Download Geometry: A Complete Course

Geometry:
A Complete Course
(with Trigonometry)
Module D – Instructor's Guide
with Detailed Solutions for
Progress Tests
Written by: Larry E. Collins
A
T
RA
ER /2010
12
Quiz Form A
Name
Class
Date
Score
Unit IV - Triangles
Part A - Basic Definitions
Lesson 1 - Triangle Parts
Lesson 2 - Triangle Types
X
N P
Use the diagram to the right to answer each of problems 1-6.
A
Y
L
R
Q
T M
1. Names the vertices of the triangle.
Point L, Point M, Point N
_______________________________________
2. Name the sides of the triangle.
LM, MN, LN
_______________________________________
3. Name the angles of the triangle.
/NLM, /NML, /MNL
_______________________________________
/XMQ (/NMQ), /PNT (/PNM), /RLQ (/RLM)
4. Name the exterior angles of the triangle.
/TML (/TMY), /XNR (/XNL), /PLY (/NLY)
_______________________________________
5. For /NLM:
a) Name the two sides of the angle
LN, LM
_______________________________________
b) Name the two side of the triangle that
include the angle.
LN, LM
_______________________________________
NM
c) Name the side of the triangle opposite the angle. _______________________________________
6. For MN:
/MNL, /NML
a) Name the two angles of the triangle that include this side. ________________________________
b) Name the angle of the triangle which is opposite this side. ________________________________
/NLM
© 2006 VideoTextInteractive Geometry: A Complete Course
1
Quiz Form A
Name
Class
Date
Score
Unit IV - Triangles
Part B - Basic Theorems
Lesson 1 - Theorem 25: “If you have any given triangle, then the
sum of the measures of its angles is 180”
Lesson 2 -
Theorem 26: “If you have a given exterior angle of a
triangle, then its measure is equal to the sum of the measure of the
two remote interior angles.”
1. The vertex angle of an isosceles triangle measures 50O.
Base angle: _______________
65O
The base angles are congruent. Sketch a figure illustrating
this triangle. Then find the measures of the base angles.
50
(x)
x + x + 50 = 180
2x = 130
x = 65
(x)
2. The measure of one angle of a triangle is three times the measure
of another and the third angle’s measure is equal to their sum.
Sketch a figure illustrating this triangle. Then find the measures of
all three angles.
x + 3x + (x + 3x) = 180
8x = 180
x = 22.5
(3x)
(x + 3x)
67.5O
First angle: _________
22.5O
Second angle: _________
90O
Third angle: _________
3x = 67.5
x + 3x = 90
(x)
3. In nABC, the exterior angle at C measures 130O, and m/A = 50O.
80O
Measure /B:_________
Sketch a figure illustrating this triangle. Find m/B.
A
m/ACD = m/A + m/B
130 = 50 + m/B
130
B
C
80 = m/B
D
© 2006 VideoTextInteractive Geometry: A Complete Course
7
Unit IV, Part B, Lessons 1&2, Quiz Form A
—Continued—
Name
C
4. The side AB of nABC is extended to K. The bisector of /A
intersects BC and D, and meets the bisector of /CBK outside
nABC at P. Also, m/A = 60O and m/ABC = 70O. Sketch a figure
25O
illustrating this situation. Find the measure of /P:_______________.
m/P is 1/2 m/C
30
How do the measures of /C and /P compare?________________
30
_________________________________________________________
A
_________________________________________________________
First, find m/C:
m/A + m/ABC + m/C = 180
60 + 70 + m/C = 180
m/C = 50
Second, find m/BDP
m/CAD + m/ACD + m/CDA = 180
30 + 50 + m/CDA = 180
m/CDA = 100
m/BDP = 100
Third, find m /CBP
m/ABC + m/CBK = 180
70 + m/CBK = 180
m/CBK = 110
/2 m/CBK = m/CBP = 55
1
•
Finally, find m/P
m/BDP + m/CBP + m/P = 180
100 + 55 + m/P = 180
m/P = 25
8
© 2006 VideoTextInteractive Geometry: A Complete Course
P
50
100
D
100
55
55
70
B
K
Unit IV, Part B, Lessons 1&2, Quiz Form A
—Continued—
Name
In problems 5-10, find the value of x in each of the accompanying figures,
using the given information.
A
A
B
40
D
(x)
5.
x = ________
6.
100
x = ________
34
(x)
C
D
34
C
AB || CD
B
m/C = 40O
m/A + m/B + m/ACB = 180
(Lines ||, Alternate Interior /’s >)
m/A + 34 + 90 = 180
m/D = 40O
m/A = 56
m/C + m/D + x = 180
40 + 40 + x = 180
x = 100
m/A + x + m/ADC = 180
56 + x + 90 = 180
x = 34
P
7.
A
A
(x)
44
E
x = ________
8.
120
T
B
78
x = ________
42
28
150
R
Q
(x)
32
D
C
m/TQR + m/TRQ + m/T = 180
m/BCA + m/ACD = m/BCD
m/TQR + m/TRQ + 150 = 180
28 + 32 = m/BCD
m/TQR + m/TRQ = 30
60 = m/BCD
m/TQR = m/TRQ (so each angle is 15)
m/DBC + m/BCD + x = 180
78 + 60 + x = 180
138 + x = 180
x = 42
m/TQR = m/TRQ = m/PQT = m/PRT
m/TQR + m/TQP = m/PQR
15 + 15 = 30
m/TRQ + m/TRP = m/PRQ
15 + 15 = 30
m/PQR + m/PRQ + x = 180
30 + 30 + x = 180
x = 120
9.
(x)
A
90
x = ________
10.
67
23
x = 67 + 23
x = 90
53
x = ________
(x)
C
164
D
164 = m/BAC + m/ACB
164 = m/BAC + 90
B
x + m/CAD + m/ACD = 180
x + 37 + 90 = 180
x = 53
74 = m/BAC
m/BAC = m/CAD + m/DAB
74 = m/CAD + m/DAB
m/CAD = m/DAB
m/CAD = 37
© 2006 VideoTextInteractive Geometry: A Complete Course
9
Unit IV, Part C, Lesson 3, Quiz Form B
—Continued—
Name
⋅
4. Two corresponding sides of two similar pentagons measure 4 and 6. If the perimeter of the larger
pentagon is 15, find the perimeter of the smaller pentagon. _______________
10
4 x
=
6 15
6 ⋅ x = 4 ⋅ 15
4 ⋅ 15 2 ⋅ 2 ⋅ 3 ⋅ 5
=
6
2⋅3
x = 10
x=
5. The sides of a pentagon measure 7, 8, 10, 11, and 12 inches respectively. Find the perimeter of
56 inches
a similar pentagon if its longest side measures 14 inches. _______________
Perimeter : 7 +8 +10 +11+12
Compare longest side to longest side.
12 48
=
14 x
14 ⋅ 48 = 12 ⋅ x
672= 12x
672 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 7 8 ⋅ 7
=
=
= 56
12
2⋅2⋅3
1
P
6. Two similar hexagons have perimeters of 80 and 120 meters respectively. If one side of the1smaller
22 meters
hexagon is 15, find the measure of the corresponding side of the larger hexagon. ___________
2
80 15
=
120 x
120 ⋅ 15 = 80 ⋅ x
120 ⋅ 15
=x
80
2⋅2⋅2⋅3⋅5 ⋅3⋅5
=x
2⋅2⋅2⋅2⋅5
45
1
or 22 = x
2
2
© 2006 VideoTextInteractive Geometry: A Complete Course
29
Unit IV, Part D, Lesson 1, Quiz Form B
—Continued—
Name
J
17.
Given: Right nHJK with right angle HJK.
JL
HK; /K = /HJL
H
Prove: nJKL ~ nHJL
K
L
STATEMENT
REASON
1. Right nHJK with Right Angle HJK.
1. Given
2. JL
2. Given
HK
3. /HLJ is a right angle
3. Definition of Perpendicular Lines.
4. /JLK is a right angle
4. Definition of Perpendicular Lines.
5. /HLJ > /JLK
5. Theorem 11 – If you have right angles,
then those right angles are congruent.
6. /K > /HJL
6. Given
8. nJKL ~ nHJL
8. Postulate Corollary 12a – If two angles of one
triangle are congruent to the two corresponding
angles of another triangle, then the triangles
are similar. (AA Postulate Corollary)
G
T
J
H
18.
Given: GH HJ; JK
Prove: HT
GT
=
JT
KT
HJ
K
STATEMENT
1. GH
HJ; JK
2. GH
JK
HJ
REASON
1. Given
2. Theorem 22 – If two lines are perpendicular to
a third line, then the two lines are parallel.
3. /GHT > /KJT
3. Theorem 16 – If two parallel lines are cut by a
transversal, the alternate interior angles
are congruent.
4. /HTG > /JTK
4. Theorem 15 – If two lines intersect, then the
vertical angles formed are congruent.
5. nGTH ~ nKTJ
5. Postulate Corollary 12a – If two angles of one
triangle are congruent to the two corresponding
angles of another triangle, then the triangles
are similar. (AA Postulate Corollary)
6. HT GT
=
JT
KT
6. If two polygons are similar, then the measures of
corresponding sides are proportional.
© 2006 VideoTextInteractive Geometry: A Complete Course
37
Unit IV, Part D, Lesson 2, Quiz Form A
—Continued—
Name
In the figure to the right, AY || EO || RB. Use this figure to complete each of the proportions in
problems 11 - 14.
11.
12.
YO AE
=
?
OB
YB
?
=
OB ER
ER
? = ____________
D
Y
AR
? = ____________
O
A
E
R
B
13.
?
YB
=
AE YO
AR
? = ____________
14.
DY DA
=
YO
?
AE
? = ____________
Using the figure to the right, and the given information in
problems 15 and 16 to determine if QT || PS. Answer yes or no.
15.
no
PR = 30, PQ = 9, RT = 12, RS = 18 Answer: ________
PR
SR
=
PQ
ST
30 18
=
9 ST
RS – RT =ST
3 ⋅10 3 ⋅6
=
3 ⋅3 1⋅6
10 3
≠
3 1
R
Q
P
T
S
NO!
18 –12=6
30 18
=
9 6
16.
RP = 13.5, RQ = 6.3, TR = 4.2, SR = 9.0
RP
RS
=
RQ
RT
13.5 9.0
=
6.3 4.2
135 90
=
63 42
yes
Answer: ________
3 ⋅45 2 ⋅45
=
3 ⋅21 2 ⋅21
45 45
=
21 21
YES!
© 2006 VideoTextInteractive Geometry: A Complete Course
41
Unit IV, Part D, Lesson 3, Quiz Form B
—Continued—
Name
Use the diagram to the right, and the given
information in problems 7 and 8, to find x.
7. nABE ~ nAEC
x=
16
=
3
__________
1
D
F
3 4
m/1 = m/2
E
BF AB
=
EB AE
3 4
=
4 AE
4 4 =3 ⋅AE
( )( )
16=3 ⋅AE
16
= AE
3
8. nCEA ~ nCBE
9
__________
5
x=
B is the midpoint of AC
D is the midpoint of CE
BD = 4, CE = 6, CB = 5, BE = x + 3
(Note: Use only these segments to solve)
DB CB
=
BE CE
4 5
=
x+3 6
x+3 ⋅5=4 ⋅6
(
)
5x+15=24
5x=9
9
x=
5
© 2006 VideoTextInteractive Geometry: A Complete Course
C
2
m/3 = m/4
AB = 4, FB = 3, EB = 4, AE = x
52
B
A
Unit IV, Part D, Lesson 4, Quiz Form A
—Continued—
Name
M
P
For problems 3 – 11, use the right triangle
shown to find the missing value.
Q
N
6
3. PQ = 9, QN = 4, MQ = ____________
4.
9 MQ
=
MQ 4
(
27 =x
QN = 3, MQ = 9, PQ = ____________
x 9
=
9 3
9 9 =x ⋅3
( )( )
(MQ )(MQ )=9 ⋅4
(MQ )2 =36
81=x ⋅3
27 =x
MQ =6
(
32
16= PN
5. PM = 12, PQ = 9, PN = ____________
6.
6 8
=
8 PN
8 ⋅8=6 ⋅ PN
64=6 ⋅PN
64
= PN
6
2 ⋅32
= PN
2 ⋅3
32
= PN
3
9 12
=
12 PN
12 12 =9 ⋅PN
( )( )
144=9 ⋅PN
16= PN
2
7. PN = 75, PQ = 72, MN =
QN = PN - PQ
QN =3
(
3 MN
=
MN 75
MN MN =3 ⋅75
)( )
(MN )2 =225
15
____________
3
MN = 8, QN = 6, PN = ____________
8.
