Download TRIANGLE CONGRUENCE

CHAPTER 7
11. a . Sample:
T RIANGLE C ONGRUENCE
LESSON 7-1
C
m/CAB = 7 0
m/CBA = 3 8
pp. 364–369
1. Triangulation is the use of triangles to determine
the location of an object.
2 . a . Sample:
Z
X
Y
A
B
b . Any opinion is allowed. The answer will be
found to be no.
12 . a . Sample: AB = 4
m/CAB = 6 0
C
t
V
W
A
Angle measures: < 22°, 50°, 108°
b . Any opinion is allowed. The answer is yes.
3 . Sample:
Z
3 cm
X
5 cm
Y
B
C
D
45°
B
6
7
8
9
10
E
80°
A
b . Any opinion is allowed. The answer is no.
. 180 2 (32 1 102) 5 180 2 134 5 46
The third angle has measure 46.
. The measure of the third angle is 180 2 (x 1 y),
or 180 2 x 2 y.
. Answers may vary. Mathematically, they should
agree.
. Sample: Yes; given the lengths of three
segments, a unique triangle is determined.
. Sample: No; given the measures of three angles,
a unique triangle is not determined.
. Sample: Yes; given the lengths of two segments
and the measure of the angle they form, a unique
triangle is determined.
10 0
b . Any opinion is allowed. The answer will be
found to be no.
13 . a . Sample:
BC = 3
C m/CAB = 4 0
m/CBA = 6 0
4 cm
4 . a . Sample:
5
B
A
b . Any opinion is allowed. The answer will be
found to be yes.
14 . a . Sample:
AB = 2
m/CAB = 4 0
m/CBA = 6 0
C
A
B
b . Any opinion is allowed. The answer will be
found to be yes.
15 . a . No unique triangle can be drawn.
b . The answer is no.
16 . a . Sample: E
D
C
182'
A
62° B
UCSMP Geometry © Scott, Foresman and Company
16. b.
CD 1.1 cm 1829
<
5
DE 2.3 cm
n
21. If m/PAB 5 40, then m/ACD 5 m/PBC 5 40;
m/D 5 m/DCB 5 m/CBA 5 m/BAD
5 m/APB 5 m/CPB 5 90;
m/DAC 5 m/BCA 5 m/PBA 5 50
22. a. Sample:
n 5 (2.3 ? 182) ÷ 1.1 < 381
DE < 3819
c . ASA; 2 angles and the side between them
17. sum of angle measures of octagon:
(n 2 2) ? 180 5 (8 2 2) ? 180 5 6 ? 180 5 1080
measure of each angle: 1080 ÷ 8 5 135
Each angle in a stop sign has measure 135.
18. a. IR > TH
b. /GHT > /BRI
19. If B, G, H, and R are not collinear, then these
points are vertices of a quadrilateral called an
isosceles trapezoid.
20. Sample: 4LMN is the image of 4FGH under
an isometry.
n
A
C
B
D
m
b. rm ◦ rn is a rotation.
23.
N
M
P
Q
S
,
24.
1
( 21], 6 2] )
book
10
9
te le p1h o ne r e ce iv e r
(2, 2] )
8
p a rro t
(7, 10)
ca r (4, 6)
r o lle r s k a te
(6, 21] )
b e ll (721], 6)
7
2
6
a ir p la ne (7, 421] )
5
b u g le (1] , 4)
2
4
mo u s e (1, 5)
r o o s te r ’s h e a d
(4 12] , 8)
3
2
w o o l ca p
(8, 9 ) b a th r o b e 1
(2, 7)
y o -y o Pu r ita n’s h a t
0
( 12] , 9 )
(8, 0)
1
UCSMP Geometry © Scott, Foresman and Company
2
3
4
5
6
7
8
101
LESSON 7-2
pp. 370–377
1. Four conditions that lead to triangle congruence
are SSS, SAS, ASA, and AAS.
2. SSS Congruence Theorem
3. Symmetry of a kite was used in the proof of the
SSS Congruence Theorem.
4. Symmetry of an isosceles triangle was used in
the proof of the SAS Congruence Theorem.
5. ASA Congruence Theorem: If two angles and
the included side of one triangle are congruent
to two angles and the included side of another
triangle, then the triangles are congruent.
6. In the AAS Congruence Theorem, the
corresponding sides are not included by the pairs
of congruent angles as they are in the ASA
Congruence Theorem.
7. AD is included by /CAD and /ADC.
8. 4ABF > 4CED by SAS
9. not enough information to know
10. not enough information to know
11. 4OY N > 4YOX by ASA
12. 4CGO > 4IPN by ASA
13. 4STE > 4BTE by AAS
14. SSS explains why all triangles are rigid.
15. a. 4DPO > 4JKL
b. If OD 5 5 yards, then LJ is 5 yards.
c. /O > /L
d . If m/P 5 73, then m/K 5 73.
16. a. /C > /E
b. /A > /F
c. CB > ED
d . AC > FE and CB > ED
17. 4ABD > 4CBD by the AAS Congruence
Theorem.
18. Yes, the triangles will be congruent by the
ASA Congruence Theorem.
19. Given: m/FDE 5 m/C9DE; G, D, and E are
collinear.
@##$ bisects /FDC9.
To prove: EG
Argument:
Conclusions
Justifications
0. m/FDE 5 m/C9DE;
Given
0. G, D, and E are collinear.
102
1. m/FDG 5
0. 180 2 m/FDE;
0. m/C9DG 5
0. 180 2 m/C9DE
2. m/FDG 5
0. m/C9DG
@##$ bisects /FDC9.
3. EG
Linear Pair Theorem
(and algebra)
Transitive Property
of Equality (and
substitution)
definition of
angle bisector
20. a. n 5 2
b.
21. a. n 5 5
b.
22. a. n 5 3
b.
23. Given: 4ABC > 4DEF. If m/A 5 32 and
m/B 5 64, then m/C 5 180 2 (32 1 64) 5
180 2 96 5 84. Since m/C 5 m/F, then
m/F 5 84.
24. Given: 4OLD > 4NEW
Sample conclusions: /O > /N, /L > /E,
/D > /W, OL > NE, LD > EW, OD > NW
Justification: CPCF Theorem
UCSMP Geometry © Scott, Foresman and Company
y
25.
