Download Ch 11 Note sheet Key

Ch 11 Note sheet Shortened L1 Key
Algebra Review: Proportion and Reasoning
Name ___________________________
If you don’t understand the notes, read page 578 in the book, including Example A.
To solve a proportion,
use “cross multiplication”.
ratio An expression that compares two quantities by division.
a
or a to b or a : b
b
proportion A statement of equality between two ratios. Proportions
are useful for solving problems involving comparisons.
Ways to write a ratio:
Example A
In a photograph, Robbie is 2.5 inches tall and his
sister Paige is 1.5 inches tall. Robbie’s actual
height is 70 inches. What is Paige’s actual height?
Photo
Paige
then solve.
Example A: Additional Notes
You can set up these proportions any way you want
as long as you match the measures and the units!
For example:
Just One Alternate Set Up:
The Set Up:
Robbie
a c
=
b d
aid = cib
Actual
2.5 in.
1.5 in.
70 in.
? in.
Actual
Photo
Robbie
Paige
70 in.
2.5 in.
? in.
1.5 in.
Now solve the proportion
Now solve the proportion
2.5 70
=
Use cross multiplication …
1.5 x
2.5 x = 105 now divide both sides by 2.5
x = 42
70
x
=
Use cross multiplication …
2.5 1.5
105 = 2.5x See, algebraically it is still the same.
x = 42
Example B (alternate)
Solve the proportion:
3x − 13 23
=
x + 40 10
First realize that the fraction bar will group the terms in the
numerator and the denominator.
( 3x − 13) = 23 Now, cross multiply.
( x + 40 ) 10
10 ( 3x − 13) = 23 ( x + 40 ) You’ll have to distribute!
30 x − 130 = 23x + 920 Add and subtract to solve.
−23x + 130 − 23x + 130 Remember, BOTH sides!
7 x = 1050
Divide both sides by 7.
x = 150
S. Stirling
Additional Example
Since ratios are fractions, you can solve a
proportion simply by making equal
fractions. Remember, to make equivalent
fraction, you must multiply the numerator
and the denominator by the same number.
(You are simply multiplying by 1.)
Example: Solve
2 x
=
7 21
If you multiply the top and bottom by 3,
you get an equal fraction (ratio).
2 3 6
i =
7 3 21
So x = 6
Page 1 of 6
Ch 11 Note sheet Shortened L1 Key
Name ___________________________
Lesson 11.1 Similar Polygons Introduction
similar Identical in shape but proportional in size.
similar polygons Polygons whose corresponding angles are congruent and whose corresponding side
lengths are proportional.
scale factor The ratio of corresponding lengths in similar figures.
The Statue of Liberty’s Nose
Consider this problem:
The Statue of Liberty in New York City has a nose that is 4 feet 6 inches long. What is the approximate
length of one of her arms?
1. Solve the problem. (Hint: Think about your own nose and arms.) Explain how you got your answer
and tell why you think your method works. Give your answer in feet.
4 ft 12 in
i
= 48 in , 48 + 6 = 54 in.
1 1 ft
Nose
Liberty
You
Arm
54 in.
? in.
2.125in. 27.5 in.
54
x
=
2.125 27.5
1485 = 2.125x
x = 698.82 inches
698.82 in 1 ft
i
= 58.24 ft
1
12 in
Scale factor: (Liberty:You)
54
≈ 25.4118
2.125
Means that she is 25.41
times larger than you. In
length, that is.
2. Pick two other body parts and find the approximate length that these parts should be on the Statue of
Liberty. Again, show how you got your answer and explain fully.
S. Stirling
Page 2 of 6
Ch 11 Note sheet Shortened L1 Key
Lesson 11.1 Similar Polygons
Name ___________________________
Read page 581 - 584 in the book, including Investigation 1.
Congruence VS Similar Notes
Definition of Congruence:
Two figures are congruent if they have
1) the same shape: the corresponding angles are
congruent
and
2) the same size: the corresponding sides are
congruent
Definition of Similarity:
Two figures are similar if they have
1) the same shape: the corresponding angles are
congruent
and
2) proportional size: the corresponding sides are
in proportion (When you compare the
corresponding sides from one figure to the other,
they are in the same ratio, scale factor.)
When writing congruent figures, you must match the
corresponding vertices. Use the ≅ symbol.
When writing similar figures, you must match the
corresponding vertices. Use the ~ symbol.
EXAMPLE (alternate)
Given: QUAD ∼ SIML
Find the missing parts:
The quadrilaterals are similar, so
corresponding angles are congruent
QUAD ∼ SIML
m∠Q = m∠S = 75° ,
m∠D = m∠L = 120° ,
m∠U = m∠I = 85° ,
m∠A = m∠M = 360 − (85 + 75 + 120) = 80°
Corresponding sides are in proportion
QU:SI UA:IM AD:ML
QUAD 20
25
AD
SIML
8
IM
ML
DQ:LS
Scale
13
LS
5
2
Set up (include scale factor)
20 5 25 13
= =
=
8 2 IM LS
Solve:
SL = 5.2
MI = 10
Scale factor = 5:2
5 5 25
i =
so IM = 10
2 5 IM
Solve:
5 13
26
=
≈ 5.2
then 5i LS = 26 , LS =
2 LS
5
m∠D = 120
m∠U = 85
m∠A = 80
S. Stirling
Can you find AD or ML? Why or Why not?