MQ = 4, PQ = 10, QN =
MN = 3 ⋅3 ⋅5 ⋅5
10 4
=
4 QN
MN = 3 2 ⋅5 2
4 ⋅4=10 ⋅QN
16=10 ⋅QN
MN = 3 2 ⋅ 5 2
MN =3 ⋅5
MN =15
16
=QN
10
2 ⋅2 ⋅2 ⋅2
=QN
2 ⋅5
8
=QN
5
MN = 225
M
9. PN = 13, PM = 12, PQ =
144
=
13
____________
10.
13 12
=
12 PQ
12 ⋅12=13 ⋅PQ
144=13 ⋅ PQ
144
= PQ
13
PQ QM
=
QM QN
PN -QN = PQ
16 -4=12
12 QM
=
QM 4
(QM )(QM )=12 ⋅4
Q
© 2006 VideoTextInteractive Geometry: A Complete Course
4
⋅
4 3
PN = 16, QN = 4, QM =2 ____________
QM 2 =48
56
8
=
5
____________
⋅
QM = 2 ⋅2 ⋅2 ⋅2 ⋅3
QM = 2 2 ⋅2 2 ⋅3
QM = 2 2 ⋅ 2 2 ⋅ 3
QM = 2 ⋅2 ⋅ 3
QM =4 3
Unit IV, Part D, Lesson 4, Quiz Form A
—Continued—
Name
A
D
13.
Given: nABC is a right triangle
BD AC
DE BC
Prove: nABC ~ nDEC
B
STATEMENT
58
E
C
REASON
1. nABC is a right triangle
2. BD AC
3. BD is an altitude
4. nADB ~ nBDC
1.
2.
3.
4.
5. /BDC is a right angle
6. nBDC is a right triangle
7. DE BC
8. nDEC ~ nBDC
9. nADB ~ nDEC
5.
6.
7.
8.
9.
© 2006 VideoTextInteractive Geometry: A Complete Course
Given
Given
Definition of Altitude
Theorem 30 - If you have the altitude to the hypotenuse of a
right triangle, then it forms two triangles that are similar to
each other, and to the original triangle.
Definition of Perpendicular Lines
Definition of Right Triangles
Given
Theorem 30
Postulate Corollary 12c - If two triangles are similar to a third
triangle, then the two triangles are similar to each other.
Unit IV, Part D, Lesson 4, Quiz Form B
—Continued—
Name
In problems 3 - 5, use the diagram shown, and find the length of the altitude to the hypotenuse.
R
3.
R
P
15
R
4.
S 3 Q
5.
12
4
Q
30
S
RS =
4 3
_______
QS RS
From
=
RS SP
3 5
SP = _______
P
6
RS SP
From
=
SP SQ
4 RS
=
RS 12
RS RS =4 ⋅12
P
Q
15 SP
=
SP 3
SP SP =15 ⋅3
( )( )
( RS )2 =48
QS = _______
12
S
From
( )( )
(SP )2 =45
SP QS 6 QS
=
=
QS RS QS 24
(QS )(QS )=6 ⋅24
(QS )2 =144
RS = 2 ⋅2 ⋅2 ⋅2 ⋅3
SP = 45
RS = 2 2 ⋅2 2 ⋅3
SP = 9 ⋅5
QS = 144
SP = 9 ⋅ 5
QS =12
RS = 2 2 ⋅ 2 2 ⋅ 3
SP =3 5
RS =2 ⋅2 ⋅ 3
RS =4 3
In problems 6 - 8, use the diagram shown, and find the length of each leg of the right triangle.
T
R
6.
7.
8.
15
M
From
3
N
10
NT = _______
NM = _______
10 3
RT NT
=
NT TM
5 NT
=
NT 20
NT NT =5 ⋅20
From
T
(
)( )
( NM )2 =300
NM = 300
NM = 3 ⋅ 100
NM =10 3
R
M
32
6R
NT = _______
NM = _______
6 3
( NT )( NT )=3 ⋅12
( NT )2 =36
From
(
NM = 2 2 ⋅ 3 2 ⋅ 3
NM =2 ⋅3 ⋅ 3
NM =6 3
T
16
NT = _______
NM = _______
16 3
NT = 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2
NT = 2 8
NT =2 4
NT =16
)( )
( NM )2 =108
NM = 2 ⋅2 ⋅3 ⋅3 ⋅3
3
8
( NT )( NT )=8 ⋅32
( NT )2 =8 ⋅32
9 NM
=
NM 12
NM NM =12 ⋅9
NM = 108
R
RT NT 8 NT
=
=
NT MT NT 32
NT = 36
NT =6
NT = 100
NT =10
15 NM
=
NM 20
NM NM =15 ⋅20
9
RT NT 3 NT
=
=
NT MT NT 12
( )( )
( NT )2 =100
N
M
N
5
(
24 NM
=
NM 32
NM NM =24 ⋅32
)( )
( NM )2 =24 ⋅32
NM = 2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅2 ⋅3
NM = 2 8 ⋅3
NM = 2 4 3
60
© 2006 VideoTextInteractive Geometry: A Complete Course
NM =16 3
Unit IV, Part D, Lesson 4, Quiz Form B
—Continued—
Name
E
Use the figure at the right for problems 9 – 12.
H
Q
G
9. EG = 3, EH = 4, HG = 5. EQ =
12
__________
5
10.
2 or 8
EQ = 4, GH = 10. GQ = __________
GQ
EQ
GQ
4
x
4
4 ⋅4
VEQG ∼VHEG
EQ
EG
=
EH HG
EQ 3
=
4
5
3 ⋅ 4 = EQ ⋅ 5
12 = 5 ⋅ EQ
EQ
HQ
4
=
10 − GQ
4
=
10 - x
= x 10 - x
=
x 2 – 10 x + 16 = 0
( x - 2)( x - 8 ) = 0
x-2=0
x=2
HG = 15, EG = 9, EQ =
VGQE ∼VGEH
EQ EG
=
EH GH
36
5 = 9
EH 15
36 15
⋅
= 9 ⋅ EH
5 1
36 ⋅ 5 ⋅ 3
= 9 ⋅ EH
5
108 = 9 ⋅ EH
108
= EH
9
9 ⋅ 12
= EH
9
12 = EH
)
16 = 10 x - x 2
12
= EQ
5
11.
(
36
12
. EH = __________
5
12.
EQ = 8,
or
or
x-8=0
x=8
HQ 2
= . HQ = __________
8 2
QG 1
HQ EQ
=
EQ QG
2x 8
=
8
x
8 ⋅ 8 = 2x ⋅ x
64 = 2 x 2
32 = x 2
32 = x
16 ⋅ 2 = x
16 ⋅ 2 = x
8 2=x
© 2006 VideoTextInteractive Geometry: A Complete Course
61
Name
Quiz Form A
Class
Date
Score
Unit IV - Triangles
Part D - Similarity – Part 2 (Triangles and Their Parts)
Lesson 5 - Theorem 31: “If you have a given right triangle,
then the square of the measure of the hypotenuse, is equal
to the sum of the squares of the measures of the two legs.”
(The Pythagorean Theorem)
Lesson 6 - Applying Pythagoras to 3-Dimensional Figures
For problems 1 – 5, find x in the given figures. Express radicals in simplest form.
2
1.
2
2
( 3) +5
2.
2
a +b =c
2
x
15
2
=x
2
3 + 25 = x
2
28 = x
2
3
2
2
2
2
a +b =c
11
x
9 + x = 11
2
2
12
x = 40
2 7 =x
2
2
( 7 ) + ( 11 ) = x
2
7
40
x=
4 ⋅ 10
x = 2 10
a +b =c
x
x=
2 10
x = _______
4 ⋅7 = x
3.
2
81 + x = 121
3
28 = x
2 7 =x
x = _______
2
2
4.
2
2
5
2
2
18 = x
2
2
2
2
2
5 + x = 13
13
7 + 11 = x
2
a +b =c
x
2
2
25 + x = 169
2
x = 144
x = 12
11
18 = x
3 2
x = _______
9 ⋅2 = x
12
x = _______
3 2 =x
5.
5
2
2
x
10
a +b =c
2
2
5 + 10 = x
2
2
2
25 + 100 = x
2
125 = x
2
25 ⋅ 5 = x
5 5 =x
x = _______
5 5 =x
© 2006 VideoTextInteractive Geometry: A Complete Course
63
Name
Quiz Form A
Class
Date
Score
Unit IV - Triangles
Part E - Congruence - Part 1 (General Geometric Relationship)
Lesson 1 - Definition
nXYZ > nUVW
Use the pair of triangles to the right, and the
given information, for problems 1 – 8.
Y
X
W
Z
U
V
1. Name the three pairs of corresponding angles. _________________________________________
/X corresponds to /U; /Y corresponds to /V;
/Z corresponds to /W. Note: /YXZ corresponds to /VUW is also a correct response. However “/YXZ corresponds to
__________________________________________________________________________________
/WUV” is an incorrect response since corresponding vertices are not in the correct order.
__________________________________________________________________________________
XY corresponds to UV; YZ corresponds to VW;
2. Name the three pairs of corresponding sides. __________________________________________
ZX
corresponds to WU. Note: XY corresponds to VU is an incorrect response, technically speaking.
__________________________________________________________________________________
Determine whether each statement is correct or incorrect.
3. nYZX > nVWU
correct (corresponding parts are congruent)
_______________________________________________
4. nXZY > nUVW
incorrect (this could be correct, but we don’t know the specific measures.)
_______________________________________________
5. XY > VU
incorrect
(because the names do not reflect the corresponding parts, this is
_______________________________________________
technically incorrect. However, the two segments are congruent, so the
answer could also be “correct”)
6. /Y > /V
(these are corresponding angles)
correct
_______________________________________________
7. Which statement is correct?
b
______________
a) nYZX > nUVW
b) nYZX > nVWU
c) nZXY > nWVU
8. Which statement is not correct?
c
______________
a) XZ > UW
b) /Y > /V
c) XZ > UV
© 2006 VideoTextInteractive Geometry: A Complete Course
71
Unit IV, Part E, Lesson 1, Quiz Form A
—Continued—
Name
A
E
Use the pair of triangles to the right,
for problems 9 – 14.
B
9. AD >
CE
_______________
D
10. BE >
BD
_______________
/CBE
11. /ABD > _______________
12. BC >
BA
_______________
nBEC
13. nBDA > _______________
nADB
14. nCEB > _______________
Given that nMOP > nHAT, complete each statement in problems 15 – 20.
15.
/H
/M > _______________
16.
AT
OP > _______________
17.
m/T
m/P = _______________
18.
HT
MP > _______________
19.
/O
/A > _______________
20.
MO
HA > _______________
21. Sketch and label a pair of obtuse triangles to illustrate the definition of congruent triangles.
Mark all six relationships on the triangles. (answers will vary)
72
© 2006 VideoTextInteractive Geometry: A Complete Course
C
Unit IV, Part E, Lesson 1, Quiz Form A
—Continued—
Name
A
22.
Given: AB > AD
AC BD
AC bisects BD
B
C
D
AC bisects /A
Prove: nABC > nADC using the definition of congruent triangles.
STATEMENT
1. AB > AD
2. AC > AC
3. AC bisects BD
4. BC > DC
5. AC BD
6. /ACB is a right angle
7. /ACD is a right angle
8. /ACB > /ACD
9. AC bisects /A
10. /BAC > /DAC
11. /ABC > /ADC
12. nABC > nADC
REASON
1.
2.
3.
4.
5.
6.
7.
8.
Given
Reflexive Property for Congruence
Given
Definition of Segment Bisector
Given
Definition of Perpendicular
Definition of Perpendicular
Theorem 11 - If you have right angles, then those right angles
are congruent.
9. Given
10. Definition of Angle Bisector
11. Corollary 25a - If two angles of one triangle are congruent to
two angles of another triangle, then the third pair of triangles
are congruent.
12. Two triangles are congruent if and only if there is a
correspondence between the vertices such that each pair of
corresponding sides and each pair of corresponding angles
are congruent.
© 2006 VideoTextInteractive Geometry: A Complete Course
73
Name
Quiz Form A
Class
Date
Score
Unit IV - Triangles
Part E - Congruence - Part 1 (General Geometric Relationship)
Lesson 2 - Postulate 13: Triangle Congruence
Lesson 3 - Congruence Postulate Corollaries
For problems 1 – 4, state the given congruence relationship in your own words and label the
diagram appropriately.
If one leg and acute angle of one right triangle are
1. Postulate Corollary 13c – LA Postulate Corollary: _______________________________________
congruent to the corresponding leg and acute angle of another right triangle, then the triangles are congruent.
___________________________________________________________________________________
C
D
A
N
C
B
F
Note: Labeling may vary, but points
must correspond
M
P
If two angles and a non-included side of one triangle
2. Postulate Corollary 13a – AAS Postulate Corollary: ______________________________________
are congruent to the corresponding angles and non-included side of another triangle, then the triangles are congruent.