1. WY > VY
2. /XY W > /ZY V
x
G
z
H
26. Sample:
C
E
A
B
LESSON 7-3
1. a.
pp. 378–383
B
A
D
F
D
C
E
b. ASA Congruence Theorem
c. Given: C is the midpoint of AE;
/BAC > /DEC; /BCA > /DCE
To prove: 4ABC > 4EDC
Argument:
Conclusions
Justifications
0. C is the midpoint of
Given
0. AE; /BAC > /DEC;
0. /BCA > /DCE
1. AC > EC
definition of
midpoint
2. 4ABC > 4EDC
ASA Congruence
Theorem
2. Given: Y is the midpoint of WV; m/Z 5 m/X
To prove: 4WXY > 4VZY
Argument:
Conclusions
Justifications
0. Y is the midpoint
Given
0. of WV; m/Z 5 m/X
UCSMP Geometry © Scott, Foresman and Company
3. 4WXY > 4VZY
definition of
midpoint
Vertical Angles
Theorem
AAS Congruence
Theorem
3. Given: AB 5 CD; BC 5 AD
To prove: /BAC > /DCA
Argument:
Conclusions
Justifications
0. AB 5 CD; BC 5 AD
Given
1. AC > AC
Reflexive Property
of Congruence
2. 4ABC > 4CDA
SSS Congruence
Theorem
3. /BAC > /DCA
CPCF Theorem
4. a. AB and BC are congruent.
b. Isosceles Triangle Base Angles Converse
Theorem
5. a. Reflexive Property of Congruence
b. AAS Congruence Theorem
c. CPCF Theorem
6. Given: XA 5 Y A, and XC 5 Y C
To prove: /XAC > /Y AC
Argument:
Conclusions
Justifications
0. XA 5 Y A, and
Given
0. XC 5 Y C
1. AC > AC
Reflexive Property
of Congruence
2. 4XAC > 4Y AC
SSS Congruence
Theorem
3. /XAC > /Y AC
CPCF Theorem
###$ bisects /JNM; NM > NK;
7. Given: NK
NL > NJ
To prove: 4JNK > 4LNM
Argument:
Conclusions
Justifications
###$ bisects /JNM;
0. NK
Given
0. NM > NK; NL > NJ
1. /JNK > /LNM
definition of
angle bisector
2. 4JNK > 4LNM
SAS Congruence
Theorem
103
8. Given: BC k AD, and BC 5 AD
To prove: 4ABC > 4CDA
Argument:
Conclusions
Justifications
0. BC k AD, and
Given
0. BC 5 AD
1. AC 5 AC
Reflexive Property
of Equality
2. /BCA > /DAC
k Lines ⇒ AIA >
Theorem
3. 4ABC > 4CDA
SAS Congruence
Theorem
9. Given: m/ADB 5 m/CDB;
Given: m/ABD 5 m/CBD
To prove: AB 5 CB
Argument:
Justifications
Conclusions
0. m/ADB 5 m/CDB;
Given
0. m/ABD 5 m/CBD
1. BD 5 BD
Reflexive Property
of Equality
2. 4ABD > 4CBD
ASA Congruence
Theorem
3. AB 5 CB
CPCF Theorem
10. a. 4ADG, 4ADB, 4BDJ, 4DJG, 4AGJ,
4AGB, 4BJG, 4BJA
b. 4ADG > 4BDJ, 4AGJ > 4BJG,
4AGB > 4BJA
11. a. SAS Congruence Theorem
b. CPCF Theorem
12. not enough information to know
13. 4EFG > 4IHG; AAS Congruence Theorem
14. a. True
b. False
15. a. Yes, assuming the units for the length
measurements are the same
b. SAS Congruence Theorem
16. If m/DAC 5 50, then m/BAC 5 m/DCA
5 m/BCA 5 50, m/D 5 m/B 5 80, and
m/DAB 5 m/DCB 5 100.
17.
B = A'
B'
A
D
104
C = D'
18. The sum of the measures of the acute angles in a
right triangle is 180 2 90 5 90.
19. Sample:
C
B' = A
D
B = A'
C'
20. The Triangle Inequality Postulate states that the
sum of the lengths of any two sides of a triangle
is greater than the length of the third side.
Because 3 cm 1 6 cm 5 9 cm , 11 cm, the
Triangle Inequality Postulate is violated.
21. Yes. Since 4ABC > 4CDA, both
/BAC > /DCA, and /BCA > /DAC (CPCF
Theorem). As a consequence, AB k DC and
BC k DA (AIA > ⇒ k Lines Theorem).
Therefore, ABCD is a parallelogram.
LESSON 7-4
pp. 384–388
1. a. There are 8 triangles in the figure: 4QUS,
4USA, 4ASD, 4DSQ, 4QUA, 4UAD,
4ADQ, 4DQU.
b. 4DAU appears congruent to 4QUA.
c. 4QUD appears congruent to 4DAQ.
2. a. Given
b. Reflexive Property of Congruence
c. SAS Congruence Theorem
d. CPCF Theorem
3. Given: Regular pentagon ALIVE
To prove: /EAV > /VIE
Argument:
Conclusions
Justifications
0. Regular pentagon
Given
0. ALIVE
1. AE > IV
definition of
regular pentagon
2. EV > EV
Reflexive Property
of Congruence
3. /AEV > /IVE
definition of
regular pentagon
4. 4AEV > 4IVE
SAS Congruence
Theorem
5. /EAV > /VIE
CPCF Theorem
C'
UCSMP Geometry © Scott, Foresman and Company
4. Given: QU > AD, and QA > UD
To prove: /1 > /2
Argument:
Conclusions
Justifications
0. QU > AD, and
Given
0. QA > UD
1. UA > UA
Reflexive Property
of Congruence
2. 4QUA > 4DAU
SSS Congruence
Theorem
3. /1 > /2
CPCF Theorem
5. Given: AD 5 AE, and m/D 5 m/E
To prove: EB 5 CD
Argument:
Justifications
Conclusions
0. AD 5 AE, and
Given
0. m/D 5 m/E
1. /A > /A
Reflexive Property
of Congruence
2. 4AEB > 4ADC
ASA Congruence
Theorem
3. EB 5 CD
CPCF Theorem
6. Given: PR 5 QS, and PS 5 QR
To prove: m/P 5 m/Q
Argument:
Conclusions
Justifications
0. PR 5 QS, and
Given
0. PS 5 QR
1. RS 5 RS
Reflexive Property
of Equality
2. 4PRS > 4QSR
SSS Congruence
Theorem
3. m/P 5 m/Q
CPCF Theorem
7. Argument: Since l, the perpendicular bisector
of AB and ED, is a symmetry line of ABCDEF
by the Regular Polygon Symmetry Theorem,
rl(A) 5 B and rl(C) 5 F. Therefore,
AC > BF because reflections preserve
distance.
