Geometry
MIDTERM EXAM REVIEW ANSWER KEY
ALGEBRA SKILLS
I.
1)
-6 6 + 6 5
2)
-48 15
II.
1)
(5a – 9) (3a + 4)
2)
2r (6r – 5) (2r – 3)
III.
1)
4
k=- ;2
3
2)
k=-
IV.
1)
x=
2)
x=
V.
1)
(-17, -3)
2)
(11, -2)
VI.
1)
x = -2
2)
y = -7x – 25
3)
y=
4)
y=5
5)
y = 9x – 25
6)
7
y = - x + 29
3
1)
2)
3)
Neither, the slopes are neither the same nor opposite reciprocals.
Neither, the slopes are neither the same nor opposite reciprocals.
Perpendicular, the slopes are opposite reciprocals
VII.
10
13
KD/MPH/KAL/CR 12/14
10 10
;
3 3
23
22
3
4
x+7
1
Geometry
MIDTERM EXAM REVIEW ANSWER KEY
UNIT 1 BASICS OF GEOMETRY
I.
1)
2)
a)
or line e
b)
Yes, they all lie on plane PSQ.
c)
Yes, they are three non-collinear points.
d)
Point P
a)
False
e)
b)
or line e
True
c)
False
d)
True
e)
True
f)
True
4)
6
II.
Sketches will be reviewed in class.
III.
1)
15 or 43
IV.
1)
Yes PQ = QR = 2 2 .
2)
PQ Midpoint = (-3, 2), QR Midpoint = (-1, 0), PR Midpoint = (-2, 1)
3)
Q (7, -3)
V.
VI.
VII.
1)
2)
2)
13
3)
PQ = 8, RP = 16, ST = 6
Answers may vary:
5)
EDF
7)
3, BDA
1)
mSQR = 50, mPQR = 100
2)
m1 = 50, m2 = 130, m3 = 50, m4 = 40
3)
15
6)
52
1)
x = 17, mKMQ = 106
4)
x= -
4
3
KD/MPH/KAL/CR 12/14
6)
3)
4)
4)
,
,
Answers may vary: BDA, GDA
x = 43, y = 77
2)
Point D
5)
7.5
mA = 57, mB = 33
3)
28
, mFMQ =
2
Geometry
VIII.
1)
a)
b)
P
e)
PA , PB
d)
AB
DA , BA
f)
2)
a)
IX
1)
(-3, 2)
X.
1)
(x + 3)2 + y2 = 121
3)
(x - 2)2 + (y – 12)2 = 29
XI.
c)
2
b)
.5
2)
c)
2 5
2.5
d)
3)
5
e)
6
f)
5.5
h)
8
inside, on, outside
2) (x - 4)2 + (y + 11)2 = 68
Standard Form: (x - 2)2 + ( y – 4)2 = 32
Center: (2, -4)
Radius: 4 2
Circumference: 8 2p u
Area: 32 u2
KD/MPH/KAL/CR 12/14
3
Geometry
MIDTERM EXAM REVIEW ANSWER KEY
UNIT 2 REASONING & PROOF
I.
1)
Sometimes
2)
Always
3)
Always
4)
Sometimes
5)
Sometimes
6)
Never
7)
Sometimes
8)
Never
9)
Always
10) Always
II.
1)
2)
3)
False. H could be between O and W.
False. 30/2 = 15 and 15 is not even.
False. 1 and 2 are congruent. 1 and 2 would only be supplementary
if they were both right angles.
III.
1)
Hypothesis: the other team arrives within 5 minutes
Conclusion: the game will start at 4:00
Statement: If the other team arrives within 5 minutes, then the game will
start at 4:00.
2)
Converse: If they are adjacent, then two angles form a linear pair.
Inverse: If two angles do not form a linear pair, then they are not adjacent.
Contrapositive: If they not are adjacent, then two angles do not form a
linear pair.
3)
Converse: If neither of two angles is obtuse, then two angles are
complementary.
Inverse: If two angles are not complementary, then either of two angles is
obtuse.
Contrapositive: If either of two angles is obtuse, then two angles are not
complementary.
4)
Conditional: If it ends in zero, then a number is divisible by five.
Converse: If a number is divisible by five, then it ends in zero.
1)
Law of Syllogism. If the newspaper club meets today, then the garbage is
going out.
No conclusion. School could be closed for Christmas.
Law of Detachment. A is an acute angle.
No conclusion. The two angles could be vertical and both right.
IV.
2)
3)
4)
KD/MPH/KAL/CR 12/14
11) Always
12) Sometimes
4
Geometry
V.
1)
3)
VI.
