# Lesson 6.4

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Lesson 6.4
```Chapter 6,
continued
Lesson 6.4
6.4 Guided Practice (pp. 382–383)
1. Because nFGH and nQRS are equiangular, all
6.4 Activity (p. 381)
angles measure 608. So, all angles are congruent and
nFGH , nQRS by the AA Similarity Postulate.
F
2. Because they are both right angles,  DFC and  EFD
T
40° 50°
E
are congruent. By the Triangle Sum Theorem,
328 1 908 1 m CDF 5 1808, so m CDF 5 588.
Therefore,  CDF and  DEF are congruent.
G
40°
R
50°
S
So, nCDF , nDEF by the AA Similarity Postulate.
1808 5 m F 1 m F 1 m G
3. Yes; if  S >  T (or  R >  U ), then the triangles are
1808 5 408 1 m F 1 508
similar by the AA Similarity Postulate.
908 5 m F
4.
EF ø 15 mm
FG ø 13 mm
72 in.
58 in.
GE ø 20 mm
1808 5 m R 1 m S 1 m T
x in.
1808 5 408 1 m S 1 508
72 in.
58 in.
72x 5 3132
RS ø 22.5 mm
x 5 43.5
ST ø 19.5 mm
The child’s shadow is 43.5 inches long.
TR ø 30 mm
Tree height
5. } 5 }}
15
22.5
FG
ST
2
3
}ø}5}
13
19.5
2
3
}ø}5}
GE
TR
20
30
2
3
}ø}5}
Corresponding angles are congruent and corresponding
side lengths are proportional, so the triangles are similar.
2. No; the corresponding sides of two similar triangles are
S
proportional, so they are not necessarily congruent.
G
3. n ABC , nFED by the AA Similarity Postulate.
40°
R
70°
40°
AC
CB
BA
4. } 5 } 5 } because the ratios of corresponding side
EF
FD
DE
T
1808 5 m E 1 m F 1 m G
lengths in similar triangles are equal.
1808 5 708 1 m F 1 408
y
AC
25
BA
5. } 5 } because } 5 } .
FD
EF
15
12
708 5 m F
EF ø 18 mm
15
18
FD
DE
6. } 5 } because } 5 }.
25
x
AC
CB
FG ø 27 mm
y
25
7. y 5 20; } 5 }
15
12
GE ø 27 mm
1808 5 m R 1 m S 1 m T
1808 5 708 1 m S 1 408
708 5 m S
TR ø 45 mm
18
30
3
5
15y 5 300
15x 5 450
y 5 20
x 5 30
congruent.
ST ø 45 mm
EF
RS
15
18
8. x 5 30; } 5 }
25
x
9. Because they are both right angles,  H and  S are
RS ø 30 mm
}ø}5}
Skill Practice
of another triangle, then the triangles are similar.
F
70°
6.4 Exercises (pp. 384–387)
1. If two angles of one triangle are congruent to two angles
E
54 in.
x in.
}5}
908 5 m S
EF
RS
54 in.
FG
ST
27
45
3
5
}ø}5}
GE
TR
27
45
3
5
}ø}5}
So, the triangles are similar.
Conjecture: Two triangles with two pairs of congruent
corresponding angles are similar.
By the Triangle Sum Theorem, 488 1 908 1 m F 5 1808,
so m F 5 428. Therefore,  F and  K are congruent.
So, nFGH , nKLJ by the AA Similarity Postulate
10. Because m YNM and m YZX both equal 458,
 YNM >  YZX. By the Vertical Angles Congruence
Theorem,  NYM >  ZYX. So, nYNM ,nYZX by the
AA Similarity Postulate.
Geometry
Worked-Out Solution Key
167
Chapter 6,
continued
11. By the Triangle Sum Theorem, 358 1 858 1 m R 5
1808 and 358 1 658 1 m V 5 1808. So, mR 5 608 and
mV 5 808.
Find BD: BD 5 BE 1 DE
20
5}
15
3
Corresponding angles are not congruent, so the triangles
are not similar.
12. Because m EAC and m DBC both equal 658,
 EAC >  DBC. By the Reﬂexive Property,  C >  C.
So, nACE , nBCD by the AA Similarity Postulate.
13. By the Reﬂexive Property,  Y >  Y. By the Triangle
Sum Theorem, 458 1 858 1 m YZX 5 1808, so
m YZX 5 508. Therefore,  YZX and  YWU are
congruent.
