1. Area of Polygons and Circles
Chapter 11
Area of Polygons and Circles

2. Chapter 11 Objectives
Calculate the sum of the interior angles of any polygon
Calculate the area of any regular polygon
Compare the perimeters of similar polygons
Compare the areas of similar polygons
Define circumference
Calculate arc length
Find the area of a circle
Define a sector
Find the area of a sector
Utilize geometric probability

3. Angle Measure in Polygons
Lesson 11.1
Angle Measure in Polygons

4. Lesson 11.1 Objectives Utilize the Polygon Interior Angles Theorem
Calculate the number of sides on a polygon know the interior angle sum
Find the sum of the exterior angles of a polygon

5. Interior Angles of a Polygon
The sum of the interior angles of a triangle is
180o
The sum of the interior angles of a quadrilateral is
360o
The sum of the interior angles of a pentagon is
???
The sum of the interior angles of a hexagon is
By splitting the interior into triangles, it should be able to tell you the sum of the interior angles.
Just count up the number of triangles and multiply by 180o.
180o
360o
540o
720o

6. Theorem 11.1: Polygon Interior Angles Theorem
The sum of the measure of the interior angles of a convex n-gon is
180o(n – 2)
n is the number of sides
This basically states that for every side that you add, you add another 180o to the interior angles.

7. Example 1 First determine the number of sides of the polygon.
Plug n into Theorem 11.1 and solve.
180(8 –2) =
1080o
Create an equation that has all interior angles equal to the answer from above.
x = 1080
And solve for x
945 + x = 1080
x = 135o
135o
145o
155o
150o xo
110o
130o
120o

8. Corollary to Theorem 11.1
The measure of each interior angle of a regular n-gon is
1/n(180)(n - 2)
It must be regular!
It basically states take the sum of the interior angles and divide by the number of sides to figure out how big each angle is.

9. Example 2
Find the measure of each interior angle in the figure at right.
n = 5
1/5(180)(5 – 2)
180/5(3)
(36)(3)
108O
108O

10. Example 3: Finding the Number of Sides
Each interior angle is 120o, name the polygon.
Use Corollary 11.1
1/n(180)(n – 2)
Set it equal the measure of each interior angle and solve for n.
1/n(180)(n – 2) = 120
Multiply both sides by n
180(n – 2) = 120n
Distribute
180n – 360 = 120n
Subtract 180n from both sides
-360 = -60n
Divided by –60
n = 6
Hexagon
Could be easier if set up as a proportion.
180(n – 2) =
120 n 1
Dividing by n is the same as multiplying by 1/n
120n =
180(n – 2)
120n =
180n – 360
-60n =
– 360
n = 6

11. Exterior Angles
An exterior angle is formed by extending each side of a polygon in one direction.
Make sure they all extend either pointing clockwise or counter-clockwise. 1 2 3 4 5

12. Theorem 11.2: Polygon Exterior Angles Theorem
The sum of the measures of the exterior angles of a convex polygon is 360o.
As if you were traveling in a circle! 1 2 3 4 5
 1 +  2 +  3 +  4 +  5 = 360o

13. Corollary to Theorem 11.2
The measure of the each exterior angle of a regular n-gon is
360/n
It must be regular!
This will be used to determine the number of sides in a polygon.

14. Example 4
The measure of an exterior angle of a regular polygon is 120o. Name it.
Plug into Corollary 11.2
360/120 = n
n = 3
Triangle

15. Homework 11.1 In Class HW Due Tomorrow 1-5 6-41, 49-54, 58-61, 63-73p HW
6-41, 49-54, 58-61, 63-73
Due Tomorrow

16. Areas of Regular Polygons
Lesson 11.2
Areas of Regular Polygons

17. Lesson 11.2 Objectives Area of Equilateral Triangle Theorem
Use the Area of a Regular Polygon Theorem
Know parts of a polygon
Define central angle

18. Theorem 11.3: Area of an Equilateral Triangle
Area of an equilateral triangle is
A = ¼ (√3) s2
Take ¼ times the length of a side squared and write in front of √3.
Be sure to simplify if possible. s

