8.4 Similar Triangles - DBCS Mrs. Marshall
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Transcript 8.4 Similar Triangles - DBCS Mrs. Marshall
8.4 Similar Triangles
Identifying Similar Triangles
In this lesson, you will continue the study
of similar polygons by looking at the
properties of similar triangles.
Ex. 1: Writing Proportionality
Statements
In the diagram,
∆BTW ~ ∆ETC.
a. Write the statement
of proportionality.
b. Find mTEC.
c. Find ET and BE.
T
E
34°
C
3
20
79°
B
12
W
Ex. 1: Writing Proportionality
Statements
In the diagram,
∆BTW ~ ∆ETC.
a. Write the statement
of proportionality.
ET
BT
=
TC
TW
=
T
E
34°
C
3
20
CE
WB
79°
B
12
W
Ex. 1: Writing Proportionality
Statements
In the diagram,
∆BTW ~ ∆ETC.
Find mTEC.
B TEC, SO
mTEC = 79°
b.
T
E
34°
C
3
20
79°
B
12
W
Ex. 1: Writing Proportionality
Statements
In the diagram,
∆BTW ~ ∆ETC.
c. Find ET and BE.
CE
ET
=
WB
BT
Write proportion.
3
ET
=
12
20
Substitute values.
3(20)
= ET
12
5
= ET
T
E
C
3
20
Multiply each side by 20.
Simplify.
34°
79°
B
12
W
Because BE = BT – ET, BE = 20 – 5 = 15. So, ET is 5 units and BE is
15 units.
Postulate 25 Angle-Angle Similarity
Postulate
If two angles of one
triangle are
congruent to the two
angles of another
triangle, then the two
triangles are similar.
If JKL XYZ and
KJL YXZ, then
∆JKL ~ ∆XYZ.
K
L
Y
J
X
Z
Ex. 2: Proving that two triangles are
similar
Color variations in
the tourmaline
crystal shown lie
along the sides of
isosceles triangles.
In the triangles, each
vertex measures
52°. Explain why the
triangles are similar.
Ex. 2: Proving that two triangles are
similar
Solution. Because
the triangles are
isosceles, you can
determine that each
base angle is 64°.
Using the AA
Similarity Postulate,
you can conclude the
triangles are similar.
Ex. 3: Why a Line Has Only One
Slope
Use the properties of
similar triangles to
explain why any two
points on a line can
be used to calculate
slope. Find the
slope of the line
using both pairs of
points shown.
D(6, 6)
6
4
C(4,3)
2
B (2, 0)
E
5
10
-2
A (0, -3)
-4
F
Ex. 3: Why a Line Has Only One
Slope
By the AA Similarity
Postulate, ∆BEC ~
∆AFD, so the ratios
of corresponding
sides are the same.
In particular,
CE
BE
=
DF
AF
By a property of
proportions,
CE
DF
=
BE
AF
D(6, 6)
6
4
C(4,3)
2
B (2, 0)
E
5
10
-2
A (0, -3)
-4
F
Ex. 3: Why a Line Has Only One
Slope
The slope of a line is
the ratio of the
change in y to the
corresponding
change in x. The
ratios
CE a DF
BE n AF
d
Represent the slopes
of BC and AD,
respectively.
D(6, 6)
6
4
C(4,3)
2
B (2, 0)
E
5
10
-2
A (0, -3)
-4
F
Ex. 3: Why a Line Has Only One
Slope
Because the two slopes
are equal, any two
points on a line can be
used to calculate its
slope. You can verify
this with specific values
from the diagram.
3-0 = 3
4-2
2
6-(-3) = 9
6-0
6
Slope of BC
= 3
2
4
C(4,3)
2
B (2, 0)
E
5
10
-2
A (0, -3)
Slope of AD
D(6, 6)
6
-4
F
Ex. 4: Using Similar Triangles
Aerial Photography. Lowlevel photos can be taken
using a remote-controlled
camera suspended from a
blimp. You want to take an
aerial photo that covers a
ground of g of 50 meters.
Use the proportion
f
h
= n
g
To estimate the altitude h that the
blimp should fly at to take the
photo. In the proportion, use f = 8
cm and n = 3 cm. These two
variables are determined by the
type of camera used.
n
f
h
g
Ex. 4: Using Similar Triangles
f
h
= n
g
8cm
h
=
Write proportion.
n
f
3cm
50 m
Substitute values.
3h = 400
Cross product property.
h ≈ 133
Divide each side by 3.
The blimp should fly at an altitude
of about 133 meters to take a
photo that covers a ground
distance of 50 meters.
h
g
Note:
In Lesson 8.3, you learned that the perimeters
of similar polygons are in the same ratio as the
lengths of the corresponding sides. This
concept can be generalized as follows: If two
polygons are similar, then the ratio of any two
corresponding lengths (such as altitudes,
medians, angle bisector segments, and
diagonals) is equal to the scale factor of the
similar polygons.
Ex. 5: Using Scale Factors
Find the length of the altitude QS.
Solution: Find the scale factor of
∆NQP to ∆TQR.
NP
=
TR
12+12
8+8
=
QS
3
=
2
Substitute 6 for QM and solve for
QS to show that QS = 4
12
M
24 = 3
16
2
Now, because the ratio of the
lengths of the altitudes is equal to
the scale factor, you can write the
following equation:
QM
12
N
P
6
Q
R
8
S
8
T