___________________________________________________________________________________
N
Note: Labeling may vary, but points
must correspond
C
B
A
3. Postulate 13 – ASA Congruence
P
M
If two angles and the included side of one triangle are congruent to
Assumption:____________________________________________
the corresponding angles and included side of another triangle, then the two triangles are congruent.
___________________________________________________________________________________
M
A
Note: Labeling may vary, but points
must correspond
N
B P
C
the three sides of a triangle are congruent to the three corresponding
4. Postulate 13 – SSS Congruence Assumption: If___________________________________________
sides of another triangle, then the two triangles are congruent.
___________________________________________________________________________________
A
M
Note: Labeling may vary, but points
must correspond
B
C
P
N
© 2006 VideoTextInteractive Geometry: A Complete Course
79
Unit IV, Part E, Lessons 2&3, Quiz Form A
—Continued—
Name
C
F
Refering to the triangles to the right, for problems 5 – 10,
name the postulate or postulate corollary which would
show nABC > nDEF.
A
5. /C > /F, AC > DF, BC > EF
Postulate 13 – SAS Congruence Assumption
________________________________
6. /C > /F, /D > /A, AB > DE
Postulate Corollary 13a – AAS Postulate Corollary
________________________________
7. /C and /F are right angles, AB > DE, BC > EF
Postulate Corollary 13e – HL Postulate Corollary
________________________________
8. /C and /F are right angles, /E > /B, DF > AC
Postulate Corollary 13c – LA Postulate Corollary
________________________________
9. /A > /D, AB > DE, AC > DF
Postulate 13 – SAS Congruence Assumption
________________________________
10. /B and /E are right angles, CB > FE, AB > DE
Postulate Corollary 13d – LL Postulate Corollary
________________________________
B D
B
A
11.
Given: BD AB; BD
AD > CB
DC
Prove: nABD > nCDB
D
STATEMENT
1. BD AB
2. BD DC
3. /ABD is a right angle
4. /CDB is a right angle
5. nABD is a right triangle
6. nCDB is a right triangle
7. BD > DB
8. AD > CB
9. nABD > nCDB
80
E
C
REASON
1.
2.
3.
4.
5.
6.
7.
8.
9.
© 2006 VideoTextInteractive Geometry: A Complete Course
Given
Given
Definition of Perpendicular Line Segments
Definition of Perpendicular Line Segments
Definition of Right Triangle
Definition of Right Triangle
Reflexive Property for Congruent Segments
Given
Postulate Corollary 13e – If the hypotenuse and one leg of one
right triangle are congruent to the hypotenuse and
corresponding leg of another right triangle, then the triangles
are congruent.
Unit IV, Part E, Lessons 2&3, Quiz Form A
—Continued—
Name
U
Repeat problem 13 using a different postulate
assumption or postulate corollary.
14.
Given: XW UW; XW
UQ > VQ
Q
XV
X
W
Prove: nUWQ > nVXQ
V
STATEMENT
1. XW UW
2. XW XV
3. UW || XV
4. /WUQ > /XVQ
5. UQ > VQ
6. /WQU > /XQV
7. nUWQ > nVXQ
REASON
1. Given
2. Given
3. Theorem 22 - If two lines are perpendicular to a third line,
then the two lines are parallel
4. Theorem 16 - If two parallel lines are cut by a transversal,
then alternate interior angles are congruent.
5. Given
8. Theorem 15 – If two lines intersect, then the vertical
angles formed are congruent.
9. Postulate 13 – If two angles and the included side of
one triangle are congruent to the corresponding angles and
included side of another triangle, then the two triangles are
congruent. (ASA Congruence Assumption)
D
15.
Given: DG is the perpendicular bisector of EF
Prove: nDEG > nDFG
E
82
F
REASON
STATEMENT
1. DG EF
2. /DGE is a right angle
3. /DGF is a right angle
4. nDFE is a right triangle
5. nDFE is a right triangle
6. DG bisects EF
7. EG > FG
8. DG > DG
9. nDEG > nDFG
G
1.
2.
3.
4.
5.
6.
7.
8.
9.
© 2006 VideoTextInteractive Geometry: A Complete Course
Given
Definition of Perpendicular
Definition of Perpendicular
Definition of Right Triangle
Definition of Right Triangle
Given
Definition of Bisector of a Line Segment.
Reflexive Property of Congruence
Postulate Corollary 13d – If two legs of one right triangle are
congruent to the corresponding legs of another right triangle,
then the two right triangles are congruent.
Unit IV, Part E, Lessons 2&3, Quiz Form B
—Continued—
Name
M
Refering to the two triangles to the right, for
problems 5 – 7, name the postulate or
postulate corollary which would show nMNQ > nRST.
R
N
Q
S
T
5. /M > /R, NQ > ST, /Q > /T
Postulate Corollary 13a – AAS Postulate Corollary
________________________________
6. MN > RS, MQ > RT, NQ > ST
Postulate 13 – SSS Congruence Assumption
________________________________
7. /N > /S, /Q > /T, NQ > ST
Postulate 13 – ASA Congruence Assumption
________________________________
H
Y
Refering to the triangles to the right for, problems
8 – 13, fill in the blanks with the required angle
or side so that the indicated congruence assumption
or postulate corollary justifies nXYZ > nGHI.
X
Z
G
I
XY
YZ
HI
GH
8. SAS Congrunece Assumption; /Y > /H; ________
> ________;
________
> ________.
YZ
ZX
IG
HI
9. LA Postulate Corollary; /Y > /H; m/Z = m/I = 90; ________
> ________
or ________
> ________.
10. ASA Congruence Assumption; /X > /G; XY > GH; ________ > ________.
XZ
GI
11. HL Postulate Corollary; m/Y = m/H = 90; YZ > HI; ________
> ________.
YZ
XY
GH
HI
12. AAS Postulate Corollary; /X > /G; /Z > /I; ________
> ________
or ________
> ________.
XZ
GI
13. HA Postulate Corollary; m/Y = m/H = 90; /X > /G; ________
> ________.
84
© 2006 VideoTextInteractive Geometry: A Complete Course
Unit IV, Part E, Lessons 2&3, Quiz Form B
—Continued—
Name
B
A
14.
Given: AB > DC; AB || DC
Q
Prove: nABQ > nDCQ
C
STATEMENT
1. AB > DC
2. AB || DC
3. /B > /C
4. /A > /D
5. nABQ > nDCQ
D
REASON
1. Given
2. Given
3. Theorem 16 – If two parallel lines are cut by a transversal,
then alternate interior angles are congruent.
4. Theorem 16 – If two parallel lines are cut by a transversal,
then alternate interior angles are congruent.
5. Postulate 13 – If two angles and the included side of one
triangle are congruent to the corresponding angles and
included side of another triangle, then the two triangles
are congruent. (ASA Congruence Assumption)
E
15.
Given: nADE is an Isosceles Triangle
EC AE; EB DE; AB > DC
Prove: nAEC > nDEB
A
STATEMENT
1. nADE is an isosceles triangle
2. AE > DE
3. EC AE
4. EB DE
5. /AEC is a right angle
6. /DEB is a right angle
7. nAEC is a right triangle
8. nDEB is a right triangle
9. AB > DC
10. BC > CB
11. AC > DB
12. nAEC > nDEB
C
B
D
REASON
1. Given
2. Definition of Isosceles Triangle
3. Given
4. Given
5. Definition of Perpendicular Line Segments
6. Definition of Perpendicular Line Segments
7. Definition of Right Triangle
8. Definition of Right Triangle
9. Given
10. Reflexive Property of Segment Congruence
11. Postulate 6 – Ruler – Fourth Assumption – Segment Addition
12. Postulate Corollary 13d – If the hypotenuse and one leg of a
right triangle are congruent to the hypotenuse and
corresponding leg of and their right triangle, then the two right
triangles are congruent. (HL Postulate Corollary)
© 2006 VideoTextInteractive Geometry: A Complete Course
85
Name
Quiz Form A
Class
Date
Score
Unit IV - Triangles
Part F - Congruence – Part 2 (Applications)
Lesson 1 - Overlapping Triangles
Lesson 2 - Using the Definiton of Congruence
Lesson 3 - Theorem 32 - “If two given triangles are
both congruent to a third triangle, then the two given
triangles are congruent to each other.”
O
A
1. Name two pairs of overlapping triangles that
appear to be congruent in the figure at the right.
E
D
nDBC and nECB; nABE and nACD
___________________________________________
B
C
U
V
W
X
N
Use the figure to the right for problems 2 and 3.
2. Name two different triangles that overlap
A
O
R
nVOY and contain OU. _________________________
nUOW, nUOX
Q
S
M
P
1
E
D
B
A
3. Name two different triangles that overlap
nVOY, nWOY
nUOX and contain OY. _________________________
B
C
U
V
W
Y
X
A
N
Use the figure to the right for problem
A 4.
E
AC – BC = AB
29 – 10 = AB
19 = AB
AC
3x + 5
3 ⋅8 + 5
B
24 + 5
29
BC
x+2
8+2
BD
N
10
Q U
Q
S
M
S
2
X
M
O
X B
A
Y
V
W
Y
X
A
B
U
D
C
V
R
T
X
DB – BC = CD
29 – 10 = CD
Q 19 = CD
Q
P
Y
C
U S
U
V
V
W
Q
W
U
U
E
X
Y
U
A
Z
N
V
1
V
Z
2
87
© 2006 VideoTextInteractive Geometry: A Complete Course
C
Q
R
X
U
V
W
A
Y
X
U
O
B
A
D
Z
P
C4x – 3U
4 ⋅8 – 3
32 – 3
29
S
M
W
P1
D
R
E
Q
AB =
CD = 19
Find AB and CD. ___________________________
AC = DB
3x + 5 = 4x - 3
8=x
C
D
O
4. AB = CD, AC = 3x+5, BD = 4x – 3, BC = x + 2.
B
T
R
D
19,
E
Q
X
T
Y
Y
A
A
B
C
YD
D
D
Unit IV, Part F, Lessons 1,2&3, Quiz Form A
—Continued—
Name
O
A
E
Q
Use the figure to the right for problem 5.
1
E
D
2
X
M
5. /1 > /2, m/AEC = 14x – 3, m/DEB = 17x – 15, AE = 4x + 7,
U > VnDEB?
W
DE = 3x + 11, BE = 3x + 7, CE = 5x B– 1. Why isCnAEC
P
Y
X
A
B
D
C
T properties of
nAEC > nDEB by S.A.S. Assumption. Since AE = DE, NCE=BE, and /AEC > /DEB, using the
__________________________________________________________________________________
angle addition.
__________________________________________________________________________________
R
m∠AEC = m∠DEB
14x – 3 = 17x – M
15
12 = 3x
4=x
AE
4x + 7
4 ⋅4 +7
16 + 7
A
Q
DE
3x + 11
3 ⋅ 4 + 11
12 + 11
23
S
23
O
A
CE
5x – 1
5 ⋅4 – 1
20 – 1O
19
BE Q
3x
P +7
3 ⋅4 +7
12 B+ 7
19
S
Q
U
E
1
E
2
Q
1
E
D
2
Prove: /NQX > /NPX
C
U
V
W
Y
X
C
X
M
Given: MQ > MP; /QMX > /PMX
6.
R
T
U
N
W
V
W
U
P
A
B
Q
T
STATEMENT
X
1. MQ > MP
2.Q/QMX
1. Given
> /PMX
XS
4. nQMN > nPMN
O
5. C.P.C.T.C.
> /PNM
6. /QNM
C
N
6. C.P.C.T.C.
V
W
A
Y
X
B
R
Z
9. /NQX
V > /PNX
V
A
65
D
O
VB
A
C
B
D
E
B
D
T Z
X
12
D
Y
© 2006 VideoTextInteractive Geometry: A Complete Course
B
C
C
8
Z
M
10 10
55
A
D
D
B 12
65
55
E
8
C
B
U
X
12
A
5
E
A
V
C
C
Y
T
N
Z
U
55
A
F
B
S C
D
88
10 10
X
R
S
A
U
P
12
O
3 4
E
7. Reflexive
Property
for Congruence
D
D
C
V
Y
Y
7 C
9. C.P.C.T.C.
V
Q
C
A
Q
U
Z
A
VS
Y
E
D C
Assumption
- If F
8. Postulate T13 - SAS Congruence
two sidesX and the included angle of one triangle
are congruent toAthe corresponding sides
B and
M
included
angle of another triangle, then the two
W
triangles are congruent.
R
U
X
Q R
S
4. Postulate 13 U- SAS Congruence Assumption - If
D
two
sides and the included angle
of one triangle
B
E V U
V
are congruent to the corresponding
sides and
6
5
A
C
included angle of another triangle, then the two
1 2
triangles
are congruent.