8. Given: A, D, E, and B are collinear; AD > BE;
AC > BC
To prove: DC > CE
UCSMP Geometry © Scott, Foresman and Company
Argument:
Conclusions
0. A, D, E, and B
0. are collinear;
0. AD > BE; AC > BC
1. /CAB > /CBA
Justifications
Given
Isosceles Triangle
Base Angles
Theorem
2. 4CAD > 4CBE
SAS Congruence
Theorem
3. DC > CE
CPCF Theorem
9. Given: QR k TU, and S is the midpoint of QU.
To prove: S is the midpoint of RT.
Argument:
Conclusions
Justifications
0. QR k TU, and S is
Given
0. the midpoint of QU.
1. QS 5 US
definition of
midpoint
2. m/QSR 5 m/UST
Vertical Angles
Theorem
3. m/Q 5 m/U
k lines ⇒ AIA >
Theorem
4. 4QSR > 4UST
ASA Congruence
Theorem
5. ST 5 SR
CPCF Theorem
6. S is the midpoint
definition of
0. of RT.
midpoint
10. 4PCA > 4PCB by the SAS Congruence
Theorem. So, PA 5 PB by the CPCF Theorem.
11. a. Vertical Angles Theorem
b. SAS Congruence Theorem
c. CPCF Theorem
d. Vertical Angles Theorem
e . Transitive Property of Congruence
12. a. Sample:
C
A
32° 110°
B
b. No, because the AA condition does not
determine a unique triangle.
105
13. sum of angle measures of a pentagon:
(n 2 2) ? 180 5 (5 2 2) ? 180 5 3 ? 180 5 540
measure of each angle: 540 ÷ 5 5 108
14. True
15. True
16. 0 , 180 2 2y
2y , 180
y , 90
The angle is acute.
17. 4ADC > 4AEB 5 AB, AC, and point F.
18. WX 1 XY 1 Y Z 5 WZ
13 1 XY 1 17 5 45
XY 5 15
19. Samples are given.
a.
b.
c.
heptagon
octagon
LESSON 7-5
pp. 389–396
1. Sample:
2. a. A right triangle has sides represented by H
and L.
b. H represents hypotenuse; L represents leg.
3. The proof of the HL Congruence Theorem
makes use of the symmetry of the isosceles
triangle.
4. Given: BD ⊥ AC; AB 5 BC
To prove: 4ABD > 4CBD
Argument:
Conclusions
Justifications
0. BD ⊥ AC; AB 5 BC
Given
1. BD 5 BD
Reflexive Property
of Equality
2. 4ABD > 4CBD
HL Congruence
Theorem
5. The SSA Condition leads to congruence when
the sides opposite the congruent angles in each
triangle are longer than the other congruent
sides.
6. a. HL Congruence Theorem
b. 4ABC > 4FED
7. a. SAS Congruence Theorem
b. 4GHI > 4KJL
8. a. HL Congruence Theorem
b. 4MNP > 4ONP
9. a.–c. reduced size:
W
WY
50° 9 cm
9 cm
11 cm
X
nonagon
decagon
Z
d. No. There are two possible triangles XZW, as
shown in the drawing.
10. Given: VX ⊥ WY , WZ 5 VY , and XZ 5 XY
To prove: /W > /V
Argument:
Conclusions
Justifications
0. VX⊥WY , WZ 5 VY ,
Given
0. and XZ 5 XY
106
UCSMP Geometry © Scott, Foresman and Company
10. (continued)
1. /WXZ and /VXY
0. are right angles.
2. 4WXZ > 4VXY
definition of
perpendicular lines
HL Congruence
Theorem
3. /W > /V
CPCF Theorem
11. Given: AP 5 AR; /P and /R are right angles.
To prove: PBRA is a kite.
Argument:
Conclusions
Justifications
0. AP 5 AR; /P and
Given
0. /R are right angles.
1. AB 5 AB
Reflexive Property
of Equality
2. 4PBA > 4RBA
HL Congruence
Theorem
3. PB > RB
CPCF Theorem
4. PBRA is a kite.
definition of kite
12. a. Given
b. definition of circle
c. definition of circle
d. SsA Congruence Theorem (Steps 1, 2, and
Given)
13. Let T be the top of the maypole, M the point on
the maypole that is the same height as June and
April’s hands, J the position of June’s hand, and
A the position of April’s hand.
T
J
A
M
ground
JA is parallel to the ground, since J and A are
equal in height (given). So, JA ⊥ TM by the
Perpendicular to Parallels Theorem. Since
TM > TM and TJ > TA, then 4TJM > 4TAM
by the HL Congruence Theorem. Thus,
JM > AM by the CPCF Theorem.
So, since JM is the distance from June’s hand to
the maypole, and AM is the distance from April’s
hand to the maypole, and JM 5 AM, their hands
are the same distance from the maypole.
14. a. definition of isosceles trapezoid
b. Isosceles Trapezoid Theorem
UCSMP Geometry © Scott, Foresman and Company
c. Reflexive Property of Congruence
d. SAS Congruence Theorem (steps 1–3)
e. CPCF Theorem
15. f . Isosceles Trapezoid Symmetry Theorem
g . Reflections preserve distance.
16. Answers may vary. Sample: “I prefer the
argument in Question 15 because it is shorter.”
17. a. reduced size:
R
80°
2"
Q
45°
S
b. Yes, by the AAS Congruence Theorem.
18. Given: AB 5 AC; BE 5 DC
a. To prove: /ADE > /AED
Argument:
Conclusions
Justifications
0. AB 5 AC;
Given
0. BE 5 DC
1. /B > /C
Isosceles Triangle
Base Angles Theorem
2. 4ABE > 4ACD SAS Congruence
Theorem
3. /ADE > /AED
CPCF Theorem
b. To prove: 4ADE is isosceles.
(continuing from the proof above)
1. AD > AE
Isosceles Triangle Base
Angles Converse Theorem
2. 4ADE is
definition of isosceles
0. isosceles.
triangle
19. 4AGY , 4DEX, 4DEY
20. a. 4ABC > 4DEF
b. rm(4DEF) 5 4ABC
c. Sample: rm(AB) 5 DE
21. The conjecture is false. Counterexample:
A
B
D
C
E
H
F
G
107
LESSON 7-6
pp. 397–403
13.
1. A tessellation is a covering of a plane with
nonoverlapping congruent regions and no holes.
2. Samples: brickwork, floor or wall tiles,
wallpaper
3. Steps 1–4: Check students’ drawings.
Step 5: ABCD is a parallelogram. Rotations
preserve angle measure, so /BAC > /DCA and
/BCA > /DAC. By the AIA > ⇒ k Lines
Theorem, AD k BC and AB k DC. Therefore,
ABCD is a parallelogram by the definition of
parallelogram.
4. Check students’ drawings.
5. Sample (reduced size):
14. a. Sample (reduced size):
Y
A
6. Sample (reduced size):
Z
C
108
O
X
Z"
A*
7. Key question: Can a given region tessellate the
plane?