So, nYZX , nYWU by the AA Similarity Postulate.
35
5}
3
35
, the correct answer is A.
Because BD 5 }
3
}
21. length of AB 5 4 2 0 5 4
}
length of AD 5 5 2 0 5 5
}
length of AC 5 8 2 0 5 8
AB
AE
AC
5
4
AE
8
}5}
}5}
14. By the Reﬂexive Property,  N >  N. By the
Corresponding Angles Postulate,  NMP >  NLQ.
So, nNMP ,nNLQ by the AA Similarity Postulate.
15. The AA Similarity Postulate is for triangles, not other
polygons.
p
24
16. B; } 5 }
12
10
240 5 12p
AB
AE
AC
7
3
AE
4
}5}
20 5 p
The length of p is 20, so the correct answer is B.
17. The proportion is incorrect because 5 is not the length of
the corresponding side of the larger triangle.
4
9
Sample answer: A correct proportion is }6 5 }x .
28
AE 5 }
3
28
} 28
, so the coordinates are E 1 }
,0 .
The length of AE is }
3
3 2
}
3 cm
The sketch shows that corresponding side lengths are not
proportional.
9cm
4cm
4cm
6cm
3 cm
The sketch shows that corresponding side lengths are not
proportional.
AB
AE
AC
4
1
AE
6
}5}
}5}
24 5 AE
}
The length of AE is 24, so the coordinates are E(24, 0).
}
24. length of AB 5 6 2 0 5 6
}
length of AD 5 9 2 0 5 9
}
length of AC 5 3 2 0 5 3
AB
AE
AC
9
6
AE
3
}5}
CE
DE
20. A; Find x: } 5 }
BE
AE
3
4
}5}
9
2
} 5 AE
5
x
}5}
3x 5 20
20
x5}
3
9
} 9
The length of AE is }2, so the coordinates are E 1 }2 , 0 2.
25. a.
A
8
B
6
E
0
1
D
168
Geometry
Worked-Out Solution Key
5
1
C
}
length of AD 5 4 2 0 5 4
}
length of AC 5 6 2 0 5 6
2 cm
3 cm
}5}
23. length of AB 5 1 2 0 5 1
2 cm
2 cm
10 5 AE
}
The length of AE is 10, so the coordinates are E(10, 0).
}
22. length of AB 5 3 2 0 5 3
}
length of AD 5 7 2 0 5 7
}
length of AC 5 4 2 0 5 4
Chapter 6,
continued
Problem Solving
 AEB >  CED by the Vertical Angles Congruence
Theorem.
31. The triangles shown in the diagram are similar by the
AA Similarity Postulate, so you can write the following
proportion.
 ABE >  CDE by the Alternate Interior Angles
Theorem.
20 in.
d in.
c. n AEB is similar to nCED. n AEB , nCED by the
800 5 26d
AA Similarity Postulate.
BE
AE
d. } 5 }
DE
CE
6
15
BE
10
30.8 ø d
BA
AE
}5}
DC
CE
8
DC
}5}
6
15
The distance between the puck and the wall when the
opponent returns it is about 30.8 inches.
}5}
BE 5 4
20 5 DC
32. a. You can use the AA Similarity Postulate to show that
the triangles are similar because you can show that
two angles of nXYZ are congruent to two angles of
nXVW.
26. Yes; Because m J and m X both equal 718,
 J >  X. By the Triangle Sum Theorem,
718 1 528 1 m L 5 1808, so m L 5 578. Therefore,
 L and  Z are congruent. So, nJKL , nXYZ by the
AA Similarity Postulate.
b.
WX
ZX
WV
ZY
xm
6m
104 m
8m
}5}
}5}
27. Yes; If m X 5 908, m Y 5 608, and nJKL contains
8x 5 624
a 608 angle, then the triangles are similar by the AA
Similarity Postulate.
x 5 78
28. No; Because m J 5 878, nXYZ needs to have an 878
The width of the lake is 78 meters.
angle in order for it to be possible that nJKL and nXYZ
are similar. This is not possible because m Y 5 948,
and the sum of 948 and 878 is 1818, which contradicts the
Triangle Sum Theorem.
c.
XY
VX
ZY
WV
}5}
10 m
8m
104 m
}
xm 5}
29. No; By the Triangle Sum Theorem, 858 1 m L 5 1808,
and m X 1 808 5 1808, so m L 5 958 and m X 5
1008. So, nXYZ needs to have a 958 angle for it to be
possible that nJKL and nXYZ are similar. This is not
possible because m X 5 1008, and the sum of 958
and 1008 is 1958, which contradicts the Triangle Sum
Theorem.
