Download Math 135 Similar Triangles Definition of Similar Triangles ABC ∆ is

Math 135
Similar Triangles
Definition of Similar Triangles
∆ABC is similar to ∆DEF (written ∆ABC ~ ∆DEF ) under the correspondence
A ↔ D, B ↔ E , C ↔ F if and only if:
1) All three pairs of corresponding angles are congruent.
2) All pairs of corresponding sides are proportional.
D
A
B
C
E
F
∆ABC ~ ∆DEF iff ∠A ≅ ∠D : ∠B ≅ ∠E : ∠C ≅ ∠F and
AB BC AC
=
=
DE EF DF
AAA Similarity Postulate
Two triangles are similar if and only if three angles of one triangle are congruent,
respectively, to three angles of the other triangle.
∆ABC ~ ∆DEF iff ∠A ≅ ∠D : ∠B ≅ ∠E : ∠C ≅ ∠F
AA Similarity Theorem
Two triangles are similar if two angles of one triangle are congruent, respectively to two
angles of the other triangle.
In ∆ABC if ∠A ≅ ∠D and ∠B ≅ ∠E , then ∆ABC ~ ∆DEF
Proof:
Given ∠A ≅ ∠D and ∠B ≅ ∠E , show that ∆ABC ~ ∆DEF
D
A
B
C
E
F
Let ∠A ≅ ∠D and ∠B ≅ ∠E ⇒ m∠A = m∠D and m∠B = m∠E
Since the some of the interior angles is 180 degrees, m∠A + m∠B + m∠C = 180° and
m∠D + m∠E + m∠F = 180° .
Use the substitution principle on the second equation with the
values m∠A = m∠D and m∠B = m∠E to get the following equation.
m∠A + m∠B + m∠F = 180° .
Subtracting the two equations gives:
m∠A + m∠B + m∠C = 180°
m∠A + m∠B + m∠C = 180°
− (m∠A + m∠B + m∠F = 180°) ⇒ − m∠A − m∠B − m∠F = −180°
m∠C − m∠F = 0°
m∠C = m∠F
⇒ ∠C ≅ ∠F
Therefore ∆ABC ~ ∆DEF by AAA Similarity Postulate
Example 1
In ∆ABC , DE is parallel to AC . If AB = 10, CE = 4, AC = 8, BD = 6 and BE = x , find
DE and BC .
B
x
6
10
D
E
4
A
8
C
Solution:
6
x
=
10 x + 4
Taking the product of the means and the extremes, get the following:
10 x = 6( x + 4 )
10 x = 6 x + 24
10 x − 6 x = 6 x − 6 x + 24
4 x = 24
4 x 24
=
4
4
⇒ x = 6 ⇒ BE = 6
Find BC
6 BC
=
10
8
10 BC = 48
BC = 4.8
Corollary 6.5
Two right triangles are similar if an acute angle of one triangle is congruent to an acute
angle of the other triangle.
D
A
B
C
F
E
Theorem 6.6 (SAS Similarity Theorem)
Two triangles are similar if two sides are proportional, respectively, to two sides of
another triangle and the angles included between the sides are congruent.
Corollary 6.7 (LL Similarity)
Two right triangles are similar if the legs of one triangle are proportional to respectively
to the legs of the other triangle.
D
A
B
C
E
F
Theorem 6.8 (SSS Similarity Theorem)
Two triangles are similar if three sides of one triangle are proportional to three sides of
the other triangle.
D
A
B
C
E
F
Example 2
Given ∆CAB ~ ∆CED , find the missing measures.
A
6
12
?
D
10
C
B
18
?
E
BC 10
=
12 18
18 ⋅ BC = 12 ⋅ 10
18 ⋅ BC = 120
⇒ BC =
120
2
=6
18
3
CE 6
=
18 12
12 ⋅ CE = 18 ⋅ 6
12 ⋅ CE = 108
⇒ CE =
108
=9
12
Example 3
Given ∆CAB ~ ∆CED , find the missing measures.
A
?
C
10
8
5
D ? B
?