8. One museum where many tessellations can be
found is the Alhambra.
9. Yes. Since the sum of the angle measures in
ABCD is 360, it is possible to have a different
angle from each of the four congruent
quadrilaterals meeting at a single point.
10. (c); the measure of each angle of a regular
pentagon is 108, and 360 is not evenly divisible
by 108.
11. No; the measure of each angle of a regular
heptagon is about 129, and 360 is not evenly
divisible by 129.
12. A new type of tessellating pentagon was
discovered as recently as 1985.
B
C'
Y' Z'
A'
C"
B"
b. They are the four angles of ABOC, so the
sum of their measures is 360.
c. Translate the figure repeatedly in all
directions.
d. Answers may vary.
15. 4ABC > 4CDA, so /BAC > /DCA and
/BCA > /DAC (CPCF Theorem). So AB k CD
and BC k AD (AIA > ⇒ k Lines), and ABCD is
a parallelogram (definition of parallelogram).
16. 4ABC > 4ADC, so AB 5 AD, BC 5 DC
(CPCF Theorem). So ABCD is a kite (definition
of kite).
17. Given: AD 5 BC; DA ⊥ AB; BC ⊥ CD
To prove: AB k DC
Argument:
Conclusions
Justifications
0. AD 5 BC;
Given
0. DA ⊥ AB; BC ⊥ CD
1. BD 5 BD
Reflexive Property
of Equality
UCSMP Geometry © Scott, Foresman and Company
17. (continued)
2. 4ABD > 4CDB
3. /ABD > /CDB
4. AB k DC
24. Sample (reduced size):
HL Congruence
Theorem
CPCF Theorem
AIA > ⇒ k Lines
Theorem
18. Given: AC > BD; AD > BC
To prove: /ADB > /BCA
Argument:
Conclusions
Justifications
0. AC > BD; AD > BC
Given
1. AB > AB
Reflexive Property
of Congruence
2. 4ABD > 4BAC
SSS Congruence
Theorem
3. /ADB > /BCA
CPCF Theorem
19. Given: m/ABD 5 m/BDC;
Given: m/ADB 5 m/DBC
To prove: AB 5 CD
Argument:
Conclusions
Justifications
0. m/ABD 5 m/BDC;
Given
0. m/ADB 5 m/DBC
1. BD > BD
Reflexive Property
of Congruence
2. 4BAD > 4DCB
ASA Congruence
Theorem
3. AB 5 CD
CPCF Theorem
20. (1) /B > /X; ASA Congruence Theorem
(2) AC > Y Z; SAS Congruence Theorem
(3) /C > /Z; AAS Congruence Theorem
21. a. No
b. There is no such segment, unless 4PQR is
isosceles with PQ 5 PR.
22. Playfair’ s Parallel Postulate: Through a point not
on a line, there is exactly one line parallel to the
given line.
23. Sample: The Moors were a North African
people of Berber and Arab descent who invaded
and conquered Spain in the eighth century. They
were driven back to Africa during centuries of
wars and inquisitions, the final Moors leaving
in 1492.
UCSMP Geometry © Scott, Foresman and Company
IN-CLASS ACTIVITY
S t ep 1.
A
D
p. 404
B
C
/B > /D; /DAB > /BCD;
/BAC > /DCA; /DAC > /BCA
S t ep 2. Argument:
Conclusions
Justifications
0. Parallelogram
Given
0. ABCD
1. AD k CB;
definition of
0. AB k DC
parallelogram
2. m/DAC
k Lines ⇒ AIA 5
0. 5 m/BCA;
Theorem
0. m/BAC
0. 5 m/DCA
3. AC 5 AC
Reflexive Property
of Equality
4. 4ADC > 4CBA
ASA Congruence
Theorem
S t ep 3. AD > BC; AB > CD; /D > /B
109
Step 4.
A
2. BD > BD
B
E
D
C
4ADE > 4CBE; 4CED > 4AEB;
4ABD > 4CDB; 4ABC > 4CDA
Step 5. /ADB > /CBD; /ABD > /CDB;
AE > CE; DE > BE
Step 6. The parallelogram possesses 2-fold rotation
symmetry.
LESSON 7-7
pp. 405–410
1. a. BC is congruent to AD.
b. /5 is congruent to /1.
c. The midpoints of AC and BD are the same
point (E).
2. a. AB, BC, and CD are congruent to AD.
b. /2, /5, and /6 are congruent to /1.
c. The midpoints of AC and BD are the same
point (E).
3. a. definition of parallelogram
b. k lines ⇒ AIA > Theorem
c. definition of parallelogram
d. k lines ⇒ AIA > Theorem
e. ASA Congruence Theorem
f. CPCF Theorem
4. Properties of a Parallelogram Theorem, part (a)
5. This image of ABCD is CDAB.
6. Given: Parallelogram ABCD
To prove: /DAB > /BCD
Drawing:
B
A
D
C
Argument: Draw auxiliary diagonal BD.
Conclusions
Justifications
0. Parallelogram ABCD
Given
1. AD > CB; AB > CD
Properties of a
Parallelogram
Theorem (a)
110
Reflexive Property
of Congruence
3. 4ABD > 4CDB
SSS Congruence
Theorem
4. /A > /C
CPCF Theorem
7. Given: Parallelogram ABCD; AC > BD 5 E
To prove: E is the midpoint of AC and BD.
Drawing:
B
A
E
C
D
Argument:
Conclusions
0. Parallelogram ABCD
1. AD k BC
2. /ADB > /CBD
3. /AED > /CEB
4. AD > CB
5. 4ADE > 4CBE
Justifications
Given
definition of
parallelogram
k lines ⇒ AIA >
Theorem
Vertical Angles
Theorem
Properties of a
Parallelogram
Theorem (a)
AAS Congruence
Theorem
CPCF Theorem
definition of
midpoint
6. AE > CE; DE > BE
7. E is the midpoint
0. of AC and BD.
8. a. CE 5 AE 5 7
b. BC 5 AD 5 12
9. a. ED 5 BE 5 x
b. AD 5 BC 5 2y
c. BD 5 2 ? BE 5 2x
10. a. RD and OP are congruent.
b. The distance between two given parallel lines
is constant.
11. (1)
D'
A'
A
B
C
C'
D
B'
UCSMP Geometry © Scott, Foresman and Company
11. (continued)
(2)
A C'
B
B'
D
D'
(3)
A'
C
B'
B
A
C'
7. m/A 1 m/B 1
0. m/BCD 1
0. m/CDA 5 360
8. 4 ? m/A 5 360
9. m/A 5 m/B 5
0. m/BCD 5
0. m/CDA 5 90
10. ABCD is a
0. rectangle.
