Transcript 11.3 Perimeters and Areas of Similar Polygons Geometry Mrs. Spitz Spring 2006 Objectives/Assignment • Compare perimeters and areas of similar figures. • Use perimeters and areas of.

11.3 Perimeters and Areas
of Similar Polygons
Geometry
Mrs. Spitz
Spring 2006
Objectives/Assignment
• Compare perimeters and areas of similar
figures.
• Use perimeters and areas of similar
figures to solve real-life problems.
• Assignment: pp. 679-680 #1-27 all
Comparing Perimeter and Area
• For any polygon, the perimeter of the
polygon is the sum of the lengths of its
sides and the area of the polygon is the
number of square units contained in its
interior.
Comparing Perimeter and Area
• In lesson 8.3, you learned that if two polygons
are similar, then the ratio of their perimeters is
the same as the ratio of the lengths of their
corresponding sides. In activity 11.3 on pg. 676,
you may have discovered that the ratio of the
areas of two similar polygons is NOT this same
ratio.
Thm 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 is a2:b2
Thm 11.5 continued
Side length of Quad I
Side length of Quad II
kb
a
=
b
ka
I
a2
Area of Quad I
Area of Quad II
II
=
b2
Quad I ~ Quad II
Ex. 1: Finding Ratios of Similar Polygons
•
a.
b.
C
Pentagons ABCDE and
LMNPQ are similar.
Find the ratio (red to
blue) of the perimeters
of the pentagons.
Find the ratio (red to
blue) of the areas of the
pentagons
D
B
5
N
A
E
M
P
10
L
Q
Ex. 1: Solution
a. Find the ratio (red to blue) of the
perimeters of the pentagons.
• The ratios of the lengths of
corresponding sides in the
pentagons is 5:10 or ½ or 1:2.
• The ratio is 1:2. So, the
perimeter of pentagon ABCDE is
half the perimeter of pentagon
LMNPQ.
C
D
B
5
A
E
N
M
P
10
L
Q
Ex. 1: Solution
b. Find the ratio (red to blue) of the
areas of the pentagons.
• Using Theorem 11.5, the ratio of
the areas is 12: 22. Or, 1:4. So,
the area of pentagon ABCDE is
one fourth the area of pentagon
LMNPQ.
C
D
B
5
A
E
N
M
P
10
L
Q
Using perimeter and area in real life
Ex. 2: Using Areas of Similar Figures
Because the ratio of the lengths of the sides of to rectangular
pieces is 1:2, the ratio of the areas of the pieces of paper is
12: 22 or, 1:4
Using perimeter and area in real life
Ex. 2: Using Areas of Similar Figures
Because the cost of the paper should be a function of its area,
the larger piece of paper should cost about 4 times as much,
or $1.68.
Using perimeter and area in real life
Ex. 3: Finding Perimeters and Areas of Similar Polygons
• Octagonal Floors. A trading pit at the
Chicago Board of Trade is in the shape of
a series of octagons. One octagon has a
side length of about 14.25 feet and an
area of about 980.4 square feet. Find the
area of a smaller octagon that has a
perimeter of about 76 feet.
Using perimeter and area in real life
Ex. 3: Solution
• All regular octagons are
similar because all
corresponding angles are
congruent and
corresponding side
lengths are proportional.
• First – Draw and label a
sketch.
Using perimeter and area in real life
Ex. 3: Solution
• FIND the ratio of the side lengths of the
two octagons, which is the same as the
ratio of their perimeters.
perimeter of ABCDEFGH
perimeter of JKLMNPQR
a
=
b

76
8(14.25)
76
=
114
2
=
3
Using perimeter and area in real life
Ex. 3: Solution
• CALCULATE the area of the smaller octagon.
Let A represent the area of the smaller octagon.
The ratio of the areas of the smaller octagon to
the larger is a2:b2 = 22:32, or 4:9.
A
=
980.4
4
Write the proportion.
9
9A = 980.4 • 4
Cross product property.
A = 3921.6
9
A  435.7
Divide each side by 9.
Use a calculator.
The area of
the smaller
octagon is about
435.7 square
feet.
Upcoming
• Quiz after 11.3. There are no other
quizzes for this chapter.
• 11.4 Friday
• 11.5 Monday
• 11.6 Wednesday
• Test Friday, May 12