1
2
U
8. nQNX > nPNX
V
V
5. QN > PN
Y
Z
B
7. NX > NXU
N
U
M
Q
B
X
V
Y
3. Reflexive Property for Congruence
P
W
N
2. Given
Q
S
U
REASON
Z
Y
U
3. MN > MN
S
D
C
D
C
R
N
BV
A
Y
X
C
Y
8 D E
12
Unit IV, Part F, Lessons 1,2&3, Quiz Form A
—Continued—
Name
B
C
V
U
W
Y
X
O
B
REASON
1. /MRS and /NRS are supplementary
1. Theorem 10 - If the exterior sides of two adjacent
angles are opposite rays, then the two angles are
supplementary.
2. /PQS and /NQS are supplementary
C two adjacent
D 10 - If the exterior sides of
2. Theorem
angles are opposite rays, then the two angles are
U
supplementary.
V
3. Given
4. /NRS > /NQS
4. Theorem 14 - If two angles are supplementary to
the sameXangle or congruent angles,
Y then they are
Z
congruent to each other.
6. NM > NP
6. Given
7. nNQM > nNRP
X
7. Postulate Corollary
13a - AAS - If two angles Y
and a non-included
side of one triangle are
R
congruent to the corresponding two angles and
U
V triangle,
U
non-included
side of another
then the
twoV triangles are congruent.
T
N
8. /NRP > /NQM
U
Z
V
U
5. Reflexive
Property for Congruence
X
V
D
V
B
S
U
D
C
R
Q
R
Q R
T
E
D
S
S
S
A
U
W
R
R
B
E
5
A
1
B
M
P
M
Q
C
C
N
N
N
6
V
X
Q
C
S
U
O
S
1 2
2
V
Y
A
8. C.P.C.T.C
Q
P
A
Y
X
U
5. /N > /N
M
W
W
3. /MRS > /PQS
2
B
S
P
A
Q
1
Q
S
Q
STATEMENT
E
Q
M
Prove: /NRP > /NQM
C
T
R
Given:/MRS > /PQS; NM > NP
B
A
N
8.
2
1
E
D
T
Z
X
7
A
Y
B
D
C
O
3 4
E
V
M
Y
D
F
F
C
Y
T
C
C
X
C
D
8
12
8
12
8 D
U
D
Z
A
A
Y
V
C
A
90
D
12
E
B
A
15
D
C
Z
G
© 2006 VideoTextInteractive Geometry: A Complete Course
10 10
B
15
A
D
E
H
B
F
B
A
15
A
1
E
D
2
M
Name
Quiz Form A
B
C
U
V
W
A
Date
Class
N
Y
X
B
D
C
Score
T
R
Q
Unit IV - Triangles
Q
S
S
Part F - Congruence M– Part 2 (Applications)
P
Lesson 4 - Theorem 33 - “If two sides of a triangle Q
O
B
A
E
A
are congruent,
then
theO angles opposite
them
Q
are congruent.”
1
2
X
E
Lesson 5 D- Theorem
34 - “If two angles
of a triangle
M
are congruent, thenD the sides opposite
them
C
P
A
B
U
V
W
Y
X
are congruent.”
B
C
Y
U
V
W
A
B
D
X
C
1
2
C
R
W
N
E
T
U
Complete each of the statements in problems 1-3, refering to the figure to the right. Then state, in
U
your own words, the theorem you are
X using to justify your
Y answer.
R
S
M
1.
Z
Q
Q
U
V
P
V
M
Q
BD
CD
If /CBD > /BCD, then __________
> __________.
X
Y
O the sides oppo34
“If
two
angles
of
a
triangle
are
congruent,
then
B
A
Theorem
______________________________________________
site
them are congruent.”
______________________________________________________
U
2.
Z
SV
U
V
1
12
5
101 1
2
If m/CBD + m/DBA > m/BCD + m/DCA,
C
B
E
U
A
A
Y
Y
R
A
Use the diagram to the right, and
Q Theorem 33 or
S Theorem 34, to
complete each statement in problems 4 and 5.
D
S
U
V
O
12
T
X
B
T
C
C
N
12
D
T
A
D
15
A
Z
65
Y E
D
55
B
B
E
X
A
C
D
8
Q
55
A
C
C
8 D
12
8
12
D
A
C
D
A
Complete
Course
©
2006
VideoTextInteractive
Geometry:
A
15
B
B
A
15
B
M
R
A
10 10
M
R
D
B
65O
m/CDB = __________.
CB =
A
C
V
12
__________.
C
B
W
R
V
E
C
V
U are congruent.”
site
them
V
M
______________________________________________________
U
D
C
3
D
D
B
Q
V
A
X
T
D
R
/ABC
/ACB
If AB > AC, then __________
> __________.
X
Y
Z
then the angles oppo33 - “If two sides of a triangle are congruent,
Z
Theorem ______________________________________________
X
7
55
B
U
5.
C
B
A
D
34 - “If two angles of a triangle areUcongruent,
then
AC
V
W the Xsides oppoY
Theorem ______________________________________________
N
site
them are congruent.” W
______________________________________________________
4.
V
A
E
AC C
D AB
then __________
> __________.
3.
U
Y
W
D
95
B
B
A
C
7
A
C
U
V
W
A
Y
B
X
Unit IV, Part F, Lessons 4&5, Quiz
Form A
W
—Continued—
U
B
F
C
X
C
U
Y
U
Z
Z
In the figure at the right, nXYZ is isosceles,
with XZ > YZ. Also, ZY =W 8, m/Y = 40O,
V
Y
ZX =
Y
8
__________.
W
Y
X
U
E
B
D
Z
Y
X
Z
U 12
40O
= __________.
V
V m/X
U
A
C
A
D
7.
V
V
V
and ZU bisects /XZY. Use this information for problems 6-9.
6.
V
F
D
X
O
O
D
D
C
Name
3 4
E
Z
V
V
10 10
Y
X
(Answers will vary) Postulate55Corollary 13 -65
ASA 55
nXZU > nYZU. Why? ___________________________________
B
C
D
B
Assumption (or SAS, AAS)A
E
___________________________________
8.
9.
C
B U
E
D
D
E
V AU
A
V
B
D
W
S
Given: RM > RN R
NQ > MQ
O
SProve: /TNQ > /TMQT
R
X
M
1. GivenQ
Y
C
D
A
5. /NRT > /MRT
C
R
B
A
5. C.P.C.T.C.
CA
8
Y
CD
B
8. C.P.C.T.C. C
9. /TNQ > /TMQ
9. Theorem G33 - If two sides of a triangle are
H
congruent, then the angles opposite them
are congruent.
F
E
A
B
8 D
12
8
12
8. NT >MT
© 2006 VideoTextInteractive Geometry: A Complete Course
M
B
15
7. Postulate 13 - SAS Congruence Assumption - If
two sides and the included angle of one triangle
A
B sides Aand
15 corresponding
are congruent
to the
included angle of another triangle, then the two
triangles are congruent.
F
X
4. Postulate 13 - SSS Congruence Assumption - If
three
12 sides of a triangle are congruent to the
corresponding sides
T of another triangle, then the
two triangles are congruent.
7. nNRT > nMRT
H
T
F
E
6. Reflexive Property forD Congruence
D
C
O
S
6. RT > RT
E
B
15
W
D
S
2. Given
8 D
12
8
B
D
2
B
V
C
3. Reflexive Property for Congruence
C
B
1
C
R
REASON
Z
M
4. nRNQ > nRMQ
12
E
D
A
3. RQ > RQ
D
D
C
A
2. NQ > MQ
C
T
B
B
STATEMENT
T
A
C
Q
A
1. RM > RN
C
N
R
10.
B
C
B
M
96
Y
D
T
A
15
A
U
X
C.P.C.T.C.
XU > YU. Why? ______________________________.
N
Q
V
Y
D
D
15
B
A
U
V
A
V
Y
X
Y
Unit IV, Part F, Lessons 4&5, Quiz Form A
—Continued—
U
V
U
C
A
Z
Name
12
D
10 10
V
55
65
55
In the figure at the right, AC is contained inBplane M, and
intersects
plane
M at
point
A.
C BD
A
D
B
E
X
Also, AC is the perpendicular bisectorC of BD.
B
N
R
12. Prove: nBAC
> nDAC
Q
T
13. Prove: /B > /D
14. Prove: nBDC is isosceles
A
T
B
A
2. Given
3. AC is the perpendicular bisector of BD
3. Given
4. /BAC is a right angle
X
M
6. nBAC is a right triangle
5. Definition of Perpendicular
6. Given
7. nDAC is a right triangle
8. BA
D > DA
D
Y
C
F
C
8. Definition of Segment Bisector
9. nBAC12>nDAC
12
8
D
B
15
A
15
8 D
12
9. Postulate Corollary 13d - If the two legs of a right
triangle are congruent to the two legs of a right
B
Atriangle are15congruent to the
B two legs of another
10. /B > /D
10. C.PC.T.C
11. BC > DC
11. Theorem 34 - If two angles of a triangle are
congruent, then the sides opposite them are
congruent. (also C.P.C.T.C.)
D
E
H
12. nBDC is isosceles
B
F
A
E
12. A triangle is an isosceles triangle, if and only if,
it has at least two congruent sides.
© 2006 VideoTextInteractive Geometry: A Complete Course
2
D
B
7. Given
C
1
C
A
Z
4. Definition of Perpendicular
T
98
A
B
5. /DAC is a right angle
A
C
1. Given
2. BD intersectsS plane M at point A O
G
B
REASON
R
S
1. AC is contained
in plane M
C
C
D
W
STATEMENT
A
E
B
V
8
Y
A
A
R
C
U
D
E
D
M
Q
E
B
B
D
C
D
U
Unit IV, Part F, Lessons 4&5, Quiz Form B
—Continued—
Q
R
V
W
A
Y
X
T
B
W Name
C
A
M
X
U
R
Y
X
U
U
V
Q
Z
S
Z
V
V
S
In the diagram at the right, RQ > RS and TQ > TS. Use this information
X
Y
for problems 8 – 11.
V
Y
C
A
T
12
D
8.
9.
If m/RQS = 6x+10, and m/RSQ
= 9x – 17, then
m/QRS =
U
V U
52
__________.
O
V
10 10
C
C
D
(Note: You might also conclude m/RSQ 8= 60O
m/SRQ
If QS > RQ, then m/RSQ = __________.
12
B
since all three sides of nRQS are congruent.)
C
C
A
15
A
B
10.
Postulate Corollary 13c
N ______________________________________.
If /QVT > /SVT, then nQVT > nSVT. Why?
11.
Q
Theorem
33 T
D
/TQS > /TSQ. Why? ______________________________________.
R
E
G
A
8
55
C
D
A
B
A
E
DC
H
B
M
A
B
D
W
R
F
R
Q
12.
S
Given: ST bisects /RTM
ST bisects /RSM
Prove: /TRM > /TMR
X
Z
M
T
1. Given
C
D 2. Given
12
8
3. /RST > /MST
D
Y
REASON
C
2. ST bisects /RSM
C
8 D
12
8
12
3. Definition of Angle Bisector
D
A
15
5. nRTS > nMTS
C
G
6. RT >MT
7. /TRM > /TMR
T
A
1. ST bisects /RTM
4. TS > TS
O
S
STATEMENT
100
B
V
4. Reflexive
for Congruence
A
15
B Property
B
7. Postulate 13 - ASA Congruence Assumption - If
two sangles and the included side of one triangle
are congruent to the corresponding angles and
D triangle, then the two
Eincluded side of another
triangles are congruent.
H
8. C.P.C.T.C.
A
A
F
B
33 - If two sides of a triangle are
9. Theorem
congruent, then the angles opposite them
are congruent.
© 2006 VideoTextInteractive Geometry: A Complete Course
15
B
A
D
8
Unit IV, Part F, Lessons 6&7, Quiz Form B
—Continued—
12
8
12
D
NameA
A
B
15
15
C
5.
Given: nABC ~ nDEF
BG bisects /ABC; EH bisects /DEF
Prove: CG FH
=
GA HD
1. nABC ~ nDEF
2. AB BC
=
DE EF
3. BG bisects /ABC; EH bisects /DEF;
108
H
A
STATEMENT
D
E
G
B
F
REASON
1. Given
2. If two triangles are similar, then corresponding
sides are proportional
3. Given
4. AB AG DE DH
=
;
=
BC GC EF HF
4. Theorem 35 - If a ray bisects an angles of a
triangle, then it divides the opposite side into two
segments that have the same ratio as the two
other sides.
5. AB DE
=
BC EF
6. AG DH
=
GC HF
7. HF GC
=
DH AG
8. GC HF
=
AG DH
5. Switch the means in propotion for statement 2.
DE
(Multiply by EC on both sides of step 2.)