17. Sample:
Quadrilateral-Sum
Theorem
Substitution
Multiplication
Property of Equality
and Substitution
definition of rectangle
A'
D
C
D'
12. (b) A is false, B is true.
13. Theorem: The distance between two given
parallel lines is constant.
14. a., b.
kite (rd)
parallelogram (R)
rectangle (rpb) (R)
rhombus (rd) (R)
sq uare (rd) (rpb) (R)
15. From the Properties of a Parallelogram
Theorem, AB 5 CD, CD 5 EF, EF 5 GH. From
the Transitive Property of Equality, AB 5 GH.
16. Given: ABCD is a parallelogram; M is the
midpoint of AB.
To prove: If MD 5 MC, then ABCD is a rectangle.
Argument:
Conclusions
Justifications
0. ABCD is a
Given
0. parallelogram;
0. M is the midpoint
0. of AB; MD 5 MC
1. AM 5 BM
definition of midpoint
2. AD 5 BC
Properties of a
Parallelogram
Theorem (a)
3. 4AMD > 4BMC
SSS Congruence
Theorem
4. /A > /B
CPCF Theorem
5. /B > /CDA;
Properties of a
0. /A > /BCD
Parallelogram
Theorem (b)
6. /A > /B >
Transitive Property of
0. /BCD > /CDA
Congruence
UCSMP Geometry © Scott, Foresman and Company
18. Given: Trapezoid ABCF; AE ⊥ CF; BD ⊥ CF;
@#$ k FC
@##$
AF 5 BC; AB
To prove: /F > /C
Argument:
Conclusions
Justifications
0. Trapezoid ABCF;
Given
0. AE ⊥ CF; BD ⊥ CF;
@#$ k FC
@##$
0. AF 5 BC; AB
1. /AEF and /BDC
definition of
0. are right angles.
perpendicular
2. AE k BD
Two Perpendiculars
Theorem
3. ABDE is a
definition of
0. parallelogram.
parallelogram
4. AE 5 BD
Properties of a
Parallelogram
Theorem (a)
5. 4AEF > 4BDC
HL Congruence
Theorem
6. /F > /C
CPCF Theorem
19. No; the congruent sides are not corresponding
sides. AC corresponds to XZ, not Y Z.
20. a. 8 triangles are formed.
©
ª ©
ª
b. ©4RIT, 4GHT , 4RTH, 4GTI
ª ,
4RIH, 4IRG, 4HGI, 4GHR
21. It cannot be justified.
22. A
D
B
C
23. Transitive Property of Congruence
24. Two Perpendiculars Theorem
111
25.
K
3. a. Answers will vary. Sample:
L
M
J
I
B
A
P
E
N
C
D
F
O
H
G
A square is formed.
26.
LESSON 7-8
pp. 411–415
1. According to the definition of parallelogram,
two pairs of parallel sides makes a quadrilateral
a parallelogram.
2. a. Answers will vary. Sample:
D
A
6.5 cm
6.5 cm
C
B
b. The quadrilateral formed is a parallelogram.
c. Yes. You can form a rectangle and a
rhombus, but they are also parallelograms.
112
b. A parallelogram is formed.
4. Sufficient conditions for parallelograms other
than its definitions:
(1) Both pairs of opposite sides are congruent.
(2) Both pairs of opposite angles are congruent.
(3) The diagonals bisect each other.
(4) One pair of sides is parallel and congruent.
5. Given: Quadrilateral PQRS with PQ > RS and
PS > QR.
To prove: PQRS is a parallelogram.
Argument:
Conclusions
Justifications
0. PQ > RS;
Given
0. PS > QR
1. Draw QS.
Unique Line
Assumption
2. QS > QS
Reflexive Property
of Congruence
3. 4PQS > 4RSQ
SSS Congruence
Theorem
4. /PQS > /RSQ
CPCF Theorem
5. PQ k SR
AIA > ⇒ k Lines
Theorem
6. /PSQ > /RQS
CPCF Theorem
7. PS k QR
AIA > ⇒ k Lines
Theorem
8. PQRS is a
definition of
0. parallelogram.
parallelogram
6. No, since the congruent angles are not opposite
each other.
7. Yes, since both pairs of opposite sides are
congruent.
8. Not necessarily. For example, ABCD could be
an isosceles trapezoid with bases AB and CD.
9. A parallelogram is formed.
10. A parallelogram, a kite, or a rectangle can be
formed.
UCSMP Geometry © Scott, Foresman and Company
11. Given: Quadrilateral ABCD; diagonals AC and
BD intersect at their midpoint, O.
To prove: ABCD is a parallelogram.
Drawing:
A
B
O
D
C
Argument:
Conclusions
0. O is the midpoint
6. of AC and BD.
1. AO > CO; DO > BO
2. /AOB > /COD
3. 4AOB > 4COD
4. /ABD > /CDB
5. AB k CD
6. /AOD > /COB
7. 4AOD > 4COB
8./ADB > /CBD
9. AD k BC
Justifications
Given
definition of
midpoint
Vertical Angles
Theorem
SAS Congruence
Theorem
CPCF Theorem
AIA > ⇒ k Lines
Theorem
Vertical Angles
Theorem
SAS Congruence
Theorem
CPCF Theorem
AIA > ⇒ k Lines
Theorem
definition of
parallelogram
10. ABCD is a
6. parallelogram.
12. (5x 1 30) 1 (7x 2 90) 1
(3x 2 10) 1 (4x 1 50) 5 360
19x 2 20 5 360
19x 5 380
x 5 20
m/N 5 5x 1 30 5 5(20) 1 30 5 130
m/S 5 4x 1 50 5 4(20) 1 50 5 130
m/E 5 7x 2 90 5 7(20) 2 90 5 50
m/T 5 3x 2 10 5 3(20) 2 10 5 50
Yes, NEST is a parallelogram because both pairs
of opposite angles are congruent.
UCSMP Geometry © Scott, Foresman and Company
13. The first scout was right. Since the diagonals
intersect at their midpoints, ABCD is a
parallelogram. So opposite angles are congruent:
/DAB > /BCD, and /ADC > /CBA. But also,
4ADC > 4BCD by SSS Theorem. So
/ADC > /BCD. Then all 4 angles of the
parallelogram are congruent, so each must be a
right angle. Therefore, ABCD is a rectangle.
14. No; WX is not parallel to ZY .
15. m/BCD 5 m/BAD 5 70
m/ADC 5 180 2 m/BAD 5 180 2 70 5 110
m/ABC 5 m/ADC 5 110
16. DE 5 BE 5 11
AD 5 BC 5 20
BD 5 2 ? BE 5 2 ? 11 5 22
17. PQRS is a parallelogram, so PQ 5 RS
(Properties of a Parallelogram Theorem).