26 in.
(66 2 26) in.
}5}
1040 5 8x
130 5 x
So, VX is 130 meters.
33. All equilateral triangles have the same angle
measurements, 608. So, all equilateral triangles are
similar by the AA Similarity Postulate.
30. Because  P >  P by the Reﬂexive Property and
E
 PST >  R by the Corresponding Angles Postulate.
nPST , nPRQ by the AA Similarity Postulate.
B
Because the triangles are similar, you can set up the
following proportion:
PT
PQ
PS
PR
PT
PQ
PS
PS 1 SR
A
}5}
}5}
x
a
}5}
8
3x
a 1 }3x
x1 a 1 }3 x 2 5 3ax
8
8
ax 1 }3 x 2 5 3ax
8
} x 2 5 2ax
3
C
D
F
m A 5 m B 5 m C 5 m D 5 m E 5 m F5 608,
so  A >  B >  C >  D >  E >  F.
34.
f
h
n
g
}5}
8 cm
hm
3 cm
50 m
}5}
400 5 3h
1
133}3 5 h
1
The blimp should ﬂy at a height of 133}3 meters to take
the photo.
4
3
}x 5 a
4
So, PS 5 }3 x.
Geometry
Worked-Out Solution Key
169
Chapter 6,
continued
U
B
E
V
S
T
A
R
C
N
D
N
Q
P
}
}
Angle bisectors SV and PN are corresponding lengths in
SV
ST
5}
by the Corresponding
similar triangles. So, }
PN
PQ
Lengths Property on page 375.
F
M
}
Let n ABC , nDEF, let BN bisect  ABC and let
}
EM bisect  DEF. Because n ABC , nDEF,
}
}
 ABC >  DEF and  A >  D. Also, BN and EM
bisect congruent angles, so  ABN >  CBN >  DEM
>  FEM.
By the AA Similarity Postulate, n ABN , nDEM.
BN
AB
AB
Therefore, }
5}
, where }
is the scale factor.
EM
DE
DE
E
A
B
A
E b
b
A
C
D
F
Because they are both right angles,  A and  D are
congruent. The acute angles  C and  F are also
congruent, so n ABC , nDEF by the AA Similarity
Postulate.
B
D
C B
a
C
a
Because  ADC >  BEC and  C >  C,
n ADC , nBEC by the AA Similarity Postulate.
b
The ratio of the hypotenuses is }a, so the ratio of the
A
b
corresponding side lengths is also }a. The altitudes
are corresponding sides, so their lengths are in the
b
ratio }a.
E
C
Mixed Review for TAKS
B
m AED ø 298, m ABC ø 298;
So, m ADE 5 m ACB and m AED 5 m ABC.
c. By the AA Similarity Postulate, n ADE , n ACB.
d. Sample answer: AB 5 3 cm, BC 5 4 cm, AC 5 2 cm,
AD 5 1 cm, DE 5 2 cm, AE 5 1.5 cm;
AC
AE
AB
DE
CB
1
2
}5}5}5}
e. The measures of the angles change, but the equalities
remain the same. Yes; the triangles remain similar by
the AA Similarity Postulate.
38. Sample answer: Given any two points on a line, you can
draw similar triangles as shown in the diagram. Because
the triangles are similar, the ratios of corresponding side
lengths are the same. So, the ratio of the rise to the run
is the same. Therefore, the slope is the same for any two
points chosen on a line.
41. C;
Length of postcard
Width of postcard
5.5 in.
3.5 in.
11
7
}} 5 } 5 }
11 in.
7 in.
11
6.6 in.
4.2 in.
11
Choice A: } 5 }
7
Choice B: } 5 }
7
5 in.
3 in.
5
Choice C: } 5 }3
4.4 in.
2.8 in.
11
Choice D: } 5 }
7
The dimensions 5 inches by 3 inches are not proportional
to the dimensions of the postcard.
42. F;
120, 138, 142, 142, 156, 158
856
2
142 1 142
5 142}3
Mean 5 }
6
Median 5 }
5 142
2
Mode 5 142
Range 5 158 2 120 5 38
The mean gives Janice the highest ﬁnal score.
170
Geometry
Worked-Out Solution Key