19. Parts of a Polygon
The center of a polygon is the center of the polygon’s circumscribed circle.
A circumscribed circle is one in that is drawn to go through all the vertices of a polygon.
The radius of a polygon is the radius of its circumscribed circle.
Will go from the center to a vertex. r

20. Apothem
The apothem is the distance from the center to any side of the polygon.
Not to the vertex, but to the center of the side.
The height of a triangle formed between the center and two consecutive vertices of the polygon. a

21. Central Angle of a Polygon
The central angle of a polygon is the angle formed by drawing lines from the center to two consecutive vertices.
This is found by
360/n
That is because the total degrees traveled around the center would be like a circle.
Then divide that by the number of sides because that determines how many central angles could be formed.
600

22. Finding the Apothem or Radius
In order to find the apothem, you must know one of the following
length of a side
length of radius
central angle
You will use trig to find the missing apothem
If given the radius, you will use cosine and half the central angle.
If given a side, you will use half the side, tangent, and half the central angle.
In order to find the radius, you must know one of the following
length of a side
length of apothem
central angle
You will use trig to find the missing radius
If given the apothem, you will use cosine and half the central angle
If given a side, you will use half the side, sine, and half the central angle. r a s

23. Example 4 CA = 72o Apothem Side length
Find the length of the apothem and the side of a regular pentagon with a radius of 5.
Apothem
Using radius
cos (1/2 CA) = a/r
cos (1/2(72) = a/5
.8090 = a/5
5(.8090) = a
a = 4.045
Side length
sin (1/2 CA) = .5s/r
sin (36) = .5s/5
.5879 = .5s/5
5(.5879) = .5s
2.939 = .5s
s = 5.878
Remember that the angle used for calculation is half the central angle.
And the bottom of the triangle is half the length of one entire side.
CA = 72o

24. Theorem 11.4: Area of a Regular Polygon
The area of a regular n-gon with side length s is half the product of the apothem and the perimeter.
A = 1/2aP
A stands for area
a stands for apothem
P stands for perimeter of the n-gon
Found by finding the side length and multiplying by the number of sides
A = 1/2a(ns)

25. Homework 11.2
In Class
1-8 p HW
9-34, 50-52, 54-64
Due Tomorrow

26. Perimeters and Areas of Similar Figures
Lesson 11.3
Perimeters and Areas of
Similar Figures

27. Lesson 11.3 Objectives Compare the perimeters of similar figures
Compare the areas of similar figures

28. Theorem 11.5: Areas of Similar Polygons
If two polygons are similar with the lengths of corresponding sides in the ratio of a:b, then the ratio of their areas are a2:b2
Remember that the ratio of side lengths a:b is the same as the ratio of the perimeters, a:b.
Theorem 8.1
Ratio of Sides
15/5 = 3
Ratio of Perimeters
15/5 = 3
Ratio of Areas 32
225/25 = 9 5 15

29. Using Theorem 11.5 First make sure the figures are similar.
They will tell you, or…
You need to use the Similarity Theorems from Chapter 8
SSS Similarity
SAS Similarity
AA Similarity
Try to find the scale factor
Ratio of side lengths
The ratio of the areas is the square of the scale factor.
So the scale factor is the square root of the ratio of the areas.

30. Homework 11.3 In Class HW Due Tomorrow Quiz Wednesday 1-6 7-28, 34-41p HW
7-28, 34-41
Due Tomorrow
Quiz Wednesday
Lessons

31. Circumference and Arc Length
Lesson 11.4
Circumference
and
Arc Length

32. Lesson 11.4 Objectives Find the circumference of a circle.
Identify arc length
Define arc measure

33. Circumference
The circumference of a circle is the distance around the circle.
For all circles, the ratio of circumference to the diameter is the same.
 , or pi C d

34. Theorem 11.6: Circumference of a Circle
The circumference (C) of a circle is
C = d or…
C = 2r
where d is diameter
and r is radius C d r

35. Using Theorem 11.6
If asked to find the circumference. identify what you know
diameter
Use C =  d
radius
Use C = 2 r
If asked to find the diameter or radius, you must work backwards.
diameter
Divide by 
radius
Divide by 2

