Lesson 6.5

Chapter 6,
continued
Lesson 6.5
4. Shortest sides
10
AB
2
}5}5}
DE
5
25
6.5 Guided Practice (pp. 389– 391)
1. Compare nLMN and nRST:
Shortest sides
LM
RS
20
24
5
6
}5}5}
Remaining sides
CA
20
2
}5}5}
FD
5
50
}5}5}
BC
EF
16
40
2
5
Longest sides
Remaining sides
All of the ratios are equal, so n ABC , nDEF by the
SSS Similarity Theorem. The scale factor of n ABC to
LN
ST
}5}5}
MN
RT
nDEF is }5.
26
33
}5}
24
30
4
5
The ratios are not all equal, so nLMN and nRST are
not similar.
Compare nLMN and nXYZ:
Shortest sides
Longest sides
Remaining sides
20
2
LM
}5}5}
30
3
YZ
LN
26
2
}5}5}
XY
39
3
}5}5}
MN
XZ
24
36
2
3
2
5. Compare n ABC and nJKL:
Shortest sides
Longest sides
Remaining sides
7
AB
}5}
JK
6
AC
12
}5}
JL
11
}5}
BC
KL
8
7
The ratios are not all equal, so n ABC and nJKL are
not similar.
All of the ratios are equal, so nLMN , nYZX.
Compare n ABC and nRST:
Because nLMN , nXYZ and nLMN is not similar to
nRST, nXYZ is not similar to nRST.
Shortest sides
Longest sides
Remaining sides
7
AB
2
}5}5}
3.5
1
RS
AC
12
2
}5}5}
RT
6
1
}5}5}
2. Scale factor:
24
12
Longest sides
33
x
2
1
}5}
Remaining sides
30
y
2
1
2
1
Shortest sides
Longest sides
Remaining sides
15 5 y
BC
14
4
}5}5}
KL
5
17.5
AC
20
4
}5}5}
JL
5
25
}5}5}
Shorter sides
Longer sides
SR
24
4
}5}5}
PN
18
3
28
RT
4
}5}5}
21
3
NQ
6. Compare n ABC and nJKL:
Ratios of the lengths of corresponding sides:
Shortest sides
Longest sides
Remaining sides
XZ
YZ
}5}5}
XW
XY
}5}5}
4
3
WZ
XZ
16
12
16
20
4
5
Compare n ABC and n RST:
4. Sample answer:
20
15
AB
JK
All of the ratios are equal, so n ABC , n JKL.
The lengths of the sides that include Ž R and Ž N are
proportional. So, nSRT , nPNQ by the SAS Similarity
Theorem.
4
3
Corresponding side lengths are proportional, so
nXZW , nYZX by the SSS Similarity Theorem.
Shortest sides
Longest sides
Remaining sides
BC
14
4
}5 } 5 }
10.5
3
ST
AC
20
5
}5}5}
RT
16
4
}5}5}
AB
RS
16
12
4
3
The ratios are not all equal, so n ABC and nRST are
not similar.
7. Both ŽW and Ž D are right angles, so Ž W > Ž D. Ratios
of the lengths of the sides that include Ž W and Ž D:
Shorter sides
Longer sides
WY
6
2
}5}5}
DE
9
3
}5}5}
XW
FD
10
15
2
3
The length of the sides that include Ž W and Ž D are
proportional. So, by the SAS Similarity Theorem,
nWXY , nDFE. The scale factor of Triangle B to
2
Triangle A is }3.
8. Both mŽ L and mŽ T 5 1128, so Ž L > Ž T. Ratios of
6.5 Exercises (p. 391–395)
the lengths of the sides that include Ž L and Ž T:
Skill Practice
1. Corresponding side lengths must be proportional, so
AC
CB
AB
} 5 } 5 }.
PX
XQ
PQ
2. You would need to know that one pair of corresponding
sides is congruent. You could then use the SAS
Congruence Postulate.
3. Shortest sides
AC
3
12
}5}5}
DF
8
2
All of the ratios are equal, so n ABC , nRST.
