Download Answers for the lesson “Prove Triangles Similar by SSS and SAS”

Answers for the lesson “Prove Triangles
Similar by SSS and SAS”
LESSON
6.4
11. nABC , nDEC; ŽACB >
Skill Practice
ŽDCE by the Vertical Angles
Congruence Theorem and
1. PX, CB, PQ
2. You would need to know that
15
12 3
18
3. } 5 } 5 }; }
10
8 2
12
16
20 2
10
4. } 5 } 5 }; }
40
50 5
25
5. nRST
BC
EC
AC
DC
3
2
} 5 } 5 }. The triangles are
one pair of corresponding sides
is congruent.
similar using the SAS Similarity
Theorem.
12. nXYZ , nDJG; ŽD > ŽX and
DJ
XY
DG
XZ
5
3
} 5 } 5 }. The triangles are
6. nJKL
similar by the SAS Similarity
Theorem.
7. similar; nFDE , nXWY; 2 : 3
8. not similar
13. Sample answer: The triangle
9. 3;
correspondence is not listed in the
correct order; nABC , nRQP.
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X
4(n 1)
P
5
Z
15.
4
R
14. D
Z
7n 1
Y
X
10. Sample answer:
34
Y
N
FH
HK
nGHJ , nFHK; }
5}
HJ
GH
KF
66
80
4
5}
5 }3 thus the triangles
JG
are similar by the SSS Similarity
Theorem.
L
34
M
They are similar by the AA
Similarity Postulate.
Geometry
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178
T
16.
27. In similar triangles the ratio of
32
16
20
S
the areas is the square of the scale
factor. Sample answer: Let the
base and height of nVWX
measure 3a and 3b and the base
and height of nABC measure 2a
and 2b. The ratio of their areas is
F
R
48
24
H
30
G
3a(3b)
2
}
2a(2b)
}
2
}
They are not similar since the
larger side in nRST would
not be opposite the largest angle.
9
5 }4.
Problem Solving
17.
E
15
D
28. AA Similarity Postulate; in
7x
nAGB ŽA and ŽAGB are
congruent to ŽA and ŽAFC in
nAFC. InnAFC ŽA and ŽAFC
are congruent to ŽA and ŽAED
in nAED.
F
25
B
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24
A
8x
C
54
29. The triangle whose sides measure
They are not similar since the
ratio of corresponding sides is
not constant for any arrangement
of side lengths.
18. 538
19. 458
20. 828
21. 24
22. 42
23. 16Ï 2
24. nNSM and nNRP,
nNSL and nNRQ,
nNLM and nNQP
3
25. }
2
}
4 inches, 4 inches, and 7 inches is
similar to the triangle whose sides
measure 3 inches, 3 inches, and
5.25 inches.
CD
30. } is the same scale factor as
CE
the other ratio.
31. ŽCBD > ŽCAE
9
26. }
4
Geometry
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179
36. Yes. Sample answer: All pairs
32.
of similar triangles have angle
pairs whose measures are in
proportion (with constant of
proportionality 1).
10
20
8
5
20
4
nXYW is not similar to nXZW.
33. a. AA Similarity Postulate
b. 80 ft
c. 70.8 ft
AB
GB
6
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so that GB 5 DE. Draw }
GH so
}
}
that GH i AC. This makes
n ABC , nGBH by the AA
Similarity Postulate. From this
similarity you have
AC
GH
} 5 }. Using this along with
34. a. 8, 24;
18
37. Sample answer: Locate G on }
AB
30
10
8
24
1
b. }
3
c. yes; yes
GH
GB
5}
what’s given, you get }
DE
DF
GH > }
DF,
which implies that }
making nGBH > nDEF.
Finally, use the definition of
congruent triangles and the
AA Similarity Postulate to
conclude n ABC , nDEF.
38. about 15.4 ft
35. Sample answer: Given that D and
} and BC
}
E are midpoints of AB
respectively the Midsegment
Theorem guarantees that
}
}. By the Corresponding
AC i DE
Angles Postulate ŽA > ŽBDE
and so ŽBDE is a right angle.
} i EF
}. By
Reasoning similarly AB
the Alternate Interior Angles
Congruence Theorem ŽBDE >
ŽDEF. This makes ŽDEF a right
angle that measures 908.
Geometry
Answer Transparencies for Checking Homework
180
Mixed Review of
Problem Solving
3. Dilation with scale factor 3 and
then a translation right.
1. Sample:
4. Yes; triangle SRQ is a dilation
and reflection of triangle LMN.
U
3 cm
V
4 cm
W
E
2.5 cm
2 cm
7. a. Use the AA Similarity
F
1
Scale factor: }2
2. 7.2;
7
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LMN, side MN would match with
RS, but side LM would not line
up with side QS.
XY
XW
6. } 5 }
XZ
XV
D
1.5 cm
5. No; After a dilation of triangle
5 cm
.
2
Postulate since ŽD and ŽB
are right angles and
ŽACB > ŽECD by the
Vertical Angles Congruence
Theorem.
b. 1.5 mi
c. 12.5 mi
Geometry
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181