Transcript Ch 7 Similarity

Ch 7
Similarity
Lesson 7-1: Using Proportions
1
Lesson 7-1
Using
Proportions
Lesson 5-1: Using Proportions
2
Ratio
Ratio: A ratio is a comparison of two numbers such as a : b.
When writing a ratio, always express it in simplest form.
Example: What is the ratio of AB to CB ?
AB 10
A

CB
6
Now try to reduce the fraction.
10 5

6 3
8
10
D
4.8
 ratio of AB to CB  5:3.
C
Lesson 5-1: Using Proportions
3.6
B
6
3
Example……….
A baseball player goes to bat 348 times and gets 107 hits.
What is the players batting average?
Solution:
Set up a ratio that compares the number of hits to the number
of times he goes to bat. Ratio: 107
348
Convert this fraction to a decimal rounded to three decimal
places.
Decimal: 107  0.307
348
The baseball player’s batting average is 0.307 which means
he is getting approxiamately one hit every three times at bat.
Lesson 5-1: Using Proportions
4
Proportion
Proportion: An equation that states that two ratios are equal.
Terms
a c

b d
First Term
Second Term
Third Term
Fourth Term
To solve a proportion, cross multiply the proportion:
a d c b
Extremes: a and d
Means : b and c
Lesson 5-1: Using Proportions
5
Proportions- examples….
Example 1: Find the value of x.
x (356 3)  1068

2
(84 3)  252
252 x  2136
2136
x
 8.5 ft
252
x
2 ft
356 yards
84 yards
Example 2: Solve the proportion.
8x = 30
x 5

6 8
8•x = 6•5
x = 3.75
8x = 30
8
8
Lesson 5-1: Using Proportions
6
Lesson 7-2
Similar
Polygons
Lesson 5-2: Similar Polygons
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Similar Polygons
Definition: Two polygons are similar if:
1. Corresponding angles are congruent.
2. Corresponding sides are in proportion.
Two polygons are similar if they have the same shape not
necessarily have the same size.
Scale Factor: The scale factor is the ratio between a pair
of corresponding sides.
Lesson 5-2: Similar Polygons
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Naming Similar Polygons
When naming similar polygons, the vertices (angles, sides)
must be named in the corresponding order.
If ABCD
PQRS
A  P ; B  Q ; C  R ; D  S
P
AB BC CD AD



PQ QR RS PS
A
D
Q
B
C
S
Lesson 5-2: Similar Polygons
R
9
Example-
A
15
The two polygons are similar.
20
Solve for x, y and z.
B
E
D
x
H
y
30
10
C
5
F
z
G
AD DC BC AB
Step1: Write the proportion of the sides. EH  HG  FG  EF
Step 2: Replace the proportion with values.15  y  30  20
x 5 z 10
Step 3: Find the scale factor between the two polygons.
Note: The scale factor has the larger quadrilateral in the numerator and
the smaller quadrilateral in the denominator.
Step 4: Write separate proportions for each missing side and solve.
15
2
y
2
30
2

 x  7.5

 y  10
  z  15
x
1
5
1
z
1
Lesson 5-2: Similar Polygons
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If ABC ~ ZYX, find the scale factor from
Example: ABC to ZYX.
Scale factor is same as the ratio of the sides. Always put the
first polygon mentioned in the numerator.
C 10 B
5
AB 18 2
Y
X


ZY
9 1
14
18 7
9
The scale factor from
Z
ABC to ZYX is 2/1.
A
What is the scale factor from ZYX to ABC? ½
Lesson 5-2: Similar Polygons
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Lesson 7-3
Proving Triangles
Similar
(AA, SSS, SAS)
Lesson 5-3: Proving Triangles
Similar
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AA Similarity (Angle-Angle)
If 2 angles of one triangle are congruent to 2 angles of
another triangle, then the triangles are similar.
E
B
A
C
Given:
Conclusion:
D
A  D
F
and
B  E
ABC ~ DEF
Lesson 5-3: Proving Triangles
Similar
13
SSS Similarity (Side-Side-Side)
If the measures of the corresponding sides of 2 triangles
are proportional, then the triangles are similar.
E
B
5
A
Given:
10
8
11
C
D
AB BC
AC


DE
EF
DF
Conclusion:
16
22
F
8
5
11


16
22
10
ABC ~ DEF
Lesson 5-3: Proving Triangles
Similar
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SAS Similarity (Side-Angle-Side)
If the measures of 2 sides of a triangle are proportional to the
measures of 2 corresponding sides of another triangle and the angles
between them are congruent, then the triangles are similar.
E
B
5
A
10
11
C
D
22
F
AB AC
Given: A  D and