36. Example 5 Find the circumference of the circle r = 5 C = 2 r
Leave it! 5

37. Example 6
C = 32
Find diameter
d = C/
d = 32 /
d = 32
32

38. Arc Length An arc length is a portion of the circumference.
denoted CD with an arc on top
It is determined by its arc measure
The measure of the angle made by joining the endpoints of the arc with the center of the circle.
denoted by placing an m in front to show we are finding the measure A B AB
mAB

39. Corollary: Arc Length Corollary
In a circle, the ratio of the length of the given arc to the entire circumference is equal to the ratio of the measure of the arc to the measure of the entire circle, 360o.
Set up a proportion using the smaller portions over the entire portions.
Arc Length
Arc Measure =
Circumference
360o

40. Example 7 Find the arc length for the following 8 AB A 60o B 1 x 16=
Arc Length
Circumference
Arc Measure
360o 6
16
Arc Length
60o
Arc Length = = 6
2 r
2 (8)
16
360o 8
60o
Arc Length =
x 16
Arc Length = 3
360o

41. Homework 11.4 In Class HW Due Tomorrow 1-14 15-35, 39-41, 48-49p
1-14 HW
15-35, 39-41, 48-49
Due Tomorrow

42. Areas of Circles and Sectors
Lesson 11.5
Areas of Circles
and
Sectors

43. Lesson 11.5 Objectives Find the area of a circle
Calculate the area of a sector of a circle
Apply the area of a circle and its sector to finding the area of complex figures

44. Theorem 11.7: Area of a Circle
The area of a circle is  times the square of the radius.
A = r2 C r

45. Sector
A sector of a circle is the region bounded by two radii of the circle and their arc.
Usually looks like a slice of pizza! A B AB

46. Theorem 11.8: Area of a Sector
The ratio of the area (A) of a sector of a circle to the area of the entire circle is equal to the ratio of the measure of the arc to the measure of the entire circle, 360o.
Sector Area
Arc Measure =
Circle Area
360o

47. Example 8
Find the area of a sector with a radius 8 and an arc measure of 75o A B AB
4800
Sector Area =
360o =
Sector Area
Circle Area
Arc Measure
360o
40
Sector Area = 3
Sector Area
75o =
 (8)2
360o
64
 r2
75o
(64)
Sector Area =
360o

48. Homework 11.5
In Class
1-9 p HW
10-37, 43, 44
Due Tomorrow

49. Geometric Probability
Lesson 11.6
Geometric Probability

50. Lesson 11.6 Objectives Recall probability
Apply probability to a line segment
Apply probability to a geometric area

51. Probability
Recall that probability is a number that represents the chance that an event will occur.
That number is a decimal or fraction from 0 to 1
0 means the event cannot occur
1 means the event will always occur
The probability is calculated by taking the number of favorable outcomes and dividing by the total number of possible outcomes.

52. Geometric Probability
The probability of finding point K on a line segment is determined by divided the length of the target segment divided by the length of the entire segment.
The probability of finding point K in a given area is determined by finding the target area and dividing by the entire area of the surface.
Any time that probability is calculated using geometric measures such as length and area, you are finding the geometric probability of the event occurring.
If K is on segment CD CD
P(K is on segment CD) = AZ
P stands for probability of the inside of parentheses happening.
If K is in area M
Area M
P(K is in area M) =
Area J

53. Example 9
Find the probability that a point randomly chosen is on line segment WX. V W X Y Z 1 WX
P(K is on segment WX) = VZ 3
P(K is on segment WX) = 11

54. Example 10
Find the probability that a point randomly chosen lies inside the circle.
s = 10
(5)2
25
Area Circle
 r2
P(K is in the circle) = = =
Area Square s2
100
102
≈ 78.5%

55. Homework 11.6 2-19, 31-34, 37-39 In Class – 8, 11, 31, 37 Due Tomorrow
skip 18 p
In Class – 8, 11, 31, 37
Due Tomorrow
Quiz Thursday
Lessons