E
Solution:
Let AC = x
10 10 + x
=
5
8
5(10 + x ) = 10 ⋅ 8
50 + 5 x = 80
50 − 50 + 5 x = 80 − 50
5 x = 30
x=
30
= 6 ⇒ AE = 16
5
Use the Pythagorean Theorem to find BE and DB
(BE )2 + 5 2 = 10 2
(BE )2 + 25 = 100
(BE )2 = 75
(BE )2 = 75
(DE )2 + 8 2 = 16 2
(DE )2 + 64 = 256
(DE )2 = 192
(DE )2 = 192
BE = 5 3
DE = 36 ⋅ 3 = 6 3
Therefore, BE = 5 3 and DB = 6 3 − 5 3 = 3
Example 4
Compute the ratios requested in the following pairs of similar figures.
Find the ratios of base: base, height: height, and area; area for the following pair of
similar rectangles.
15"
5"
2"
6"
Bases: 2:6 or 1:3
Heights: 5:15 or 1:3
Areas: 10:90 or 1:9 (See below)
Area of smaller rectangle
A1 = 2 ⋅ 5 = 10
Area of larger rectangle
A1 = 6 ⋅ 15 = 90
Example 5
Show that the area of the larger triangle shown in red is equal to the sum of the smaller
two triangles
C
B
5 cm
3 cm
E
4 cm
Triangle with Hypotenuse = 3
Find the length of the leg using Pythagorean Theorem
a2 + b2 = c2
x 2 + x 2 = 32
2x 2 = 9
9
2
9
3
x=
=
2
2
x2 =
Length of leg =
3
2
1
1 3 3
9
Area1 = bh = ⋅
⋅
=
2
2 2 2 4
Triangle with Hypotenuse = 4
Find the length of the leg using Pythagorean Theore
a2 + b2 = c2
x 2 + x 2 = 42
2 x 2 = 16
16
2
x= 8=2 2
x2 =
Length of leg = 2 2
Area 2 =
1
1
1
bh = ⋅ 2 2 ⋅ 2 2 = ⋅ 8 = 4
2
2
2
Triangle with Hypotenuse = 5
Find the length of the leg using Pythagorean Theore
a2 + b2 = c2
x 2 + x 2 = 52
2 x 2 = 25
25
2
25
5
x=
=
2
2
x2 =
Length of leg =
5
2
1
1 5 5
15
Area3 = bh = ⋅
⋅
=
2
2 2 2
4
Check:
Area1 + Area 2 = Area3
9
25
9 16 25
+4=
⇒ +
=
4
4
4 4
4
Example 6
In the figure shown, ∠A ≅ ∠C , M and N are midpoints of the sides AB and
BC and MNPQ is a rectangle . Show that ∆MQA ≅ ∆NPC .
B
N
M
A
Q
P
C
Statement
∠A ≅ ∠C
M and N are midpoints of the sides
AB and BC
CB ≅ AB
Reason
Given
MQ ≅ NP
Opposite sides of a rectangle are
congruent.
∠AQM and ∠CPN are right angles
The exterior angle of a rectangle are right
angles
If two angles of one triangle are congruent
to two angle of another triangle, then the
remaining angles are congruent.
ASA
∠AMQ ≅ ∠CNP
∆MQA ≅ ∆NPC
∆ABC is a isosceles triangle
Find the area of MNPQ where MN = 14 m and BC = 22 m
B
11
14 m
M
N 22
7
11
j
A
Q
c2 = a2 + b2
112 = 7 2 + b 2
121 = 49 + b 2
73 = b 2 ⇒ b = 72 = 6 2
NP = 6 2
Area = 14 • 6 2 = 84 2 ≈ 118.79 m 2
P
C
Example 7
Explain how the figure can be used to find
a
B
x
D
A
C
1
a
a x
=
x 1
x ⋅ x = a ⋅1
x2= a
x2 = a
x= a
Example 8
In the figure shown A,B,C, and D are midpoints of the sides of parallelogram PQRS.
If the area of ∆AQB = 5 in 2 , find the area of hexagon PABRCD.
B
Q
R
A
C
P
D
S
The parallelogram can be divided into 8 congruent triangles and the area of the hexagon
can be found by adding up six of these congruent triangles. See below:
B
Q
5
T
A
5
P
R
5
5
C
5
5
D
S