You would need one of the measures. You only know
that they would be in a 20:8 or 5:2 ratio.
Page 3 of 6
Ch 11 Note sheet Shortened L1 Key
Name ___________________________
EXAMPLE (alternate)
Are the triangles similar?
The corresponding angles are congruent
m∠A = m∠A same angle
m∠ADE = m∠DBC DE BC so corresponding angles equal.
m∠AED = m∠ECB same reason
The corresponding sides are in proportion
AB:AD BC:DE CA:EA
ABC
4
8
6
ADE
2
4
3
See if they all equal the scale factor (or that the cross
products are equal).
2 4 8 6
= = =
1 2 4 3
They all equal the scale factor 2:1!
So ΔABC ∼ ΔADE
Lesson 11.2 Similar Triangles (the short cuts)
Read pages 589 – 591 in the book if necessary.
Remember that to show that two triangles are congruent, by the definition of congruence, you would
need to show that all six corresponding parts are congruent. So, we investigated and found 4 congruence
short cuts. There are 3 shortcuts for showing that two triangles are similar.
Congruence Short Cuts:
ASA Congruence If two angles of one triangle
and the included side are congruent to two
angles and the included side of another
triangle, then the two triangles are congruent.
AAS Congruence If two angles of one triangle
and an non-included side are congruent to two
angles and the non-included side of another
triangle, then the two triangles are congruent.
SAS Congruence If two sides of one triangle and
the included angle are congruent to two sides
and the included angle of another triangle, then
the two triangles are congruent.
SSS Congruence If three sides of one triangle are
congruent to three sides of another triangle,
then the two triangles are congruent.
Similarity Short Cuts:
AA Similarity Postulate If two angles of one
triangle are congruent to two angles of
another triangle, then the two triangles are
similar.
SAS Similarity Theorem If two sides of one
triangle are proportional to two sides of
another triangle and the included angles
are congruent, then the triangles are
similar.
SSS Similarity Theorem If the three sides of
one triangle are proportional to the three
sides of another triangle, then the two
triangles are similar.
Remember that the “included” part is important! The only difference between congruence and similarity
properties is that the corresponding sides are proportional (not congruent).
S. Stirling
Page 4 of 6
Ch 11 Note sheet Shortened L1 Key
Name ___________________________
EXAMPLE 1
EXAMPLE 2
Is ΔXYO ∼ ΔMKO ?
Are the triangles similar? If so, explain why
they are similar.
Shoot for SAS Similarity since m∠O = m∠O .
See if the corresponding sides are in proportion.
OX:OM OY:OK
XYO
30
70
48
108
Since the lines are parallel, the alternate
interior angles are congruent.
m∠A = m∠E
m∠B = m∠D
MKO
So ΔABC ∼ ΔEDC by AA Similarity.
ratios are not equal, the triangles are not similar.
Simplify each ratio:
OR
Remember to match corresponding vertices!
Cross multiply
EXAMPLE 3
30 3
48 4
= and
= , since the
70 7
108 9
30 48
=
, 3240 ≠ 3360 .
70 108
Since you are not given any sides, shoot for
SSS Similarity
Match corresponding sides
BA:JO AK:OL
BAK
60
38
JOL
50
28
KB:LJ
36
26
See if they all equal the scale factor (or
that the cross products are equal).
60 6
38 19
=
=
50 5
28 14
So the triangles are not similar.
OR
60 38
≠ , 1680 ≠ 1900
50 28
S. Stirling
Page 5 of 6
Ch 11 Note sheet Shortened L1 Key
Name ___________________________
Lesson 11.3 Indirect Measurement with Similar Triangles
Read pages 598 – 599 in the book.
indirect measurement Finding a distance or length by using properties of similar triangles or trigonometry.
Typical problems include shadows and mirrors.
EXAMPLE 1
At a certain time of day, a 6 ft man casts a 4 ft shadow. At the same time of day, how tall is a tree
that casts an 18 ft shadow?
Make a drawing:
Since the triangles will be similar.
Height
Shadow
Man
Tree
6 ft
4 ft
? ft
18 ft
Solve the proportion:
6 x
=
4 18
4 x = 108
x = 27
So the tree is 27 feet tall.
EXAMPLE 2
Samantha has her friend Sophie place a mirror between her and a tall building so that she can see
the very top of the building in the mirror. Sam is 5 feet tall. She is standing 79 feet from the
building and the mirror is 4 feet in front of her. Find the height of the building.
Make a drawing:
Since the triangles will be similar.
Height
Distance
S. Stirling
Sam
Building
5 ft
4 ft
? ft
75 ft
Solve the proportion:
5 x
=
4 75
4 x = 375
x = 93.75
So the building is 93.75
feet tall.
Page 6 of 6