Chapter 11

Area of Polygons and Circles

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

Lesson 11.1

Angle Measure in Polygons

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

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.

360o

180o

540o

720o

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.

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.

– 155 + 135 + 145 + 150 + 110 + 120 + 130 + x = 1080

• And solve for x– 945 + x = 1080– x = 135o

130o 120o

110o

150o

145o135o

155o

xo

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.

Example 2Find 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

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)

n

120=1

Dividing by n is the same as multiplying by 1/n

120n =180(n – 2)

120n =180n – 360

-60n = – 360

n = 6

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

34

5

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

34

5

1 + 2 + 3 + 4 + 5 = 360o

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.

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

Homework 11.1• In Class

– 1-5• p665-668

• HW– 6-41, 49-54, 58-61, 63-73

• Due Tomorrow

Lesson 11.2

Areas of Regular Polygons

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

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

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

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

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

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

Example 4• 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• Using radius

– sin (1/2 CA) = .5s/r

– sin (36) = .5s/5

– .5879 = .5s/5

– 5(.5879) = .5s– 2.939 = .5s– s = 5.878

CA = 72o

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.

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)• n stands for the number of sides• s stands for the length of one side

Homework 11.2• In Class

– 1-8• p672-675

• HW– 9-34, 50-52, 54-64

• Due Tomorrow

Lesson 11.3

Perimeters and AreasofSimilar Figures

Lesson 11.3 Objectives• Compare the perimeters of

similar figures• Compare the areas of similar

figures

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

5 15

Ratio of Sides

15/5 = 3

Ratio of Perimeters15/5 = 3

Ratio of Areas

225/25 = 932

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.

Homework 11.3• In Class

– 1-6• p679-681

• HW– 7-28, 34-41

• Due Tomorrow• Quiz Wednesday

– Lessons 11.1-11.3

Lesson 11.4

CircumferenceandArc Length

Lesson 11.4 Objectives• Find the circumference of a

circle.• Identify arc length• Define arc measure

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

Theorem 11.6:Circumference of a Circle

• The circumference (C) of a circle is– C = d or…– C = 2r

• where d is diameter• and r is radius

C

d

r

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

Example 5Find the circumference of

the circle• r = 5

• C = 2 r• C = 2 (5)• C = 10

– Leave it!

5

Example 6• C = 32

• Find diameter

– d = C/

– d = 32 /– d = 32

32

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 measureA

B

ABmAB

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

Circumference

Arc Measure

360o

Example 7• Find the arc length for the

followingA

B

AB60o

8

=Arc Length

Circumference

Arc Measure

360o

=Arc Length

2 r

60o

360o2 (8)

Arc Length =60o

360ox 16

Arc Length =16

x 16

Arc Length =16

6

Arc Length =8 3

16

Homework 11.4• In Class

• p686-689

– 1-14

• HW– 15-35, 39-41, 48-49

• Due Tomorrow

Lesson 11.5

Areas of CirclesandSectors

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

Theorem 11.7:Area of a Circle

• The area of a circle is times the square of the radius.– A = r2

C

r

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!

AB

A

B

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

Circle Area

Arc Measure

360o

Example 8• Find the area of a sector with a

radius 8 and an arc measure of 75o

AB

A

B

=Sector Area

Circle Area

Arc Measure

360o

=Sector Area

r2

75o

360o

(8)2

Sector Area =(64)

75o

360o

Sector Area =4800 360o

Sector Area =40 3

64

Homework 11.5• In Class

– 1-9• p695-698

• HW– 10-37, 43, 44

• Due Tomorrow

Lesson 11.6

Geometric Probability

Lesson 11.6 Objectives• Recall probability• Apply probability to a line

segment• Apply probability to a

geometric area

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.

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

P(K is on segment CD) =CDAZ

If K is in area M

P(K is in area M) =Area MArea J

P stands for probability of the inside of parentheses happening.

Example 9• Find the probability that a

point randomly chosen is on line segment WX.

0 1

V W X Y Z

P(K is on segment WX) =WXVZ

P(K is on segment WX) =3

11

Example 10• Find the probability that a

point randomly chosen lies inside the circle.

s = 10

Area CircleP(K is in the circle) =

Area Square r2

=s2

=(5)2

102

25

100

≈ 78.5%

Homework 11.6• 2-19, 31-34, 37-39

– skip 18– p701-704

• In Class – 8, 11, 31, 37• Due Tomorrow• Quiz Thursday

– Lessons 11.4-11.6