30 5 2y
Ratios of the lengths of the sides that include
Ž R and Ž N:
4
3
2
1
33 5 2x
The lengths of the other sides are 16.5 and 15.
12
9
8
4
}5}
3. Both Ž R and Ž N are right angles, so Ž R > Ž N.
}5}5}
BC
ST
}5}
16.5 5 x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Longest sides
Longest sides
Remaining sides
RC
18
3
}5}5}
EF
12
2
}5}5}
AB
DE
15
10
Shorter sides
Longer sides
10
5
KL
}5}5}
8
4
ST
}5}5}
JL
RT
24
18
4
3
Because the lengths of the sides that include Ž L and
Ž T are not proportional, Triangle B is not similar to
Triangle A.
3
2
All of the ratios are equal, so n ABC , nDEF by the
SSS Similarity Theorem. The scale factor of n ABC to
3
nDEF is }2.
Geometry
Worked-Out Solution Key
171
Chapter 6,
continued
9.
15.
X
Z
X
4(n 1 1)
668
348
Y
N
P
808
4
5
Q
7n 2 1
Z
Y
PQ
XY
QR
YZ
5
7n 2 1
}5}
16.
T
32
168
4(7n 2 1) 5 20(n 1 1)
S
28n 2 4 5 20n 1 20
20
48
n53
248
H
10. Ratios of the lengths of the corresponding sides:
Shortest sides
15
15 1 5
15
20
GJ
FK
3
4
18
24
16.5
22
3
4
Shorter sides
Longer sides
BC
3
21
}5}5}
14
2
CE
}5}5}
AC
CD
27
18
3
2
The corresponding side lengths are proportional. The
included angles Ž ACB and Ž DCE are congruent because
they are vertical angles. So, n ACB , nDCE by the SAS
Similarity Theorem.
12. Ratios of the lengths of the corresponding sides:
21
35
3
5
16
x
48
16
15
x 5 51.2
So, RT would be the longest side of nRST, but this
cannot be true because RT is not opposite the largest
angle. So, the triangles cannot be similar.
11. Ratios of the lengths of the corresponding sides:
XY
DJ
RT
16
}5}
All the ratios are equal, so the corresponding side lengths
are proportional. So, nGHJ , nFHK by the SSS
Similarity Theorem.
}5}5}
32
5}
Find RT: }
FH
15
}5}5}5}
Shorter sides
ST
= }, mŽ T 5 248, and mŽ R 5 1408.
5}
would be }
30 15
GH
3
4
}5}5}
Remaining sides
16.5
16.5 1 5.5
G
30
If nRST and nFGH were similar, then the scale factor
Longest sides
}5}5}5}
HJ
HK
F
R
8n 5 24
GH
FH
M
Because mŽ Y and mŽ M both equal 348, Ž Y > Ž M.
By the Triangle Sum Theorem, 668 1 348 1 mŽ Z 5 1808
so, mŽ Z 5 1808. Therefore, Ž Z and Ž N are congruent.
So, nXYZ , nLMN by the AA Similarity Postulate.
}5}
4
4(n 1 1)
348
L
Find the value of n that makes corresponding side lengths
proportional.
Longer sides
XZ
DG
30
50
3
5
}5}5}
17.
E
25
15
7x
D
F
B
24
A
8x
54
8
AB
24
}5}5}
DE
15
3
C
BC
8x
}5}
EF
25
n ABC , nRQP by the SAS Similarity Theorem
MN
MP
14. D; Because } 5 } and Ž M > Ž M,
MR
MQ
nMNP , nMRQ by the SAS Similarity Theorem. The
correct answer is D.
54
7x
8x
54
5}
If n ABC and nDEF are similar, }
7x
25
The corresponding side lengths are proportional.
Both mŽ X and mŽ L equal 478, so Ž X > Ž L. So,
nXYZ , nDJG by the SAS Similarity Theorem.
13. The student named the triangles incorrectly.
AC
DF
}5}
56x 2 5 1350
x ø 4.9.
8x
54
8
and }
do not equal }3 when x ø 4.9. So, the
But }
7x
25
triangles cannot be similar.