18. No; each angle of a regular pentagon measures
108, but 360 is not evenly divisible by 108. So a
regular pentagon will not tessellate.
19. Given: 4PTS is isosceles with vertex angle T,
and m/PTQ 5 m/STR.
a. To prove: 4TPQ > 4TSR
Argument:
Conclusions
Justifications
0. 4PTS is
Given
0. isosceles with
0. vertex angle T,
0. and m/PTQ
0. 5 m/STR.
1. PT 5 ST
definition of
isosceles triangle
2. m/P 5 m/S
Isosceles Triangle
Base Angles Theorem
3. 4TPQ > 4TSR
ASA Congruence
Theorem
b. To prove: 4TQR is isosceles.
Argument:
Conclusions
Justifications
0. 4TPQ > 4TSR
from part a
1. TQ 5 TR
CPCF Theorem
2. 4TQR is isosceles.
definition of
isosceles triangle
20. Yes; sample path:
Q to T to P to Q to R to S to T to R
113
21. The conjecture is false. Sample counterexample:
A
D
B
C
LESSON 7-9
pp. 416–422
1. a. x 1 90 1 50 5 180
x 1 140 5 180
x 5 40
b. x 1 y 5 180
40 1 y 5 180
y 5 140
c. z 5 x 5 40
2. m/ABD 5 m/C 1 m/D, by the Exterior Angle
Theorem
3. /2 and /1
4. /1 and /3
5. /2 and /3
6. a. Exterior Angle Theorem: In a triangle, the
measure of an exterior angle is equal to the
sum of the measures of the interior angles at
the other two vertices of the triangle.
b. Exterior Angle Inequality: In a triangle, the
measure of an exterior angle is greater than
the measure of the interior angle at each of
the other two vertices.
7. a. /B is the largest angle, since the side
opposite /B, AC, is the longest.
b. /C is the smallest angle, since the side
opposite /C, AB, is the shortest.
8. a. m/D 1 m/E 1 m/F 5 180
61 1 60 1 m/F 5 180
121 1 m/F 5 180
m/F 5 59
EF is the longest side, since the angle
opposite EF, /D, is the largest.
b. DE is the shortest side, since the angle
opposite DE, /F, is the smallest.
9. a. 77 1 135 1 148 5 360
114
b. Exterior angle at C: 80
at B: 180 2 87 5 93
at A: 180 2 90 5 90
at D: 180 2 83 5 97
Sum: 80 1 93 1 90 1 97 5 360
10. Sample:
U
X 70°
W
V
a. m/V , 70
b. m/U , 180 2 70, so m/U , 110
c. m/UWX , 180 2 70, so m/UWX , 110
11. True
12.
L
shortest
longest
V
middle
U
If m/U 5 60, then m/L . 60 and m/V . 60,
and the sum of the measures of the interior
angles of 4LUV . 180. If m/V 5 60, then
m/L , 60 and m/U , 60, and the sum of the
measures of the interior angles of 4LUV , 180.
If m/L 5 60, then m/U , 60 and m/V . 60,
and it is possible for the sum of the interior
angles of 4LUV to equal 180.
So /L has measure 60.
13. m/GHJ 5 18 1 15 5 33
m/G 5 180 2 (40 1 33) 5 107
In 4GHJ, /GHJ is the smallest angle, so GJ is
the shortest side.
14. a. /3 is an exterior angle of 4PRT. Therefore,
by the Exterior Angle Inequality,
m/3 . m/1.
b. m/2 , m/Q 5 90. Therefore, by the
Unequal Angles Theorem, PQ , PS.
15. a. m/ACB 5 m/BCD 5 x
b. m/CBD 5 m/A 1 m/ACB 5 y 1 x
c. m/A 1 m/ACD 1 m/D 5 180
y 1 2x 1 m/D 5 180
m/D 5 180 2 y 2 2x
UCSMP Geometry © Scott, Foresman and Company
16. m/BCD and m/ACE each equal m/A 1 m/B.
/BCD and /ACE also form vertical angles, so
they have the same measure.
17. ADCE is a parallelogram; Sufficient Conditions
for a Parallelogram Theorem, part (c)
18. Yes; Sufficient Conditions for a Parallelogram
Theorem, part (d)
19. No; congruent sides are not opposite each other.
20. VS 5 QV 5 3x
TS 5 QR 5 y
QS 5 2 ? QV 5 2 ? 3x 5 6x
21. Sample:
a. AB 5 DE
b. HL Congruence Theorem
22. Given: M is the midpoint of AB; /N > /MCB
To prove: M is the midpoint of NC.
Argument:
Conclusions
Justifications
0. M is the midpoint
Given
0. of AB; /N > /MCB
1. AM 5 BM
definition of
midpoint
2. /NMA > /CMB
Vertical Angles
Theorem
3. 4ANM > 4BCM
AAS Congruence
Theorem
4. NM 5 CM
CPCF Theorem
5. M is the midpoint
definition of
0. of NC.
midpoint
23. a. Sample:
sum 5 72 1 72 1 72 1 72 1 72 5 5(72)
sum 5 360
b. Sample:
sum 5 60 1 60 1 60 1 60 1 60 1 60
sum 5 6(60) 5 360
UCSMP Geometry © Scott, Foresman and Company
c. The sum of the measures of the exterior
angles of a convex polygon, one at each
vertex, is 360.
CHAPTER 7
PROGRESS SELF-TEST
pp. 426–427
1. a. m/CBD 5 180 2 m/ABD 5 180 2 60
5 120
b. m/C , 60, by the Exterior Angle
Inequality
c. m/D , 60, by the Exterior Angle
Inequality
2. a. Sample (reduced size):
M
10 cm
L
30°
70°
N
b. Yes
c. by the ASA Congruence Theorem
3. a. Sample: A
1"
100° 45°
B
C
Yes
by the AAS Congruence Theorem
No
The given information is the SSA condition
but not SsA Congruence, so it does not yield
a unique triangle.
5. a. Yes
b. SSS Congruence Theorem
6. AAS Triangle Congruence Theorem: If two
angles and a non-included side of one triangle
are congruent to two angles and the
corresponding non-included side of a second
triangle, then the triangles are congruent.
7. a. 4ABC > 4CDA by the ASA Congruence
Theorem.
b. 4WVU > 4ZXY by the HL Congruence
Theorem.
b.
c.
4. a.
b.