6. Substitution
7. Multiplication Property for Equality
8. Symmetry Property of Equality
© 2006 VideoTextInteractive Geometry: A Complete Course
B
Name
Quiz Form A
Class
Date
Score
Unit IV - Triangles
Part G - Congruence – Part 3 (Triangle Inequalities)
Lesson 1 - Theorem 37 - “If you have a given exterior
angle of a triangle, then the measure of that angle is greater
than the measure of either remote interior angle.” (Exterior
Angle Inequality Theorem)
Lesson 2 - Theorem 38 - “In a given triangle, if two sides
are not congruent, then the angles opposite those sides are
not congruent.”
Lesson 3 - Theorem 39 - “In a given triangle, if two
angles are not congruent, then the sides opposite those
angles are not congruent.”
Lesson 4 - Theorem 40 - “In a given triangle, the sum of
the lengths of any two sides, is greater than the length of the
third side.”
For each of problems 1 and 2, write an inequality that relates only x and y.
1. x = 47 + y
x>y
_________________________
2. y – x = 25
x<y
_________________________
For each of problems 3 and 4, write an inequality that relates only m/1 and m/2
3. m/1 – 160 = m/2
m/1 > m/2
_________________________
4. m/2 = m/1 + 65
m/1 < m/2
_________________________
5. In the figure to the right, AB > BC > AC and r = t.
Fill in each of the following blanks with >, <, or =.
C
<
y ____________
x
(x)
>
t ____________
y
O
(r)
>
r ____________
y
O
(t)
O
D
(y)
O
A
C
B
C
© 2006 VideoTextInteractive Geometry: A Complete Course
111
17
Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A
—Continued—
Name
In problems 6 and 7, state one or more conclusions that can be drawn from the given information.
Give a reason for each conclusion.
6. Given:
Quadrilateral RABC with diagonal
RB
C
CB
>
CR + RB_______
_______
Conclusion:
(x)
O
(t)
D
O
O
(y)
(r)
RB
>
RA + AB_______
_______
Conclusion:
U
A
R
O
A
B
C
B
S
N D
G
C
12
1
17
17
Theorem 40 - If you have the sum of the measures of two sides of a triangle,
then that sum is7greater
____________________________________________________________________
Reason:
1 2
A
88 92
H
0
B
E
8
0
2
F
12
than the measure of the third side of the triangle.
____________________________________________________________________
____________________________________________________________________
t
nUNS and nAYD; SU > DA;
y
> DY;z m/S < m/D
xSN
A
B
A
R
C
S
A
B
O
(x)
O
(y)
A
10
10
42
41
X 31 32 X
X
O
(r)
A
Y
N D
T U
AYD
UN > _______
(t)
Conclusion:
P
D
r
7. Given:
A
U
C
x
O
AY
_______
Conclusion:
<
UN_______
A
Reason:
14
14
S of a second
C
U measure
Corollary 39a - If Cthe measure of one angle
ofW a triangle
is greaterR1than the
angle of
____________________________________________________________________
40
C
B
Y
B
48 G
S
17
17
Y
N D
46
12
Z
130o
E
46
V
D
7
8
X
92
2
88
1
the
of Ethe side opposite the
larger angle is greater2than the measure of theS side
____________________________________________________________________
A triangle, then the measure
B
F
R
T
H
12
0
0
T
opposite the smaller angle.
____________________________________________________________________
t
r
A
P
y
B
8. What is wrong with this picture?
S
A
Y
N D
T
T
X
128 o
y
x
A
10
B
42
41
X 31 32 X
10
E
46
46
Z
D
R
14
R
S
S
C
U
remote interior Sangle. (Theorem 37)
____________________________________________________________________
T
V
R
T
U
48
14 X
Y8
2
12
W
____________________________________________________________________
A
A
P
P
112
N D
Y
X
B
41
2
1
42
10
E
Q
7
R
© 2006
VideoTextInteractive Geometry:
A Complete Course
A
T
T
A
10
Y
V (with a measure of 128O) should be greater than (not less than) the measure of either
The exterior 40
angle
S
____________________________________________________________________
7
F
S
130o
Y
X
N D
12
Explain.
1
17
2
1
z
A
B
A
P
D
x A
U
A
U
C
Y
M
4
11
N
T
Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A
—Continued—
UA
A U
A diagram given
R information R
What conclusion
and the
C can you makeCfrom the given
D
in problems 9 - 12, by
37-40.
D applying Theorem
(x)
O
(t)
O
(x)
O
(t)
O
O
O
(y)
(y) AC > BC
____________________
(r)
O
9. m/1 (r)
> m/2
O
A
A
B
C
C
C
G
C
z x
A
UA
A
RA
11.
B
C
17
88 92
H
0
F
E
12
Y
W
48
1
14
17
7
B
NS AD
T
14
R
X
A
Y
2
1
A
10 10
B
42
46
A
E
41
46
CD
10
42
T
S
V
T
TU
Y
R
M
X
Q
7
2
1
D
V
U
A
T
T
S46
R D
4
1
Y
E
S
R
T
S
T
7
N
T 11 M
© 2006 VideoTextInteractive Geometry: A Complete Course
46
R
46 46
C D
S
PR
T
A
Y
46
A
P
X
S
P
X
S
T
TR
P
X 31 32 X
C
U
R
A
U C
X
S
T
S R
S
P
A
N YD
RS
12 R
YN D A
46
12
A
V
Z
U
8
P
41
2
F
12
T
130o
V
12
2
A
10
0
UA
B
AS U
48
Z
14 14 1 W
7
8
12 F
A U
B
X
2
z
2
0
N DS
8
8
A 130o10
10
10
10
41 E 42
B
41 B 42
o
31 32 X
X 31 32 X y
X
y
m/1 > m/2X
128 o
x128/VRT > /VTR
____________________X
12. SR =x ST; VX = VT ____________________
YN D
B N DS 40
Y
40
S
G
Y
7
y
r
x
12
1
1
17
N
D
t
y
r
E F
U
D
t
7
88 92
H
0
0
NS D
S
12
17
0
88 92
B
H
C
C
B
G
17
1 2
E
B
1 A2
A
EH > FH
____________________
B
10.
B
17
17
B
Name
113
A
U
A
Unit IV, Part G, Lessons 1,2,3&4, Quiz Form A
—Continued—
Name
130o
B
S
128 o
y
x
Y
N D
V
12
S
1
Given: TU > US > VS
8
13.7
X
S
2
Prove: ST > SV
F
R
12
T
STATEMENT
3. m/STU = m/TSU
4. US > VS
5. /SUV > /SVU
A
Y
D
10
B
6. m/SUV = m/SVU
A
7. m/SUV
> m/STU
10
41
42
T
X
E
8. m/SVU > m/STU
9. ST > SV
46
C
114
46
D
U
V
REASON
1. TU > US
P
2. /STU > /TSU
A
T
R
1. Given
A
Q
2. Theorem 33 - If two sides of a triangle are congruent,
then the angles opposite them are congruent
7
4
2
P 3. Definition
R Angles
of Congruent
1
4. Given
5. Theorem 33 - If two sides of
M a triangle are congruent,
N
T
then the angles opposite them are congruent 11
Y of Congruent Angles
6. Definition
7. Theorem 36 - If you have a given exterior angle of a triangle,
then the measure of that angle is greater than the measure of
either remote interior angle.
8. Substitution
9. Theorem 38 - If the measure of one angle of a triangle is greater
then the measure of a second angle of the triangle, then the
measure of the side opposite the larger angle is greater than the
S measure of the sideT opposite the smaller angle.
© 2006 VideoTextInteractive Geometry: A Complete Course
Unit IV, Part G, Lessons 1,2,3&4, Quiz Form B
(x)
—Continued—
(r)
U
A
R
C
D
Name
(t)
O
O
O
(y)
O
A
B
C
B
S
G
C
N D
12
1
17
17
7
In problem 6, state one or more conclusions that can be drawn from the given information.
Give a
88 92
2
1 2
reason for each conclusion.
A
E
B
F
H
/D
/S > _______
Conclusion:
A
R
C
O
<
m/S___
O
z
A
m/D
_______
B
130o
N D
S
T
O
Reason:
B
C
B
P
A
x
D
(t)
(x)Conclusion:
(y)
(r)
D
t
Uy
r
A
U
C
A
12
nUNS and nAYD; SU > DA; SN > DY; UN < AY
6. Given:
O
8
0
0
S
x
Y
N D
A
Y
y
10
128 o
A
B
Theorem
38 - If two sides of a triangle are not congruent, then the angles opposite those sides
are not congruent
42
____________________________________________________________________
41
G
C
12
1
VX 31 32 X
X
10
E
S
40
17 Theorem17(Side-to-Angle Version)
Hinge
7
____________________________________________________________________
8
1 2
A
0
E
A
U
C
7.
X
46
46
48
S
Y
Z
88 92
2
T
V
14
W
D
R
14
S
C
F
U
R
T
U
H
12
____________________________________________________________________
0
B
t
A
P
Given:
nATP
D
P
y
r
x
Prove:z AT
> TPS – PA
B
A
A
Y
N D
40
Y
48
14
W
Z
46
R
3. AT > TP – PA
116
41
1. nATP
2. AT + PA > TP
14
U
S
A
10
X 31 32 X
X
T
X
B
C
42
R
T
T
STATEMENT
2
1
10
Q
7
M
Y
REASON
E
1. Given
2. Theorem 40 - If you have the sum of the measures of two sides
46
of a D
triangle, thenR that sum is greater than the measure of
S
T
the third side of the triangle.
3. Subtraction Property for Inequalities (i.e., Addition Property
where the Additive Inverse is added to both sides)
© 2006 VideoTextInteractive Geometry: A Complete Course
4
11
G
C
V
12
S
1
17
17
7
8
X
Unit IV,
G, Lessons 1,2,3&4,
92
88 Part
2 Quiz Form B
E
B
F
R
H
12
—Continued—
2
A
U
Given: PA > PT; m/1 > m/2
8.
S Name
0
0
B
Prove: NRTD > AR
S
A
Y
2
1
X 31 32 X
Z
14
R
U
B
41
STATEMENT
1. PA > PT
2. m/1 > m/2
S
C
3. PR > PR
4. RT > AR
10
42
46
M
4
11
N
Y
E
REASON
1.
2.
3.
4.
46
D
R
Given
Given
S
Reflexive Property Tfor Congruence
Hinge Theorem (Side-to-Angle Version) - If two sides of one
triangle are congruent to two sides of a second triangle and the
measure of the included angle of the first triangle is greater than
the measure of the included angle of the second triangle, then
the length of the side opposite the included angle of the first
triangle is greater than the length of the side opposite the
included angle of the second triangle.
A
U
A
T
X
A
Q
7
R
T
T
10
V
U
A
P
P
z
T
T
130o
S
128 o
y
x
B
Y
N D
V
12
S
1
7
8
X
2
S
R
12
A
T
T
A
P
2
1
P
X
A
10
41
42
10
Q
7
R
9.A What is wrong Twith this picture?
Y
V
U
T
M
4
11
N
Y
Theorem 40 - If you have the sum of the measures of two sides of a triangle, then that sum is greater
____________________________________________________________________
E
Explain.
than the measure of the third side of the triangle.
____________________________________________________________________
46
C
46
D
R
S
T
____________________________________________________________________
© 2006 VideoTextInteractive Geometry: A Complete Course
117
G
C
12
A
17
17
7
88 92
H
1 2
0
0
t
D
V
S1
R7
T
0
8
T
2
A
C Y
B r
S y– 12, useNeach
D
For problems
10
accompanying
figure, and fillU in the blanks
with <,P=, or >.P1
x
G
z
12
A
B
1
tV
S
r
>
10. XY _______XZ;
XW _______
7 >
8
17
A X
88 92
CH
0
(x)
O
(t)
0
F
12R X
40
D
O
O
48
14
C
Y
A
B
W
Z
14
R
B
P
48
14
1
17
17
A
88 92
E
A H
10
42
41
7
Y
0
B
B
31 32 X
46
D
S
y
W P12 14
41
A
T
2
1
U
Y
R
R
T
X
R
46
S
T
D
R
S
T
2
1
P
A
Y
N D
W
14
A
10
41
10
42
T
M
4
11
Y
E
Given: XS > YS; RX > TY
S is midpoint of RT
Z
46
R
U
S
Prove: m/R > m/T
C
46
D
R
STATEMENT
1. S is midpoint of RT
2. RS > TS
3. RS = TS
4. RX > TY
5. RX = TY
6. XS > YS
7. m/R > m/T
118
Q
7
R
T
X
X 31 32 X
13.