18. Because Ž LSN and Ž QRN are supplementary,
@##$ i @##$
LM
PQ by the Consecutive Interior Angles Converse.
So mŽ NLM 5 mŽ NQP 5 538 by the Alternate Interior
Angles Theorem.
19. Because Ž LSN and Ž QRN are supplementary,
@##$ i @##$
LM
PQ by the Consecutive Interior Angles Converse.
So mŽ QPN 5 mŽ LMN 5 458 by the Alternate Interior
Angles Theorem.
172
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
R
Chapter 6,
continued
20. By the Triangle Sum Theorem, 538 1 458 1 mŽ PNQ
5 1808, so mŽ PNQ 5 828.
28. AA Similarity Postulate; You know Ž A > Ž A by the
} }
} }
Reflexive Property. Because BG i CF and CF i DE,
you know that Ž ABG > Ž ACF > Ž ADE and
Ž AGB > Ž AFC > Ž AED by the Corresponding Angles
Postulate. So, n ABG , n ACF, n ABG , n ADE,
n ACF , n ADE by the AA Similarity Postulate.
21. nLSN , nQRN by the AA Similarity Postulate, so
QR
LS
RN
SN
18
12
RN
16
Problem Solving
}5}
}5}
29. Compare the first piece to the second piece:
24 5 RN.
22. nLMN , nQPN by the AA Similarity Postulate, so
PQ
ML
RN
SN
PQ
28
24
16
}5}
Remaining sides
5
}
7
}
3
4
Compare the second piece to the third piece:
PQ 5 42.
23. LM 5 LS 1 SM
PQ 5 PR 1 RQ
28 5 12 1 SM
42 5 PR 1 18
16 5 SM
24 5 PR
Longest sides
Remaining sides
4
}
3
7
4
}5}
5.25
3
}
4
3
The second and third pieces are similar, but the first and
second pieces are not similar, so the first and third pieces
are not similar.
NM
NP
}5}
NM
16
Shortest sides
All of the ratios are equal, so the second and third pieces
are similar.
nMSN , nPRN by the AA Similarity Postulate, so
}5}
}
24
24Ï 2
DC
30. You need to know } is also equal to the other two ratios
EC
}
16Ï 2 5 NM.
of corresponding side lengths.
24. nLSN , nQRN, nMSN , nPRN, and
31. You need to know that the included angles are congruent,
VX
51
3
25. Scale factor of nVWX , n ABC: } 5 } 5 }
34
2
AC
32.
nLMN , nQPN, by the AA Similarity Postulate.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Longest sides
3
}
4
The ratios are not all equal, so the first and second pieces
are not similar.
}5}
SM
RP
Shortest sides
WY
BD
3
YX
26. Find bases: } 5 }
2
DC
45
DC
Ž CBD > Ž CAE.
10
3
2
}5}
WY
12
3
2
208
A
3
2
}5}
}5}
DC 5 30
WY 5 18
Use the Pythagorean Theorem to find the height of each
triangle.
(AD)2 1 (DC)2 5 (AC)2
(VY )2 1 (YX )2 5 (VX )2
(AD)2 1 302 5 342
(VY )2 1 452 5 512
(AD)2 5 256
(VY )2 5 576
AD 5 16
Area of nVWX
Area of n ABC
1
2
}
1
} (42)(16)
2
} (63)(24)
756
5
9
5}
5 }4
336
5 } 5 }4 . Notice that }4 5 1 }2 2 .
9ab
4ab
9
9
3 2
F
BC
AB
You can see that }
5}
and Ž A > Ž D, but it is
DE
EF
AC
AB
obvious that }
Þ}
. So, there is no SSA Similarity
DF
DE
Postulate.
33. a. The triangles are similar by the AA Similarity
x in.
66 in.
Sample answer: Let the base and height of nVWX be 3a
and 3b. Let the base and height of n ABC be 2a and 2b.
The ratio of their areas is
9ab
2
}
4ab
}
2
4
(95 1 7) ft
7 ft
}5}
the square of the scale factor.
