Transcript 360 o - Mona Shores Blogs

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
180o
360o
– ???
• 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.
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
135o
145o
155o
150o
xo
110o
130o
120o
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 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
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
Could be easier if
set up as a
proportion.
180(n – 2)
n
=
– Hexagon
1
Dividing by n is the
same as multiplying
by 1/n
120n = 180(n – 2)
120n = 180n – 360
– Divided by –60
• n=6
120
-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 counterclockwise.
4
3
5
2
1
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!
4
3
5
2
1
 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
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
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 Areas
of
Similar 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
Ratio of Sides
15/
5
=3
Ratio of Perimeters
15/
5
=3
Ratio of Areas
225/
25
5
15
=9
32
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
Circumference
and
Arc 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 = 2r
• 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 5
Find 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 measure
A
AB
mAB
B
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
=
360o
Circumference
Example 7
• Find the arc length for the
following
A
AB
60o
8
Arc Length
Circumference
Arc Length
2
16(8)
r
Arc Length =
=
=
B
Arc Measure
Arc Length =
1
6
Arc Length =
16
6
Arc Length =
8
3
360o
60o
360o
60o
360o
x 16
x 16
Homework 11.4
• In Class
• p686-689
– 1-14
• HW
– 15-35, 39-41, 48-49
• Due Tomorrow
Lesson 11.5
Areas of Circles
and
Sectors
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!
A
AB
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
Arc Measure
=
360o
Circle Area
Example 8
• Find the area of a sector with a radius
8 and an arc measure of 75o
A
AB
B
Sector Area
Circle Area
Sector Area
64
(8)
r2 2
Sector Area =
=
=
Sector Area =
4800
360o
Sector Area =
40
3
Arc Measure
360o
75o
360o
75o (64)
360o
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.
If K is on segment CD
P(K is on segment CD) =
P stands for
probability of
the inside of
parentheses
happening.
CD
AZ
If K is in area M
P(K is in area M) =
Area M
Area J
Any time that
probability is calculated
using geometric
measures such as
length and area, you
are finding the
geometric
probability of the
event occurring.
Example 9
• Find the probability that a point
randomly chosen is on line
segment WX.
V
0
W
X
1
P(K is on segment WX) =
WX
VZ
P(K is on segment WX) =
3
11
Y
Z
Example 10
• Find the probability that a point
randomly chosen lies inside the
circle.
s = 10
Area Circle
 r2
P(K is in the circle) =
=
Area Square
s2
=
252
(5)
100
102
≈ 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