115
8. a. 4ABC > 4CDA
b. ASA Congruence Theorem
@#$ k CD
@##$
9. Given: M is the midpoint of AC; AB
To prove: 4MBA > 4MDC
Argument:
Justifications
Conclusions
0. M is the midpoint
Given
@#$ k CD
@##$
0. of AC; AB
1. AM > CM
definition of
midpoint
2. /AMB > /CMD
Vertical Angles
Theorem
3. /BAM > /DCM
k Lines ⇒ AIA >
Theorem
4. 4MBA > 4MDC
ASA Congruence
Theorem
10. Given: WX 5 WY ; /WUY > /WVX
To prove: 4WUV is isosceles.
Argument:
Conclusions
Justifications
0. WX 5 WY ;
Given
0. /WUY > /WVX
1. /W > /W
Reflexive Property
of Congruence
2. 4UWY > 4VWX
AAS Congruence
Theorem
3. WU > WV
CPCF Theorem
4. 4WUV is isosceles.
definition of
isosceles triangle
11. Since AB 5 AC, AD 5 AD, and /ADB and
/ADC are both right angles, 4ADB > 4ADC
by the HL Congruence Theorem. Thus,
BD 5 CD by the CPCF Theorem.
12. ABCD is a parallelogram because one pair of
opposite sides is parallel and congruent.
13. OR 5 PO 5 x
LR 5 AP 5 y
PR 5 2 ? PO 5 2x
14. Yes, ABCD is a parallelogram because the
diagonals bisect each other.
15. Sample (reduced size):
16. (f ) Y Z;
m/XWY 5 180 2 (34 1 97) 5 180 2 131 5 49,
so the shortest segment in 4WXY is WY .
m/WY Z 5 180 2 97 5 83, and m/ZWY 5
180 2 (83 1 55) 5 180 2 138 5 42, so the
shortest segment in 4WZY is Y Z.
So Y Z must be the shortest segment in the
figure.
CHAPTER 7
REVIEW
pp. 428–432
1. a. No
b. There are many noncongruent triangles that
fit the given information.
2. a. Yes
b. SSS Congruence Theorem
3. a. Yes
b. ASA Congruence Theorem
4. a. Yes
b. AAS Congruence Theorem
5. a. Sample: O
O
N
M
5 cm
b. No. There are many noncongruent triangles
that fit the given information.
6. a. Sample:
Q
1"
P
1"
P
1.5"
30° R
b. No. There are two noncongruent triangles
that fit the given information.
7. a. Sample:
T
2 cm
S 60°
4 cm
U
b. Yes, by the SsA Congruence Theorem.
116
UCSMP Geometry © Scott, Foresman and Company
8. a. Sample:
W
2" 55°1"
V
X
b. Yes, by the SAS Congruence Theorem.
9. a. m/QSR 5 180 2 m/QST 5 180 2 132 5 48
b. m/Q , 132, by the Exterior Angle
Inequality
c. m/Q 1 m/R 5 m/QST 5 132, by the
Exterior Angle Theorem
10. a. q 1 2q 5 138
3q 5 138
q 5 46
m/X 5 q 5 46
b. m/Y 5 2q 5 2(46) 5 92
11. /1 is the largest.
/1 is an exterior angle of 4TWX, 4TWY, and
4TWZ. Therefore, m/1 . m/2, m/1 . m/3,
and m/1 . m/4.
12. m/2 5 360 2 (m/A 1 m/B 1 m/C) 5
360 2 (110 1 100 1 90) 5 360 2 300 5 60
m/1 5 180 2 m/2 5 180 2 60 5 120
13. 4MOP > 4MNP by the AAS Congruence
Theorem
14. not enough information to tell
15. 4KLM > 4KJM by HL Congruence Theorem
16. not enough information to tell
17. 4ACB > 4DFG by AAS Congruence Theorem
18. not enough information to tell
19. Sample:
a. AC > DF
b. SSS Congruence Theorem
20. Sample:
a. HI > KL
b. ASA Congruence Theorem
UCSMP Geometry © Scott, Foresman and Company
21. Given: /DAC > /BAC; /DCA > /BCA
To prove: 4ADC > 4ABC
Argument:
Conclusions
Justifications
0. /DAC > /BAC;
Given
0. /DCA > /BCA
1. AC > AC
Reflexive Property
of Congruence
2. 4ADC > 4ABC
ASA Congruence
Theorem
22. Given: WZ ⊥ ZY ; WX ⊥ XY ; WZ 5 WX
To prove: 4WZY > 4WXY
Argument:
Conclusions
Justifications
0. WZ ⊥ ZY ; WX ⊥ XY ;
Given
0. WZ 5 WX
1. /Z and /X are
definition of
0. right angles.
perpendicular
2. WY > WY
Reflexive Property
of Congruence
3. 4WZY > 4WXY
HL Congruence
Theorem
23. Given: (P and (Q intersect at A and B.
To prove: 4APQ > 4BPQ
Argument:
Conclusions
Justifications
0. (P and (Q
Given
0. intersect at A and B.
1. PA 5 PB; QA 5 QB
definition of circle
2. PQ 5 PQ
Reflexive Property
of Equality
3. 4APQ > 4BPQ
SSS Congruence
Theorem
####$
24. Given: UW bisects /Y UV; UW > UY ;
/V > /UXY
To prove: 4UVW > 4UXY
Argument:
Conclusions
Justifications
####$ bisects /Y UV;
0. UW
Given
0. UW > UY ;
0. /V > /UXY
1. /Y UX > /WUV
definition of angle
bisector
2. 4UVW > 4UXY
AAS Congruence
Theorem
117
25. Cannot be proved. This is the SSA condition,
but the longer congruent sides are not opposite
the congruent angles. So the SsA Congruence
Theorem does not apply.
26. Given: AB > DC; /ABC > /DCB
To prove: 4ACB > 4DBC
Argument:
Conclusions
Justifications
0. AB > DC;
Given
0. /ABC > /DCB
1. BC > BC
Reflexive Property
of Congruence
2. 4ACB > 4DBC
SAS Congruence
Theorem
27. Given: AB > AC; BD > DC
To prove: /BAD > /CAD
Argument:
Conclusions
Justifications
0. AB > AC; BD > DC
Given
1. AD > AD
Reflexive Property
of Congruence
2. 4ABD > 4ACD
SSS Congruence
Theorem
3. /BAD > /CAD
CPCF Theorem
28. Given: ABCDEFGH is a regular octagon.
To prove: AC > BD
Argument:
Conclusions
Justifications
0. ABCDEFGH is a
Given
0. regular octagon.