DE DF
Conclusion:
ABC ~ DEF
Lesson 5-3: Proving Triangles
Similar
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Proving Triangles Similar
Similarity is reflexive, symmetric, and transitive.
Steps for proving triangles similar:
1. Mark the Given.
2. Mark …
Shared Angles or Vertical Angles
3. Choose a Method. (AA, SSS , SAS)
Think about what you need for the chosen method and
be sure to include those parts in the proof.
Lesson 5-3: Proving Triangles
Similar
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Given : DE FG
Pr ove : DEC
FGC
Problem #1
Step 1: Mark the given … and what it implies
Step 2: Mark the vertical angles
Step 3: Choose a method: (AA,SSS,SAS)
Step 4: List the Parts in the order of the method with reasons
Step 5: Is there more?
Statements
Reasons
G
Given
1. DE FG
AA
D
2. D  F
C
E
F
Alternate Interior <s
3. E  G Alternate Interior <s
4. DEC FGC AA Similarity
Lesson 5-3: Proving Triangles
Similar
17
Problem
Given : IJ  3LN JK  3NP
#2 Pr ove : IJK LNP
IK  3LP
Step 1: Mark the given … and what it implies
Step 2: Choose a method: (AA,SSS,SAS)
Step 4: List the Parts in the order of the method with reasons
Statements
Reasons
Step 5: Is there more?
1. IJ = 3LN ; JK = 3NP ; IK = 3LP
Given
SSS
J
K
N
I
L
IJ
JK
IK
2.
=3,
=3,
=3
LN
NP
LP
P
IJ
JK
IK
3.
=
=
LN NP LP
4.
IJK~
LNP
Lesson 5-3: Proving Triangles
Similar
Division Property
Substitution
SSS Similarity
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Given : G is the midpo int of ED
Problem #3
H is the midpo int of EF
Pr ove : EGH
EDF
Step 1: Mark the given … and what it implies
Step 2: Mark the reflexive angles
SAS
Step 3: Choose a method: (AA,SSS,SAS)
Step 4: List the Parts in the order of the method with reasons
Next Slide………….
E
Step 5: Is there more?
G
H
D
Lesson 5-3: Proving Triangles
Similar
F
19
Statements
Reasons
1.
G is the Midpoint of ED
Given
H is the Midpoint of EF
2. EG = DG and EH = HF
Def. of Midpoint
3. ED = EG + GD and EF = EH + HF Segment Addition Post.
4. ED = 2 EG and EF = 2 EH
Substitution
ED
EF
Division Property
5.
EG
=2 and
EH
=2
Substitution
ED EF
6.
=
EG EH
7. GEHDEF
Reflexive Property
8. EGH~ EDF
SAS Postulate
Lesson 5-3: Proving Triangles
Similar
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Lesson 7-4
Proportional
Parts
Lesson 5-4: Proportional Parts
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Similar Polygons
Two polygons are similar if and only if their corresponding angles
are congruent and the measures of their corresponding sides are
proportional.
AB AC BC


DE DF EF
F
C
A
B
D
E
Lesson 5-4: Proportional Parts
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Side Splitter Theorem
If a line is parallel to one side of a triangle and intersects the other
two sides in two distinct points, then it separates these sides into
C
segments of proportional length.
If BD AE , then
CB CD

BA DE
B 1
Converse:
If a line intersects two sides of a triangle and
A 4
separates the sides into corresponding segments
of proportional lengths, then the line is parallel
to the third side.
CB CD
If
BA

DE
Lesson 5-4: Proportional Parts
2 D
3 E
, then BD AE
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Examples………
Example 1: If BE = 6, EA = 4, and BD = 9, find DC.
6 9
9

4 x
D
6x = 36
x
x=6
C
Example 2: Solve for x.
2x + 3
E
5
A
B
4x + 3
D
9
C
Lesson 5-4: Proportional Parts
B
6
E
4
A
2x  3 4x  3

5
9
5(4 x  3)  9(2 x  3)
20 x  15  18 x  27
2 x  12
x6
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Midsegment Theorem
A segment that joins the midpoints of two sides of a triangle is
parallel to the third side of the triangle, and its length is one-half
the length of the third side.
R
If L is the midpo int of RS and
M is the midpo int of RT then
1
LM ST and ML  ST .
2
M
L
T
S
Lesson 5-4: Proportional Parts
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Extension of Side Splitter
If three or more parallel lines have two transversals, they cut off the
transversals proportionally.
If three or more parallel lines cut off congruent segments on one
transversal, then they cut off congruent segments on every
E
transversal.
D
AB DE AC BC AC DF

,

,

, etc.
BC EF DF EF
BC EF
Lesson 5-4: Proportional Parts
A
B
C
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F
Forgotten Theorem
An angle bisector in a triangle separates the opposite side into
segments that have the same ratio as the other two sides.
AD AC
If CD is the bi sec tor of ACB, then

DB BC
C
A
D
Lesson 5-4: Proportional Parts
B
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If two triangles are similar:
(1) then the perimeters are proportional to the measures of the
corresponding sides.
(2) then the measures of the corresponding altitudes are proportional
to the measure of the corresponding sides..
(3) then the measures of the corresponding angle bisectors of the
triangles are proportional to the measures of the corresponding sides..
A
B
G
H
AB BC AC Perimeter of



DE EF DF Perimeter of
D
AG (altitude of

DI (altitude of
F
E
C
J
I
AH (anglebi sec tor of

ABC ~ DEF
DJ (anglebi sec tor of
Lesson 5-4: Proportional Parts
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ABC
DEF
ABC )
DEF )
ABC )
DEF )
Example:
Given: ΔABC ~ ΔDEF, AB = 15, AC = 20, BC = 25, and DF = 4.
Find the perimeter of ΔDEF.
The perimeter of ΔABC is 15 + 20 + 25 = 60.
Side DF corresponds to side AC, so we can set up a proportion as:
AC Perimeter of

DF Perimeter of
20 60

4
x
20 x  240
x  12
ABC
DEF
E
B
25
15
A
D
F
4
C
20
Lesson 5-4: Proportional Parts
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