circle
= ½ x π x r2
= ½ x 22/7 x 14 x 14
= 1078 cm2
** This is generally the method to solve over-lapping circles :
- 'shifting'/'moving' parts to make up to segment of a circle.
Formula for Area of Segment = Angle of segment x π r2
360o
VIDEO
Circle Content : Circumference and Area, Semicircle and Quarter Circle
----- Do you know? ----------
= 3.14159265359
~ 3.14 = 22/7
Circumference = πd = 3.14 x d
Ratio of circle Circumference to its Diameter
Circumference : Diameter
3.14 : 1
Diagrammatically,
~~~~ END ~~~~ :)