S
V
U
A
P
B
40
R
N
T
T
X
D
E
A
S
T
46
C4
z
B
E
128 o
S
y
11
S
10
42
41
R
7S
M
A
130o
Q
UV
T
X
x
D
A
10
X 31 32 X
46
A
P
Y
Y
10 N D
E
V
42
T
Y
2
46
C
C
N D
Z
8
X
12
F
10
S
T
0
U
C
Ut
10 S
UT
40
U
S
<
12. BC _______
DE
T
B
B
46
Y
N D
1 2
X
<
11. RU _______
US
A
z
M
T
T
B
G
C
A S
Y
y N D
x T
14
X
(y)
U
AR
7
R
D
S
X 31 32 X
2
T
A
P
x
S
R
12
128 o
y
S
T
REASON
1.
2.
3.
4.
5.
6.
7.
© 2006 VideoTextInteractive Geometry: A Complete Course
V
U
X
2
F
o
A130
U
C
88 92
H
E
B
A
U
A
YS
N D
12
X
2 17
17
Name12
A Part G, Lesson 1,2,3&4,
E
B
F B
Unit IV,
Quiz Form
—Continued—
1 2
A
R
8
G
C
0
V
S
B
C 1
B
Given
Definition of Midpoint
Definition of Congruence
Given
Definition of Congruence
Given
Hinge Theorem (Side-to-Angle Version) - If two sides of one
triangle are congruent to two sides of a second triangle and the
measure of the included angle of the first triangle is greater than
the measure of the included angle of the second triangle, then
the length of the side opposite the included angle of the first
triangle is greater than the length of the side opposite the
included angle of the second triangle.
N
Unit Test Form A
Name
Date
Class
Score
Unit IV - Triangles
1. Write true or false for the following statements
True
(A-2) ____________
a. An equilateral triangle is also an isosceles triangle.
(A-2) ____________
False
b. A right triangle may also be an acute triangle.
False
(A-2) ____________
c. All isosceles triangles are equiangular triangles.
(A-1) ____________
False
d. A right triangle has only one altitude.
False
(B-1) ____________
e. If nDEF is a right triangle and the measure of one acute angles is 65,
then the measure of the other acute angle is 35.
(A-2, B-1)
/B
/C
2. nABC is isosceles with AB = AC. The base angles are ______________
and _______________.
(A-2, B-1, F-4)
3. nDEF is isosceles with base angles measuring 25 degrees. The measure of the
130
vertex angle is ______.
(F-4)
60
4. nMNQ is equiliateral. m/Q = _______________.
(A-1)
vertex
perpendicular
5. An altitude of a triangle is a segment drawn from a _______________
and _______________
to
the opposite side of the triangle.
(E-3)
hypotenuse
6. The HL Congruence Postulate states that two right triangles are congruent if the ____________
leg
and _______________
of one are congruent to the corresponding parts of the other.
(B-1)
Find the measure of the indicated angles in problems 7 - 9.
x
x
x
7.
35
35
55
x = _______________
A
C
y
t
H
y
E
93
8.
x
3535
A
D
W
AH
H
V
V
E
C
W
A B
E
A
A
U
9.
x
A D
x
E
EB || DC
x
3
AB
P
D 104
D
Ax = _______________
7
C
EB || DC
E
38
A
35
A
E
B D
D
60
C
A
P
C
7
B
D
PM
60
7 A 60
B
45
Q
S
Q S
90 – 52 = x
45
C
60
B
C
6060
B
C
B
60
B24
A 38 = x 24 B
24 C
C
C
R
R D
R
© 2006 VideoTextInteractive Geometry: A Complete Course
C
C
D
D
D
F N
F
N
R
N
A
1
3
D
180
– 35 – D
93 = 52
10
7
180 – 38 – 38 = x
S 10 Q
104 = x 10
119
CV
F
U
38
93
x
x
A38
x = _______________
90 – 35 = x
W
55 = x
y
93
35
C
C
t
D
38
3
Unit IV, Unit Test Form A
—Continued—
Name
D
Use the figure to the right for problems 16 - 20.
38
(D-2)
E
EB || DC
93
2
AB = 6, BC = 4. AE = 3, ED = _______________.
x
x
16.
AB AE
=
BC ED
6 3
=
4 x
C (4 )(3 )= (6 )(x )
D
35
35
A
A
A
G
7
M
10
60
A
B
S
Q
C
45
45
60
60
C
R
AB AE
C=
BC ED
9 x
=
6 4
(6 )(x )= (9 )(4 )
C
P
7
6
B
17. ABA= 9, BC = 6, ED= 4, AEB = _______________.
24
C
B
D
A
12=6x
2=x
W (D-2)
E
B
D
D
F
N
R
3
Q
E
6x=36D
x=6
1
C
B
(D-2)
F
H
1
T
1
18. BC =3, AC = 10, AE = E2 , ED = _______________.
M
K
B
3
BC ED
=
AB AE
3 ED
=
1
7
2
R 3
2
D
A
(7 )( ED ) =  31   73 
A
1
B
  
21
3
ED =1
7 ⋅ ED =
10
(D-2)
1
19. AE = 4, ED = 2, BE = 7, DC = _______________.
2
B
(D-2)
G
20.
(4 )( DC ) = (6 )(7 )
(4 )( DCA) = 42
AD DC
=
AE EB
AE+ ED DC
=
AE
EB
4+2 DC
=
EB
4
6 DC
=
4 7
GAC =F 12,
A
AB = 8, AE =
1
2
3
_______________,
B
6
4
T
2 ⋅2 ⋅2 ⋅114
0 20
x AE= 2 ⋅2 ⋅3 = 3
R AE=6 2 8
12
T3
B
B
M
a
3
C
AD = 10
63
D
x
6
3
12
8
A
8
10
B
N
Q
6
5
C
A
61
2
42 2 ⋅ 21 21
DC =
=
=
4
2⋅2
2
1
DC
C =10
2
AB AE
=
AC AD M
10
8 AE
=
K
12 10
(12 )( AE )= (8 )(10 )
Q
M
4
6
E
R
8
© 2006 VideoTextInteractive Geometry: A Complete Course
C
M
C
K
D
121
E
Unit IV, Unit Test Form A
—Continued—
Name
D
38
B
E
EB || DC
93
E
x ABC is a right triangle, as shown to the right,
For
x problems 21-23, triangle
with /C as the right angle. DE AB, AE A
= 4, ED = 3, AD = 5, and CD = 7. C
4
3
C
B
7
A
5
D
P
P
D
N
Postulate Corollary 12b - If an acute angle of one right triangle is M
congruent to an acute
State why nADE ~ nABC.________________________________________________________________
7
60
A
angle of another right triangle, then the
________________________________________________________________________________________
10 right triangles are similar.
B
45
S
Q
(D-1)
A
21.
________________________________________________________________________________________
B
B
C
24
60
R
(D-3)
45
60
C
D
AD AE ED
=
=
AB AC
CB
22. Fill in each box to complete the following extended
D proportion:
F
N
R
Q
E
12
23. Find AC:
F ____________
H
AE A
ED
=
AC BC
4 3
=
12 BC
(12 )(3 )= (4 )(BC )
36= (4 )( BC )
4 ⋅4
= BC
4
G
2
C
4
D
AC = 12 ( AD+
2 DC or 5+7
1 )
D
9
Find BC: ____________
15
Find AB: ____________
E
1
3
(D-3)
Q
A
C
AE AD
= B
AC AB
4 5
=
12 AB
(12 )(5 )= (4 )( AB )
x
60= (4 )( AB )
4 ⋅1
15
= AB
4
15= AB
35
35
9 = BC
C
A
For problems
24 and 25, HW bisects /H, as shown in the figure
to the right.
A
61
(F-6)
y
A
B
1
56 = 10 WA
53.6 = WA
C
( )
A
M
a
63
V
C
AC
U
G
B
(F-6)
10
K
4; 6
25. If HE = 8, HA = 12, and EA = 10, Find EWDand WA. ____________
x
B
14
12
8
N
B
N
Q
A
E
(14 )(4 )=(10 )(WA)
M
W
H
5.6
24. If HE = 10, EW = 4, and HA = 14, Find WA. ____________
HE
EW
=
HA
WA
10 4
=
14 WA
B
R
HE EW 6
5 HA = WA
A
x
8
= 8
C 12 10 - x
(12 )(x )= (8 )(10 - x )
EW =4
122
3
R 6
C
4
6
WA=10 - 4
WA=6
M
8
E
M
C
K
© 2006 VideoTextInteractive Geometry: A Complete Course
D
T
S
12
8
10
WA=10 -E
x
12x=80 -8x
20x=80
x=4
D
M
t
2
C
A
R
K
B
D
D
F
B
3
x
93
x
Unit IV, Unit Test Form A
—Continued—
Name
35
35
C
A
A
C
D
t
For problems 26–29,
nABC is a right triangle with altitude AD
W
and hypotenuse BC, as shown
in the figure to the right.
y
H
(D-4)
B
A
E
V
27
26. CD = 3, AD = 9. Find BD. ____________
CD AD
=
AD BD
3 9
=
9 BD
(9 )(9 )= (3 )(BD )
B
C
F
A
U
Q
G
B
D
81= (3 )( BD )
F
H
3 ⋅27
= BD
3
27 = BD
R
T
S
E
C
M
(D-4)
2 3
27. CD = 1, CB = 12. Find AC.____________
CD AC
=
AC CB
1 AC
=
AC 12
( AC )( AC )= (1 )(12 )
R
K
A
( AC )2 =12
B
D
B
A
2
AC = 12 = 4 ⋅ 3
AC =2 3
(D-4)
32
28. AB = 8, BD = 6. Find BC. ____________
3
BC AB
=
AB BD
BC 8
=
8 6
(8 )(8 )= (BC )(6 )
G
E
B
10
3
2 ⋅32
= BC
2 ⋅3
32
= BC
3
K
4
S
2
B
5
T
14
x
P
C
R
8
12
4 5
8 5
29. CD = 4, DB = 16, Find AC____________.
Find AB____________.
Find AD ____________.
T 8B
CD
AC
=
AC
CB
CD
AC
=
AC CD+ DB
4
AC
=
AC 4+16
( AC )( AC )= (4 )(4 +16 )
( AC )2 =80
1
M
Q
64= ( BC )(6 )
(D-4)
A
F
G
DB AB
=
AB BC
DB
AB
=
AB BD+ DC
16
AB D
=
AB 16 + 4
( AB )( AB )= (16B)(16+4 )
A
( AB )2 = (16 )( 20 )= 320
AC = 80 = 16 ⋅ 5
AB= 320 = 64 ⋅ 5
AC =4 5
AB=8 5
CD
AD
=
AD
DB
4 AD
=
AD 16
R
C
( AD )( AD )= (4E)(16 )
( AD )2 =64
AD = 64
C
D 25 R
F
30
AD=8
© 2006 VideoTextInteractive Geometry: A Complete Course
123
C
Unit IV, Unit Test Form A
—Continued—
Name
(D-5)
No
30. If the measures of the three sides of a triangle are 10, 12, and 15, is the triangle a right triangle? ____
The given triangle is not a right triangle. a + b > c . So, the triangle is an acute triangle.
If not, explain. ______________________________________________________________________
2
2
a +b
2
=c
2
2
38
x
2
E
93
x
x
10 2 +12 2 =15 2
35
35
A
100+144=225
244 > 225
C(D-5)
A
D
A
C
7
D
26
31.t In rectangle ABCD, as shown to the right, find AC. _______________
10
Q
W
( DCy )2 + ( AD )2 = ( AC )H2
(10 )2 + (24 )2 = ( AC )2
676 = AC
E
4
⋅
169
= AC
V
100+576= ( AC )2
2 ⋅13= AC
676= ( AC )2
26 = AC
B
A
B
C
24
C
F
U
(D-5)
N
A
Q
G
D
B
E
D
2
32. The length of the hypotenuse of a 45-45-90 triangle is _______________
times the length of a leg.
(D-5)
x
R
T
93
S
33. The length of the longer leg of a 30-60-90 triangle
is
x
C
35
of the hypotenuse.
35
F
38
H
3
E
2
_______________
xtimes the
A
(D-5)
38
A
7
1
W
PS = ⋅7 or
2
2
x
H
7E 3
3
⋅7 or
SQ =
V2
2
D
A
PR.
D If PQ = 7, Find PS
x
D
EB
A
B
C
A
10
U
Q
G
A
E
C
B
Find
Find
Q
2
DC_______________.
K
2
M
E
R
F
60
F
D
AB = BD = 1
T
1
2
C a
A
B
1
D
A
Q
32
C
1
A
x
B
1
5
14
12
8
B
4
B
C
R
G
2
K
A
2
C
D
x
6
C
8
8
A
10
R
M
E
Q
B
A
© 2006 VideoTextInteractive Geometry: A Complete Course
D
E
D
T
124
MR
103
C
A
B
45
Q
45
4
S
D
D
N
start with
Q
3
C
R
A
5
D
A
P
2
7
60
4
A
B
D
3N
CB_______________.
P
D
C
M
N
60
G H C G
R
E
T
B
B
S
Q
C
3
24
S
D
B
24 1
Find
BD_______________.