1. AB > BC > CD
definition of
regular polygon
2. /ABC > /DCB
definition of
regular polygon
3. 4ABC > 4DCB
SAS Congruence
Theorem
4. AC > BD
CPCF Theorem
##$ bisects /MJL; MJ > LJ
29. Given: JK
To prove: /M > /L
Argument:
Conclusions
Justifications
##$
0. JK bisects /MJL;
Given
0. MJ > LJ
1. /KJM > /KJL
definition of angle
bisector
2. JK > JK
Reflexive Property
of Congruence
118
3. 4KJM > 4KJL
SAS Congruence
Theorem
4. /M > /L
CPCF Theorem
30. Given: AD ⊥ DC; AB ⊥ BC; AD > BC
To prove: AB > CD
Argument:
Conclusions
Justifications
0. AD ⊥ DC; AB ⊥ BC;
Given
0. AD > BC
1. /D and /B are
definition of
0. right angles.
perpendicular lines
2. AC > AC
Reflexive Property
of Congruence
3. 4ADC > 4CBA
HL Congruence
Theorem
4. AB > CD
CPCF Theorem
31. Given: N is the midpoint of OE; l k m
To prove: AE > UO
Argument:
Conclusions
Justifications
0. N is the midpoint
Given
0. of OE; l k m
1. ON > NE
definition of
midpoint
2. /NUO > /NAE;
k lines ⇒ AIA >
0. /NOU > /NEA
Theorem
3. 4NUO > 4NAE
AAS Congruence
Theorem
4. AE > UO
CPCF Theorem
32. Given: /SPQ > /RQP; /S > /R
To prove: QS > PR
Argument:
Conclusions
Justifications
0. /SPQ > /RQP;
Given
0. /S > /R
1. PQ > PQ
Reflexive Property
of Congruence
2. 4SPQ > 4RQP
AAS Congruence
Theorem
3. QS > PR
CPCF Theorem
33. DE 5 BE 5 12
AD 5 BC 5 19
BD 5 2 ? BE 5 2 ? 12 5 24
34. AE 5 CE 5 3x
AD 5 BC 5 y 2 2
AC 5 2 ? CE 5 2 ? 3x 5 6x
UCSMP Geometry © Scott, Foresman and Company
35. m/DCB 5 m/DAB 5 130
m/ADC 5 180 2 m/DAB 5 180 2 130 5 50
m/ABC 5 m/ADC 5 50
36. Z
W
49. Sample:
C
Y
X
37. Yes, since both pairs of opposite sides are
congruent.
38. Not necessarily, since the diagonals may or may
not bisect each other.
39. Not necessarily. For example, ABCD could be
an isosceles trapezoid with bases AB and CD,
and AB Þ CD.
40. Yes, since both pairs of opposite angles are
congruent.
41. /A is the largest angle, since BC is the
longest side.
42. /C is the smallest angle, since AB is the
shortest side.
43. m/F 5 180 2 (m/D 1 m/E) 5
180 2 (47 1 68) 5 180 2 115 5 65
sides from shortest to longest: EF, ED, DF
44. In 4GHJ, HJ is the shortest side. But HJ is
also in 4HJI, where HI is the shortest side. So
the shortest segment in the figure is HI.
45. 4ARP > 4ABP by the HL Congruence
Theorem. So, by the CPCF Theorem, RP 5 PB.
46. The SSS Congruence Theorem states that
triangles with all three sides congruent are
congruent.
47. 4XBD > 4XBA > 4XBC by the AAS
Congruence Theorem, so DB 5 AB 5 CB by the
CPCF Theorem.
48. a. QR and ZP have lengths equal to the width
of the river.
b. 4ZXP > 4RXQ by the ASA Congruence
Theorem, so QR > ZP by the CPCF
Theorem.
UCSMP Geometry © Scott, Foresman and Company
50. Sample:
51. Sample:
52. No; each angle in a regular octagon measures
135, but 360 is not evenly divisible by 135. So a
regular octagon will not tessellate.
53. ABCD is approximately in the shape of a
parallelogram.
54. Let their intersection be the midpoint of both PQ
and MN.
119
55. AB 5 CD, and CD 5 EF by the Properties of a
Parallelogram Theorem (opposite sides are
congruent). Therefore, AB 5 EF by the
Transitive Property of Equality.
56. Yes; by the Properties of a Parallelogram
Theorem (opposite angles are congruent).
CHAPTER 7
REFRESHER
p. 433
1. 6(2 1 x) 5 12 1 6x
2. x(4 2 y) 5 4x 2 xy
3. (h 1 l)a 5 ah 1 al
1
2
1
2
1
2
4. b(w 1 t) 5 bw 1 bt
5. (a 2 2b)b 5 ab 2 2b2
6. 2xy(y 1 x) 5 2xy2 1 2x2y
7. ax 1 2x 5 x(a 1 2)
8. 9c2 2 12c 5 3c(3c 2 4)
9. 4b 2 20h 5 4(b 2 5h)
10. pr 2 1 ph2 5 p(r 2 1 h2)
1
2
1
2
1
2
11. bh 1 lh 5 h(b 1 l)
12. 12ar 2 1 6ar 5 6ar(2r 1 1)
13. (y 1 2)(y 1 4) 5 y2 1 2y 1 4y 1 8
5 y2 1 6y 1 8
14. (a 2 7)(3b 1 2) 5 3ab 2 21b 1 2a 2 14
15. (2r 1 t)(2r 2 t) 5 4r 2 1 2tr 2 2rt 2 t2
5 4r 2 2 t2
16. (s 1 4)2 5 (s 1 4)(s 1 4) 5 s2 1 4s 1 4s 1 16
5 s2 1 8s 1 16
2
17. (m 1 n) 5 (m 1 n)(m 1 n) 5
m2 1 mn 1 mn 1 n2 5 m2 1 2mn 1 n2
120
1
1
2
2
1
1
1
1
ey 1 eh 1 f y 1 f h
2
2
2
2
18. (e 1 f )(y 1 h) 5 (ey 1 eh 1 f y 1 f h) 5
p
19. 2 < 1.41
p
20. 4 3 < 6.93
p
p
p
21. 42 1 52 5 16 1 25 5 41 < 6.40
22.
s
17
< 2.06
4
p
23. 100 5 10
p
p
p
24. 40 5 4 ? 10 5 2 10
p
p
p
25. 18 5 9 ? 2 5 3 2
p
p
p
p
26. 2 45 5 2 9 ? 5 5 2 ? 3 5 5 6 5
27. x2 5 25
x 5 65
28. y2 1 9 5 25
y2 5 16
y 5 64
29. 40 5 2z2
2z2 5 40
z2 5 20p
p
z 5 6 20 5 62 5 < 64.47
30.
3 5 pr 2
pr 2 5 3
r2 5
3
p
r56
s
3
< 60.98
p
UCSMP Geometry © Scott, Foresman and Company