}
5
208
b. Let x be the height of the tree.
27. Conjecture: In similar triangles, the ratio of the areas is
3a(3b)
2
}
2a(2b)
}
2
C
Postulate.
VY 5 24
Ratios of areas: }} 5
8
E
D
So, BC 5 12 1 30 5 42 and XW 5 45 1 18 5 63.
}
B
7x 5 6732
x ø 962
The height of the tree is about 962 inches, or about
80 feet.
c. Let x be the distance from Curtis to the tree.
962 in.
75 in.
(6 1 x) ft
6 ft
}5}
5772 5 450 1 75x
70.96 5 x
Curtis is about 71 feet from the tree.
Geometry
Worked-Out Solution Key
173
Chapter 6,
continued
Also notice that n ADC , nEFC by the AA Similarity
Postulate because Ž A > Ž A and Ž ADC > Ž EFC. So,
you can write the proportion:
34. a. Using the Pythagorean Theorem:
a2 1 b2 5 c2
a2 1 b2 5 c2
62 1 b2 5 102
182 1 b2 5 302
b 5 64
b2 5 576
b58
b 5 24
2
8y 1 5y
5y
40
EF
13
5
40
EF
}5}
8
30
18
AD
EF
}5}
10
6
AC
EC
}5}
200
13
} 5 EF
} 200
feet, or about 15.4 feet.
So, the length of EF is }
13
24
8
1
b. } 5 }
24
3
Mixed Review for TAKS
c. Ratios of corresponding side lengths:
Shortest sides
Longest sides
Remaining sides
6
18
}5}
10
30
}5}
1
3
}5}
1
3
8
24
39. C;
22 cm j dimension on blueprint
x jdimension of actual parking lot
48 cm
42 m
}5}
1
3
All of the ratios are equal, so the two triangles are
similar. This suggests a Hypotenuse-Leg Similarity
Theorem for right triangles.
35. Sample answer: Because D, E, and F are midpoints,
}i}
} }
DE AC and EF i AB by the Midsegment Theorem.
So, Ž A > Ž BDE by the Corresponding Angles Postulate
and Ž BDE > Ž DEF by the Alternate Interior Angles
Theorem. Therefore, mŽ DEF 5 908.
48x 5 924
x 5 19.25
The actual width of the parking lot is 19.25 meters.
40. H;
y
1
A(0, 0)
B(4, 0)
1
x
36. Yes; All pairs of corresponding angles are in proportion
AC
GH
AB
GB
AB
DE
AC
DF
AC
GH
AC
DF
C(2, 24)
D(22, 24)
}
Equation of the line that includes AB:
y2 2 y1
2
1
} 5 }. But } 5 } and GB 5 DE, so } 5 }.
y 5 mx 1 b
Therefore, GH 5 DF. Because Ž BGH > Ž EDF,
n GBH > nDEF by the SAS Congruence Postulate. So,
Ž B > Ž E, and n ABC , nDEF by the AA Similarity
Postulate.
0 5 0(0) 1 b
} }
38. Because Ž ADC and Ž BCD are right angles, AD i BC
by the Consecutive Interior Angles Converse. So,
Ž ADE > Ž B and Ž A > Ž ACB by the Alternate Interior
Angles Theorem. Therefore, n AED , nCEB by the AA
Similarity Postulate.
AD
40
8
The scale factor of n AED to n CEB is }
5}
5 }5.
25
CB
Let AE 5 8y, EC 5 5y, DE 5 8x, and BE 5 5x.
1 Notice that the ratios of corresponding side lengths are }85. 2
020
m5}
5}
50
x 2x
420
05b
y50
}
Equation of the line that includes BC:
y2 2 y1
24 2 0
2
1
y 5 mx 1 b
0 5 2(4) 1 b
28 5 b
y 5 2x 2 8
}
Equation of the line that includes CD:
y2 2 y1
24 2 (24)
1
y 5 mx 1 b
24 5 0(2) 1 b
24 5 b
y 5 24
Geometry
Worked-Out Solution Key
0
m5}
5}
5}
50
x 2x
22 2 2
24
2
174
24
m5}
5}
5}
52
x 2x
224
22
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
when the two triangles are similar.