E
F
B
F
60
7
C
F
60
7
7 3
10
2
_______________.
P
2
Find AD_______________.
H
Find SQ
D
S
P
E
C
R
EB || DC
B
35. AB = 1 in the figure to the right.B
B
B
7
D
_______________.
B 2
R
K
A
A
(D-5)
A
length
M
34. In nPRQ shown to the right,
A
C QS
DEB || DC
E
M
K
61
E
C
A
C
D
t
Unit IV, Unit Test Form A
—Continued—
y
W
Name
H
A
E
V
C
A
(E-3)
40.
U
RT; VR > UT; /V > /T
Given: VS
G
D
B
Prove: RS > US
H
R
STATEMENT
T
S
C
REASON
1. VS RT
2. /VSR and /VST are right angles
3. nVSR and nVST are right triangles
4. VR > UT
5. /V > /T
6. nRSV > nUST
E
M
1.
2.
3.
4.
5.
6.
Given
Definition of Perpendicular
R
K
Definition of Right Triangles
A
B
Given
Given
Postulate Corollary 13b - HA Congruence Postulate - If the 38
93
hypotenuse andDacute angleof one
right triangle are
B congruent
x
x angle of another
to the hypotenuse and corresponding acute
right triangle,
then the
35 triangles are congruent
35
7. C.P.C.T.C.
x
7. RS > US
C
A
D
A
C
D
10
t
W
y
H
G
E
V
C
(D-1)
41.
Given: /A > /D; /B > /E;
U of DE
G is the midpoint of AB; H is the midpoint
Prove: CG = AC
FH DF
C
24
M
Q
3
A
G
T
S
C
STATEMENT
B
10
F
NK
4
Q
S
2
R
F
G
B
A
E
A
T
D
B
F
H
P
R
E
REASON
M
14
x
12
8
T
B
A
C
1. /A > /D
2. /B > /E
3. nABC ~ nDEF
4. G is the midpoint of AB
5. H is the midpoint of DE
6. CG is a median
7. FH is a median
CG
1. Given
2. Given
D
E
R
K
3. Postulate Corollary 12a - If two angles of one triangle are
B
congruent to the Btwo corresponding angles of another
C triangle,
D 25 R
then the two triangles are similar.
4. Given
B
5. Given
A
6. Definition of a Median
2
7. Definition of a Median
A
D
AC
8. FH = DF
8. Corollary 29b - If two triangles are similar, then the measures
of corresponding medians are in the same ratio
1 as the measures
A
G
B
F
of correspondingG sides.
E
30
3
C
M
Q
10
126
© 2006 VideoTextInteractive Geometry: A Complete Course
S
2
3
K
4
P
C
T
x
B
6
S 5
14
C
A
8
1
A
C
A
Unit IV, Unit Test Form A
—Continued—
y
t
H
7
D
Name
10
B
A
E
B
C
24
60
C
C
D
F
N
A
Given: nFDN Uin which /F > /NDQ
Q
G
Prove: nNQD ~ nNDF
F
H
T
S
E
D
B
R
S
R
(D-1)
42.
60
Q
W
V
2
D
1
A
E
C
STATEMENT
B
REASON
M
1. nFDN in which /F > /NDQ
2. /N > /N
K
3. nNQD ~ nNDF
A
1. Given
2. Reflexive Property for Congruence
R 3. Postulate Corollary 12a - If two angles of one triangle are
congruent to the two corresponding
angles of another triangle,
D
B
then
the
two
triangles
are
similar
38
E
EB || DC
B
x
93
D
A
C
B
S
T
A
Prove: ED > CD
K
N
Q
x
R
F
B
S
2
1
E
A
1
2
D
6
B
C
K
1. Given
34 - If twoFangles
R the
C of aDtriangle are congruent, then
D 25 R 2. Theorem
30
sides opposite them are congruent
3. Given
H
4. Given
5. Given
M
6. Definition of Perpendicular
P A
61
7. Definition of Perpendicular
8. Definition of Right Triangles
G K
J
9. Definition of Right Triangles
M
a
10. Postulate Corollary 13e - HI Congruence Postulate - If a
3 hypotenuse and one leg of a right triangle
63 are congruent to the
B
N
Q
C
hypotenuse and corresponding leg of another right triangle,
R
then the triangles
are congruent.
A
A
11.C.P.C.T.C.
C
R
A
11. EDC> CD
A
2
B
M P
10
K
4
1
S
B
6
5
T
14
R
8
12
25
M
M
C
R
R
83
C
10
M
REASON
A
E
F
x
2
D
R
M
D
x
8
C
P
45
Q
45
A
C
3. AE > BC
4. DE AE
5. DC
BC
B
6. /E is a right angle
7. /C is a right angle
8. nAED is a right triangle
9. nBCD is a right triangle
10.nAED > nBCD
3
5
D
D
A
B
Q
E
12
8
T
7
60
D
6
B
14
E
D
1. /1 > /2
2. AD >
B BD
K
G
B
C
5
T
M
STATEMENT
G
60
R
F 4
H
C
3
C
S
M
Given: G/1 >
2 /D2; AE > BC;
B
DE
AE; DC BC
P
S
10 1
B Q
C
24
a
60
10
3
A
B
Q
C
(E-3)
F
G
C
P
D
A
C
B
7
G
V
M E
4
3
61
A
A2
D
A
E
43.
x
A
E
W
U
B
x
35
35
D
P
D
A
C
B
P
8
D
6
x
3
Course
© 2006 VideoTextInteractive
Geometry: A Complete
12
x
8
A
B
C
B 4
6
K 10
E
127
A
R
Q
S
N
3
A
Unit IV, Unit Test Form A
—Continued—
D
B
B
B
7
10
60
Name
P
N
A
B
S
Q
A
5
D
M
7
C
24
C
P
D
A
C
45
45
60
60
C
R
D
C
Q
(F-4)
D
F
N
R
Given: /ARG > /CGR; DG > DR
44.
Q
E
D
Prove:
m/3 = m/4
F
H
1
3
C
2
D
4
D
A
1
A
E
G
2
C
B
STATEMENT
REASON
x
1. /ARG > /CGR
2. m/ARG > m/CGR
3. m/1 + m/3 = m/ARG
4. m/2 + m/4 = m/CGR
5. m/1 + m/3 = m/2 + m/4
6. DG > DR
7. /1 > /2
R
C
t
A
8.2 m/1 = m/2
9. m/3 = m/4
1.
2.
3.
4.
5.
6.
7.
M
Given
Definition of Congruent Angles
35
35
Postulate 7 - Protractor - Fourth Assumption - Angle Addition
Postulate 7 - Protractor - Fourth Assumption - Angle Addition
A
C
Substitution
D
Given
Theorem 33 - If two sides of a triangle areWcongruent, then the
y
H are
A congruent.E
A
61 angles opposite them
8. Definition of Congruence V
C
9. Subtraction Property for Equality
M
a
A
B
3
C
1
93
63
B
F
N
A
U
B
N
Q
x
C
G
D
B
H
R
M
10
K
B
6
(G-3)
D
x
C
12
8
T
B
4
E
8
A
B
E
C
M
STATEMENT K
D
REASON
D
25 R
B
F
1. m/B > m/A
2. AC >
F BC C
D
30
R
K
R
Prove:
AC > BC
C
M
6
10
E
C
12
8
A
Given: m/B8 > m/A
45.
6
3
5
14
T
S
1. Given
2. Corollary 38a - If the measure of one angle of a triangle is
A
B
greater than the measure of a second angle of the triangle, then
C
of the third side opposite the larger angle is greater
the measure
than
the
measure
of the side opposite the smaller angle.
R
R
H
G
E
P
K
S
J
B
D
3
E
M
Q
10
K
4
S
T
2
R
F
x
A
14
x
P
R
A
M
12
8
T
B
A
128
K
B
C
B
5
5
© 2006 VideoTextInteractive Geometry: A Complete Course
x
25
B
Q
A
G
A
F
G
C
y
39
D
7
z
N
49
E
C
7
Q
Unit Test Form B
x
Name
Date
Class
35
Score
C
A
t
Unit IV - Triangles
x
H
E
In all of problems 1–50, choose the letter for the correct response.
c
1. A triangle with no congruent sides is _______________.
C
b. Obtuse
d. Isosceles
A
H
E
G
A
R
V
b. three
d. one
B
T
S
CC
F
A
U
G
(A-2)
D
B
K
A
c
An equiangular triangle is also _______________.
a. right
c. equilateral
U D
W
y
b
2. An acute triangle is one with _______________
acute angle(s).
3.
A
C
t
(A-2)
a. two
c. at least one
A
x
35 V
35
(A-2)
a. Equiangular
c. Scalene
W
93
y
R
T
S
b. obtuse
d. scalene
B
H
E
C
M
D
B
(A-2)
c
4. nDEG, shown to the right, is a right triangle. DG is called the _______________.
a. altitude
c. hypotenuse
A
b. leg
d. angle
R
K
B
G
E
(A-1)
D
d
5. In nABF, shown to the right, BG is a(n) _______________
of the triangle.
a. side
c. hypotenuse
A
Q
b. leg
d. altitude
3
2
G
E
c
6. An equilaeral triangle must also be _______________.
b. obtuse
d. right
A
F
G
P
A
Q
B
T
1 x
M
10
3
(A-2)
S
d
7. In an isoceles triangle the base angles are _______________.
a. right
c. congruent
4
S
(A-2)
a. scalene
c. acute
G
B
b. acute
d. both b and c
K
4
T
2
D
E
14
x
B
P
R
12
8
T
B
C
R
A
(D-2)
8.
C
a
An auxilary line is _______________.
a. a line added to a figure to help solve a problem
c. a hypotenuse
D
E
b. an altitude
C
B that may be omitted from
d. a line
aD figure
25 R
© 2006 VideoTextInteractive Geometry: A Complete Course
C
P
30
129
S
A
D
R
K
y
M
7
WA
Unit IV, Unit Test Form B
H
E
—Continued—
P
D
A
C
B
10
B
B
C
24
A
B
S
Q
Name
A
60
45
60
C
D
D
B
M
F
(A-1)
60
C
R
V
45
N
A
D
A
2
61
R
b Q
9. In nABC, shown to the right, G/3 is called
a(n) _______________.
D
E
C
D
B
a. remote interior angle
R
ET
S
b. exterior angle
C
c. acute
a
F
H
G
A
F
EG
B
2
3
C
D
1
A
B
D
Q
M
K
3
4
|| DC
c B
6
10. The HA theorem
states that two right
triangles
are Econgruent ifEB_______________.
93
R
K
A
D
(B-1)
7
A
2
10
B
a.A 61
c. 45
E
C
B
G
E
D
b. 63
B d. 34
(C-2)
A
A
B
(B-1)
T
60
C
303
C
12
B
2 B
5
(D-2)
35
25
1
⋅ BC
A
2
F
G
1
c. AM = ⋅ AB
2 M 10
G
Q
N
H
4
2
E
D
B
1
R
the
d. AN
AN
T
=
MN NC
B
(D-2)
a. 6R
12
8
T
B
R
C
E
M
R
C
130
D
T
H
A
E
A
B
B
C
KC
B
M
D
N
P
Q
B
N
G
K
3E
R
8
S
J
6
R
B
C A
E
G
A
F
G
12
4
x
M
10
A
E
Q
5
2 M M8
or
3
3
x
K
3
8
B
C
C
K
E
BS
T
2
D
K
4
C
y
39
7
14
x
P
R
B
A
F
C
R
D
A
B
© 2006 VideoTextInteractive Geometry: A Complete Course
S
T
U
V
C
C
H
R
D
E
12
8
T
F
X
30
D 25 R
E
C
R
W
R
S
4
J
G
H
D
x
b. 2
d. 2
Q
V
U
25
c. 1.5
A
D
M
A
A
A
A6
d
S right,
15. Ifx STC|| PR14in theP figure to the
then x10= _______________.
8
C
C 6
12
C
R
B
K
61
Q
5
T
3P
F
63
6
R
C
measure
of the
S
A
a
C
S
A
D
G6 K
A
G
E
D
M
b. MN || XBC
3
K
4
W
DR
B
63
Q
MB NC
a. MN =
C
D
U
10
M
A
8
M
C
x
A
K
35
8
B A
8
b
14. If AM = AN in the figure to the right, then _______________.
R
93
B
30
M
R
M N
61
R
B
D 25 R
A
2
E
A
W
Ax
16
A
measures
B
P
1
3
V
of the angles of a triangle are in the ratio 2:2:5, then
M
C
R
R
c
K
largest angle is _______________.
P
C
S
E b. 40O
a. 80O D
c. 100OB
d.C 90O
x
F
C
K
C
Q
R
H
C
C
5
a
D
F
E
A
D
6
4
10
45
y
D
E
M
45
8
A
3
M
A
R
D 25 R
t
D
N
y
bQ
12. GIf P = x , then =Q_______________.
M
D
9
9B y
10
P
x
K
3
4
a.
