Aim: How do we find the area of a sector of a circle?
Objectives: use ratio to find the area of a sector of a circle.
Lesson Development: We should be familiar with the area formula for the area of a circle: A   r 2 and
circumference C   D . In this lesson, we will see how the area formula is derived.
Divide just one of the sectors into two equal parts. We now have thirteen sectors – number them 1 to 13:
Rearrange the 13 sectors like this:
Notice this shape resembles a rectangle
The height is the circle's radius: just look at sectors 1 and 13 above. When they were in the circle they were
"radius" high.
The width (actually one "bumpy" edge) is half of the curved parts around the circle ... in other words it is about
half the circumference of the circle.
A  r ( r )   r 2 . If we had done more slices, the shape will be more “rectangular”. If we have infinitely many
slices, then A   r 2
Definition: sector of a circle is a part of the circle with the same center.
Since we know the sector is a part of circle, we just need to figure out the fraction and multiply it by A   r 2 to

get its area. Asec tor 
*  r 2 , where  is the central angle of the sector. Similarly, the length of the arc:
360

Arc length 
* 2 r .
360
EX1: Find the area of the sector and its arc length
EX2: The area of a sector of a circle is 4 and the radius
of the circle is 5. What is the arc length of the sector and
the measure of the arc?
4
The arc measures 90 degrees.
* 360  90
16
1
Its length is 2( )*(4)  2
4
EX3: The area of the sector AOB is 10 and
mAOB  100 . Find the radius of circle O.
A
60
64
 (8)2 
360
6
1
8
arc  (2 )(8)  
6
3
100
*  ( r 2 )  10
360
5 2
r  10
18
50 25
r2 

18 9
5 5
r  ,  ( reject )
3 3
EX4: What is the area of the shaded region?
60
8
 (4)2  
360
3
3
 4(4)( )  8 3
2
Asec tor 
Atriangle
Area of shaded region:
8
 8 3
3
EX5: What is the area of the shaded region?
The area of the entire sector:
120
*  (242 )  192 . But
360
we need to subtract that from the area of the triangle.
A
1
(24)(24)sin120  96 3 (or we can divide it into
2
two 30-60-90 triangles and use the ratios to find out the
base and height and therefore its area)
Area of the shaded region: 192 - 96 3
EX6: Three circles with radii 6 are tangent to each other. Find the area of the region enclosed between them.
In our prior examples, we needed triangles. In this case, we will construct a triangle from the center of each
circle. What kind of triangle do we get?
area 
3
1
(12) 2 3(  (6)2 )  36 3  18
4
6
EX7: If time permits, #29 on P455. Hint: construct a right triangle with hypotenuse, XY .
HW#27: P453 – 455: 12 – 14, 15*, 16, 19, 29**
*draw radii to endpoints of chord
**EC: If correct solution is handed in before unit test, +1 pt on your test.
HW#27 Solutions
12) r = 2
13) shaded area = 4  8
15) 12  8
16) shaded area= 18  18 3
Aim: How do we find the area of a sector of a circle?
3
3 2 6  9 3

(3) 
2
4
4
2
19) 100  192 cm
14) shaded area=
Area of a circle: A   r 2 and circumference: C   D .
Divide just one of the sectors into two equal parts. We now
have thirteen sectors – number them 1 to 13:
Rearrange the 13 sectors like this:
Notice the shape resembles a rectangle
Definition: sector of a circle is a part of the circle with the same center.
EX1: Find the area of the sector and its arc length
EX2: The area of a sector of a circle is 4 and the radius
of the circle is 5. What is the arc length of the sector and
the measure of the arc?
EX3: The area of the sector AOB is 10 and mAOB  100 . Find the radius of circle O.
EX4: What is the exact area of the shaded region?
EX6: Three circles with radii 6 are tangent to each other. Find the EXACT area of the region enclosed between
them.