}
37. Sample answer: Locate G on AB so that GB 5 DE.
}
}i}
Draw GH so that GH AC. Because Ž A > Ž BGH and
Ž C > Ž BHG by the Corresponding Angles Postulate,
n ABC , nGBH by the AA Similarity Postulate. So,
Chapter 6,
continued
}
Equation of the line that includes AD:
Lesson 6.6
24 2 0
24
m5}
5}
52
22 2 0
22
Investigating Geometry Activity 6.6 (p. 396)
Explore 1: Answers will vary.
y 5 mx 1 b
Explore 2: Answers will vary.
0 5 2(0) 1 b
1. When two sides of a triangle are cut by a line parallel to
05b
the triangle’s third side, the ratios of the lengths of the
segments formed to the triangle’s third side will be equal.
y 5 2x
The function y 5 2x describes a line that would include
an edge of the parallelogram shown in the diagram.
2. The ratio of the lengths of the two sides of a triangle is
equal to the ratio of the lengths of the segments formed
when an angle bisector is drawn to the third side.
Quiz 6.3–6.5 (p. 395)
6.6 Guided Practice (pp. 398–399)
60
5
AD
1. } 5 } 5 }
KN
36
3
1.
5
The scale factor of ABCD to KLMN is }3.
DC
5
2. Find x: } 5 }
NM
3
5
AB
Find y: }
5 }3
KL
70
5
}5}
x
3
45
5
}5}
y
3
210 5 5x
36
YZ
1260
44
315
11
} 5 YZ
27 5 y
90
9
NP
2. } 5 } 5 }
5
50
PQ
NS
SR
858 5 z8
72
40
85 5 z
}
}
PS is parallel to QR by the Converse of the Triangle
Proportionality Theorem.
5 27 1 10 1 42 1 36 5 115
15 + AB 5 288
5
3
}5}
AB 5 19.2
AB
4. DB 5 DC 5 }
AC
3x 5 575
2
x 5 191 }3
4
2
4. Both Ž P and Ž D are right angles, so Ž P > Ž D. Ratios
of the lengths of the sides that include Ž P and Ž D:
Shorter sides
Longer sides
WP
ZD
}5}5}
10
3
}5}5}
YP
ND
36
12
Longest sides
Remaining sides
AC
XR
}5}5}
CF
RS
}5}5}
4
5
Skill Practice
FA
SX
to one side of a triangle intersects the other two sides,
then it divides the two sides proportionally.
BE
BD
} }
If DE i AC, then }
5}
.
EC
DA
D
Shortest sides
32
40
6.6 Exercises (pp. 400–403)
A
5. Ratios of corresponding side lengths:
4
5
}
4Ï2 5 AB
1. The Triangle Proportionality Theorem: If a line parallel
3
1
Because the lengths of the sides that include Ž P and Ž D
are not proportional, nWPY and nZDN are not similar.
}5}5}
AB
}5}
}
4
4Ï2
The perimeter of ABCD is 191 }3 and the perimeter of
KLMN is 115.
20
25
18
15
AB
16
}5}
3.
Perimeter of ABCD:
Use Theorem 6.1 to find the perimeter x.
30
9
9
5
}5}5}
3. Perimeter of KLMN 5 KL 1 LM 1 MN 1 NK
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
44
35
} 5 YZ
Find z: mŽ A 5 mŽ K
x
115
XY
YZ
}5}
135 5 y + 5
42 5 x
XW
WV
}5}
28
35
4
5
All of the ratios are equal, so n ACF , nXRS by the
SSS Similarity Theorem.
B
E
C
BD BE
5
DA EC
6. Both mŽ M and mŽ J equal 428, so Ž M > Ž J.
Ž LGM > Ž HGJ by the Vertical Angles Congruence
Theorem. So, nLGM > nHGJ by the AA Similarity
Theorem.
Geometry
Worked-Out Solution Key
175