F b.
H
x
T
P
S
E
T
y
14
9
2
c.
d. x
M
x P
8
yR
13. If the
C 1
B
7
x
60
C
S
60
C
F
G
F
CE
24
C
R
Q
d
11. In nMNQ, shown to the right, find a. _______________
W
8
C
C
D
x
5
S
T
x
a. twoA angles andx the
them
are congruent
B side included between
14
2
x
35
b. each hypotenuse and aP corresponding leg areA congruent
R
12B
8
TP B
c. eachD hypotenuse and Aa corresponding
angle
are
congruent
D
B
C
Q
B
10
38
A
A
1
M
(E-3)
35
1
3
U
y
H
B
A
E
C
E
93 U
S
A
b
16. KB || MT. KR = _______________.
C
A
C
D
a. 20
c. 5
H
(D-2)
b. 15
d. 14
W
AC
R
G
7
R
8
60
C
BF
HJ
BF
GJ
c. CF = HJ
(C-3)
A
K
B
D
D 25 R
E
A
2
D
2
B
P
G
A
F
G
C
T
2
1
R
B
2
D G
D
D
H
T
a
M
Q
x
R
B
E
C
R
A
D
C
8M
10
4
B
B
K
a
M
63
C
S
6
2
E
6
H 4
3 3
12
T
4
Px
P
R
B
R
Q
8
10
F
G
D
x
8
BG
N
F
V
M
K
E
30
U
A
B
A
x
25
M
6
E
R
B
X D 25 R
8
A
61
b. two rectangles
3
C
octagonsQ
W
Rd. two regular
1
D
x
6
5
12
8
T
R
8G
A
a. Postulate 12 - SAS Similarity Assumption
E
b. Postulate Corollary 12a - AA Similarity Postulate
P
c. Postulate 12 - SSSC Similarity
F S C
DC
25 R Assumption
30
d. Definition of Similar Triangles
K
K
B
K
A
C
b
M because of the ______.
C
22. In the figure to the right, DE
|| BC. So, nABC ~ nADE
K
D
K
A
1
C
SB
A
B
5 8
d.
=
14
y S10
P
a. Postulate Corollary 12a - AA Similarity Postulate
S
5 T
S
T
2b. Postulate 12 - SSS xSimilarity14Assumption
C
c. Postulate 12 - SAS Similarity
R
12
8 Assumption
(D-1)
A
B
10
M
T
10
D
A
b
10 are similar by the _______________.
21. The triangles shown to the right
D
K
P
3
F
M
Q
B
3
P
C
C
U
1
B
d
20. Which of these polygons must be similar? _______________
R
CM
Q
5
G
2
(D-1)
M
D
3
C
A
E
x
352
D
45
W
45
C
(C-3)
a. two isosceles triangles
A
G
F
G
c. two parallelograms
with
65DO angles
A
E 60
C
30
A
c. 6 = 8
C
10 A x
b. (6)(10) = 8xS
R
7
A
MA
A
V
c K
19. If in the figure to the right, nABC 3~ nAED,4 then _______________.
a. 6 = x
8 10
8
C
B
R
C
B Q
F
BF HJ
d. CF = JH
JG
R BF
E
5
N
b. CF = GH
M
H
R
6
C
D
d E
18. IfS nBCF ~ nGHJ, then _______________.
a. CFC = GJ
BS
T
y
C
12
E
B
H
T
Q
B
C35D
P
14
60
R
D
F
b. 25
D
d.D 28
B
3
A
10
A
24
E
3
B
E
D T
x
t
B
D
E
K
F
A
B
c
17. KB || MT. MT = _______________.
a. 28
U
c. 70
(C-3) 3
4H
C
K
V
C
3
C
B
10
M
B
1
B
Q
M
S
2
P
A
E
60
N
x EB || DC
QD
B x
3
T
A
E
G
35
R
D
FF
G
38
A
Use the figure to the right for problems 16 and 17.
35
G
Name
x
(D-2)
C
24
R
V
Unit IV, Unit Test Form B
x
—Continued—
B
C
D
E
B A
A
R
T
E
D
R
D
J
F
C
B
A
D 25 R
C
H
(C-3)
25
23. nPQR ~ nMNO. PQ = 18,
a. 92
3
(C-2)
R
QR
= 12, PR
=M16,
b. 36
and
c. 46
M
MN
= 12.
x
B
K
The perimeter
d. 40
R
C
B
a
of nMNO is _________.
P
Q
A
W
R
G
K
S
J
B
D
E
3 and 4? _______________
X
c
24. What is the geometric mean between
b. 3 2
2S
a. 3
2
C
P
8
U
R
T
c. 1 6
V
2
d. 3 6
A
S
e.
A C
35
16
F
x
P
131S
© 2006 VideoTextInteractive Geometry: A Complete Course
5
25
x
B
C
B
C
y
39
z25
61
Q
2
M
10
3
Unit IV, Unit Test Form B
E
—Continued—
S
A
2
F
G
T
1
B
3 14
C
x
P
R
T
(E-3)
K
4
Q
12
8
MA
10
Q
3
C
B
a. 15 5
c. 15 3
S
T
b. 52 30
Pd. 5 55
5
D
x
D 25 R
C
26. The length of DK, in the figure to the right, is the geometric
mean between
M
(E-3)
12
8
T
R
B
6
10
C
R B
M
8
8
C
8
C
D
x
A
E
14
6
5
R
B
b 6
In right triangle CFD, shown to the right, find the length of the altitude CR. ______
25.
B
a
Name
G
K
4
E
F
30
C
A
c
the lengths of _______.
D
a. DR and MR
b. MK and MD
B
c. MK and KR
d. CD and DR
K
E
C
D 25 R
F
30
C
R
D
(D-5)
A
H
R
c
27. The hypotenuse and one leg of a right triangle are 29 and 21. The other leg measures ___________.
a. 1282
c. 20
R
b. 36
d. 5 2
P
C
P
A
S
G
K
25
a
28. The legs of a right triangle are 6 and 9. The hypotenuse measures
_______________.
M
M
R
a. 117
c. 54
b. 3 6
Cd. 117
S
A
W
R
A
X
c
29. A triangle whose sides are 5, 12, and 13 is a(n) _______________.
25
a. obtuse triangle
c. right triangle
M
K
R
b. acute
triangle M
T
S
d. given measure cannot form a triangle
R
(F-7)
x
B
U
V
W
a
30. A triangle whose sides are 6, 11, andX 15 is a(n) _______________.
a. obtuse triangle
c. right triangle
b. acute triangle
V
U cannot
T measure
form a triangle
Sd. given
(D-5)
a
31. The longer leg of a 30-60-90 triangle is 7. The hypotenuse measures _______________.
a. 14 3
3
b. 7 3
c. 14
d. 3 7
(D-5)
a
32. One side of a square is x. The measure of a diagonal is _______________.
b. x 3
c. x 3
2
d. x 2
3
(D-5)
c
33. Each side of an equilateral triangle is 18. The measure of an altitude is _______________.
132
B
R
P
(F-7)
a. 9
x
K
b. 9 3
2
c. 9 3
© 2006 VideoTextInteractive Geometry: A Complete Course
d. 18 3
C
B
Q
S
J
(D-5)
a. x 2
D
V
K
4
C
Unit IV, SUnit Test Form
B
T
2
—Continued—
x
P
B
t
14
y
R
x
T
93
A
x
E
B
C
B
M x
U A
C
A
E
a.DnABC > nDEF
Bc. nABC > nEDF
y
C
b. nABC
> nFED
R
H
E
VnEDF
d.
n
BCA
>
D 25 R
30
T
D
A
FC
C
B
CD
B
EB 24
C
R
PA
M
(E-1)
a. /S > /W
c. TU > VX
H
D
E
BB
A
b. /T >
/W
R
S
d. US > WX C
x
R
G
F
(E-2)
A
t
a. ASA
c. SAS
(E-1)
C
K
J
B
A
E
A
U
G
S
25
b. AC
M
M
R
T
x
C
2
M
A R
F
G
(E-2)
38.
T
B
A
12
W
R
3
b. SSS
S
d. none of2 these
K
1
30 A P
a
B
B
14 C A
x
T
U
A
V
12
8
(E-2)
T
D 25 R
2
R
E
c
nGHK > nJHK by __________.
a. ASA
b. SSS
D
F
G
3
B
F
S
R
D 25 R
T
B
R
12
8
T
M
6
M
Q
R
B
R
M
D8
x
F
30
5
C
A
14
x
8
C
1061
C
M
prove nPRQ >
a. ASA
E
b. SSS
c.R SAS
D 25 R
M
25
C
F
30
M
A 6
K
R
T
A
N8
R
A C
B
N
R
V
UM H
4
C
D
P
Q
x
25
D
K
3
12
G 6 KB
R
SC
J
A
B
4
C
E
H
E
C
A
A
DP
Q
A
F
V
G
x
R
D
B
K
(F-2)
R
8
K
U
T
S
C
B
6
D
E
A
C
H
3 G
X
6
10
W
B
N
M
K
40. In the figure to the right, given that PR > CRS, and PQ
bisectsS RS,X you couldR
P
b
nPSQ _________.
D
7
12
63
S
10
C
c. SAS
P
3
C
6M
63
W
8
S
a
P
E
M
R
C
1
C B
K
4
B
10 39
C
D
K
D
M
30 8
M
E
Q
2
(E-2)
A
C8S
25
x R
39. In the figure to the right, given that GK > JK and HK GJ, you could prove
G
5
Q
A
C
4
D
F
y
C
6
5
T
P
S
A
K
G
2
D
F
Q
M
x
61
1
x
B
M
B
E
D
X
Q
C
6
M
4
D
45
M
C
a
B
The two triangles shown to the right are congruent
by _______________.
10
a. ASA
c. SAS
10
a
60
R
R
D
AN
8
D
A
QK
A
P
D
8
3 C
R C
61 x
A
5
D
45
3
B
7
CA
B
R
3
C
B
D 25 52RC
6
M
R
E
B
8
T
C
5
C
M
B
4
3
B
C
12
D 25 R
E
E
B
60
CD
1C
A
8S
1
14
x
3
A
1
CB
A
F
E
c.BAD
G
P
K
E
B
AM N
10
2 EK
Q
H
a. DC
A C
E
2 B
14
60
R
C
c B
37. In nADC, the side included by /A and /D is _______________.
S
T
R
7
x
Q
3C
C
EB || DC A
B
R
D
4
3 D
B
R
P
10
24
A
F
Q
R
1
B
SK
C
A
S
BF
A
10
DT
P
V
C
AR
2
D
Q
B
M
A
2
A
D
WB
H
E
4
SA
b. SSS
G
E
G
R
d. Not enough
information
to
tell.
D
B
y
P
C
A
D
D
2 c
36. The two triangles shown to the right are congruent
by ___________.
R
K
C
F
P
3
G
2
F
A
x
C
E
R
x
35
35
A
Q
B
x
M
B
T
60
B
C
10
K
45
1
45
60 14 B
D
8
T
B
E
4
D
S
DC
C
G
38
E
93
D
E
M
S
2
Q
H
T
D
3
Q
N
A
b G
If nSTU is congruent to nVWX, then _______________.
K
7
P
Q
60
D
F
U
Q
10
F
H
S
F
G
B
N
C
R
G
A
C
D
B
S
EB || DC
60
8
R D
K
W
4
Q
C
E
A
M
60
E
24
KG
DA
bDB and _____.
d
34. Two statements of t congruence for the two triangles shown
are
_____
B
35.
F
D10
F
C
7
H
E
C
35
35
12
38
P
QB
3D
6
T
C
6
D
D
A
B
8
10
V
A
(E-1)
D
S
E
R
B
C
G
A
WR
8
C
H
12
8
A
Name
5
N
A
DA
x
U
6
B
x
F
35
35
10
3
C A
x
M
Q
K
S
J
A
C
B
E
B
C
H
R
41. In the figure to the right, nABCC> nDEF.
Therefore, /CBA >
P
W
R
c Q
/FED because _____.
R
P
A
SX
a. they are right triangles b. they are acute triangles
25 V
c. CPCTC
d. not enough information
to
U tell
T
S
M
M
R
P
C
(Appendix A-1)
G
K
A
A
S
J
5
x
R
B
K
C
B
x
A
x
25
B
C
B
C
b
42. The statement “If nABC > nDEF, then /DEF > /ABC” illustrates _______________.
a. the reflexive property of congruence
T
S
b. the symmetric property of congruence
c. the transitive property of congruence
R
M
5
5
W
X
M
M
K
y
39
z
7
W
V
U
X
S
T
U
39
y
C
F
A
S
R
R
B
D
E
V
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