Download Answers for the lesson “Prove Triangles Congruent by SAS and HL”

LESSON
4.5
Answers for the lesson “Prove Triangles
Congruent by SAS and HL”
18. n KMN, n KLN. Sample answer:
Skill Practice
}
MK > }
MN > }
LN > }
LK since
1. included
they are all radii of the same size
circle. It is given that }
MK > }
MN
and }
LK > }
LN. Ž KMN and Ž KLN
are right angles and since all
right angles are congruent,
Ž KMN > Ž KLN. Therefore
n KMN > n KLN by SAS.
2. SAS requires two sides and the
included angle of one triangle to
be congruent to the corresponding
two sides and the included angle
of a second triangle. SSS requires
that three sides of one triangle be
congruent to the corresponding
sides of a second triangle.
19. HL;
3. Ž XYW
4. Ž WZY
5. Ž ZWY
6. Ž WXY
7. Ž XYZ
8. Ž XWZ
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9. not enough
A
B
C
E
H
F
D
F
10. enough
B
11. not enough
12. not enough
20. not enough information
13. enough
14. not enough
21. SAS
15. B
16. n BAD, n DCB. Sample answer:
Ž A > Ž C because they are
both right angles, }
AB > }
CD and
}
AD > }
CB because the sides of a
square are congruent, therefore
n BAD > n DCB by SAS.
17. n STU, n RVU; }
ST > }
TU >
}
UV > }
VR and ŽT > ŽV
because it is a regular pentagon,
therefore n STU > n RVU
by SAS.
G
22. HL
23. Yes; they are congruent by the
SAS Congruence Postulate.
24. }
YX and }
YW should have the same
length since it can be shown that
n XZY > nWZY;
4x 1 6 5 5x 2 1,
2x 5 27, x 5 7.
25. }
AC > }
DF
27. }
BC > }
EF
26. }
BA > }
ED
28. Yes; they are congruent by SAS.
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105
29. Because }
RM > }
PQ, ŽRMQ and
ŽRMP are right angles and thus
are congruent. }
QM > }
MP and
}
}
MR > MR. nRMP > nRMQ
by SAS.
triangles are congruent by SAS.
32. SAS
33. Two sides and the included angle
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of one sail need to be congruent
to two sides and the included
angle of the second sail; the two
sails need to be right triangles
with congruent hypotenuses
and one pair of congruent legs.
34. Definition of a right triangle;
Given; }
LM > }
NM; HL
35. Statements (Reasons)
1. }
PQ bisectsŽ SPT, }
SP > }
TP.
(Given)
2. ŽSPQ > ŽTPQ (Definition of
angle bisector)
(Reflexive Property
of Congruence)
4. n SPQ > nTPQ
36. Statements (Reasons)
(SAS)
1. }
JM > }
LM
(Given)
2. Ž KJM andŽ KLM are right
angles.
(Given in diagram)
Problem Solving
3. }
PQ > }
PQ
3. nVXW > n XYZ
37. Statements (Reasons)
30. Since Ž DAC > Ž FAB the
31. SAS
2. ŽVXW > Ž XYZ
(Corresponding Angles
Postulate)
(SAS)
1. }
VX > }
XY, }
XW > }
YZ, }
XW i }
YZ
(Given)
3. n JKM and n LKM are right
triangles.
(Definition of right triangle)
} > KM
} (Reflexive Property
4. KM
of Congruence)
5. n JKM > n LKM
(HL)
38. Statements (Reasons)
1. D is the midpoint of }
AC .
(Given)
2. }
BD >}
AC
(Given in diagram)
3. Ž BDA andŽ BDC are right
angles.
(Definition of
perpendicular lines)
4. Ž BDA > Ž BDC (Right Angle
Congruence Theorem)
5. }
DA > }
DC
(Definition of
midpoint)
6. }
BD > }
BD (Reflexive Property
of Congruence)
7. n ABD > nCBD
(SAS)
Geometry
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106
MP > }
NO
(Slopes of
5. }
Perpendicular Lines Postulate)
39. D
40. Statements (Reasons)
1. }
CR > }
CS, }
QC > }
CR, }
QC > }
CS
(Given)
6. Ž PMO andŽ PMN are right
angles.
(Definition of
perpendicular lines)
2. Ž SCQ andŽ RCQ are right
angles.
(Definition of
perpendicular lines)
7. Ž PMO > Ž PMN
(Right
Angles Congruence Theorem)
3. Ž SCQ >Ž RCQ (Right Angle
Congruence Theorem)
4. }
QC > }
QC (Reflexive Property
of Congruence)
5. nQCR > nQCS
(SAS)
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41. Find the length of each side of the
two triangles and show that pairs
of corresponding sides have the
same length and therefore
are congruent.
Statements (Reasons)
}
}
1. MO 5 4Ï 2 , MN 5 4Ï 2
(Distance formula)
2. }
MO > }
MN
(Definition of
congruent segments)
3. }
MP > }
MP (Reflexive Property
of Congruence)
4. slope of }
MP 5 21,
}
slope of NO 5 1
(Slope formula)
8. n PMO > n PMN
(SAS)
Sample answer: Both methods
involve the use of the distance
formula, with the first method
involving calculating the
lengths of all the sides of the two
triangles. The second method also
requires finding the slopes of two
of the sides in order to show they
are perpendicular and that right
triangles are formed.
Mixed Review of
Problem Solving
1. a. n ACE and n DCF are obtuse
triangles, and n ECB, n FCG,
n ABC, and n DGC are
acute triangles.
b. All triangles are scalene.
2. Sample answer: Using the
distance formula it can be shown
that n PQR > nSTR by SSS.
Geometry
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107
3. 708;
Statements (Reasons)
7
1. }
DF > }
EH, mŽEHB 5 258,
mŽBFG 5 658, }
DF > }
AG at
point F
(Given)
0
2. nFGB > nHGB
(Problem 5a)
3. }
FB > }
HB
4. ŽDFG is a right angle.
(Definition of
perpendicular lines)
4. Yes; Because }
AC > }
GE and
}
AB > }
FE, }
BC > }
GF. Also, since
}
}
}
AG > CE and AH > }
DE,
}
}
then HG > CD.
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(Corr. parts of
s are >.)
>n
Ž G and ŽC are right angles, so
they are congruent. Therefore,
nBCD > nFGH by SAS.
5. a. Statements (Reasons)
1. }
BG > }
FH, }
GF > }
GH
(Given)
5. mŽ DFG 5 908 (Definition
of right angle)
6. mŽDFB 1 mŽ BFG 5
mŽ DFC (Angle Addition
Postulate)
7. mŽ DFB 1 658 5 908
(Substitution Property
of Equality)
8. mŽ DFB 5 258 (Subtraction
Property of Equality)
2. ŽBGF andŽ BGH are right
angles.
(Definition of
perpendicular lines)
9. mŽ DFB 5 mŽEHB
(Transitive Property
of Equality)
3. ŽBGF > Ž BGH
(Right
Angle Congruence Theorem)
10. ŽDFB > Ž EHB
4. }
BG > }
BG
(Reflexive
Property of Congruence)
5. nFGB > n HGB
(Definition of congruent
angles)
11. nBDF > nBEH
(SAS)
(SAS)
b. yes;
Geometry
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108
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
6. 23;
2
3
Geometry
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109