Addition Property of Equality
Transitive Property of Equality
2)
4)
Statements
1)
2)
3)
4)
5)
6)
VI. 1)
2 (3x – 7) – 8x = 8 – 2 (9x – 5)
6x – 14 – 8x = 8 – 18x + 10
-2x – 14 = 18 – 18x
16x – 14 = 18
16x = 32
x=2
Reasons
1)
2)
3)
4)
5)
6)
Statements
1)
2)
3)
4)
2)
AD
AD = AB + BD
BD = BC + CD
AD = AB + BC + CD
2)
3)
4)
5)
6)
7)
8)
3 is supplementary to 1
4 is supplementary to 2
m3 + m1 = 180
m4 + m2 = 180
1  2
m1 = m2
m3 + m1 = m4 + m2
m3 + m2 = m4 + m2
m3 = m4
3  4
KD/MPH/KAL/CR 12/14
Given
Distributive Property of Equality
Substitution
Addition Property of Equality
Addition Property of Equality
Division Property of Equality
Reasons
1)
2)
3)
4)
Statements
1)
Symmetric Property of Equality
Multiplication Property of Equality
Given
Segment Addition Postulate
Segment Addition Postulate
Substitution
Reasons
1)
Given
2)
Definition of Supplementary
Angles
Vertical angles are congruent
Definition of Congruent Angles
Substitution
Substitution
Subtraction Property of Equality
Definition of Congruent Angles
3)
4)
5)
6)
7)
8)
5
Geometry
3)
Statements
1)
VIII.
AC  BC
2)
3)
4)
5)
6)
7)
8)
9)
3 is complementary to 1
ACB is a right angle
mACB = 90
mACB = m1 + m2
m3 + m1 = 90
m3 + m1 = mACB
m3 + m1 = m1 + m2
m3 = m2
3  2
1)
3)
5)
7)
9)
11)
13)
Def. of Congruent Segments
Def. of Right Angle
Def. of Supplementary Angles
Def. of Angle Bisector
Addition Property of Equality
Def. of Complementary Angles
Angle Addition Postulate
KD/MPH/KAL/CR 12/14
Reasons
1)
Given
2)
3)
4)
5)
6)
7)
8)
9)
Definition of Perpendicular Lines
Definition of a Right Angle
Angle Addition Postulate
Definition of Complementary s
Substitution
Substitution
Subtraction Property of Equality
Definition of Congruent Angles
2)
4)
6)
8)
10)
12)
Vertical angles are congruent.
Transitive Property of Equality
Def. of Midpoint
Def. of Perpendicular Lines
Symmetric Property of Equality
Segment Addition Postulate
6
Geometry
MIDTERM EXAM REVIEW ANSWER KEY
UNIT 3 PARALLEL & PERPENDICULAR LINES
I.
1)
a)
2)
a)
d)
g)
alt ext s
vert s
no relationship
b)
e)
II.
1)
1)
67
x = 15
y = 40
III.
1)
j || k because if a line is perpendicular to two coplanar lines, then those lines
2)
3)
4)
are parallel.
m || n because alternate interior angles are congruent.
m || n because consecutive interior angles are supplementary.
j || k || l because corresponding angles and alternate exterior angles are
congruent.
1)
120
IV.
V.
1)
b)
c)
d)
Answers may vary for a - d
2)
2)
cons int s
cons ext s
3)
ABC
c)
f)
corr s
alt int s
x = -3.5, 10
30
3)
Statements
e)
4)
x=9
y = 33
15
Reasons
1)
1  2
1)
Given
2)
JL ||KM
2)
Alt int s   || lines
3)
If a line is  to one of two parallel
lines, then it is  to the other line.
3)
2)
Statements
1)
2)
3)
1  2, 3  4
l || n, n || m
l || m
KD/MPH/KAL/CR 12/14
Reasons
1)
2)
3)
Given
Alt int s   || lines
Transitive Property of Parallelism
7
Geometry
3)
Statements
Reasons
1)
1)
Given
2)
3)
FRA  RAY
4  5
6  7
4) m4 = m5
m6 = m7
mFRA = mRAY
5) mFRA = m4 + m5
mRAY = m6 + m7
6) m4 + m5 = m6 + m7
7) m5 + m5 = m6 + m6
8) 2 m5 = 2 m6
9) m5 = m6
10) 5  6
2)
3)
|| lines  alt int s 
Definition of Angle Bisector
4)
Definition of Congruent Angles
5)
Angle Addition Postulate
6)
7)
8)
9)
10)
Substitution
Substitution
Substitution
Division Property of Equality
Definition of Congruent Angles
11)
11) Alt int s   || lines
KD/MPH/KAL/CR 12/14
8
Geometry
MIDTERM EXAM REVIEW ANSWER KEY
UNIT 4 CONGRUENT TRIANGLES
I.
II.
III.
IV.
V.
1)
1)
3)
2)
Vertex  = 104 and base  = 38
3)
3
1)
m1 = 30, m2 = 120
2)
m1 = 37, m2 = 42, m3 = 132, m4 = 73, m5 = 73, m6 = 30
3)
-2
1)
a)
2)
95
3)
2
1)
y = 14; 23, 23, 134
2)
x = -8 or x = 3; 24
3)
x = 1.4; y = -1; 4
4)
x = 13; 45, 45
5)
z = 12; SR = 9
6)
x = 54
7)
mCAD = 44, mACD = 44, mACB = 136, mABC = 22, mCAB = 22
1)
NMQ  STR
by ASA
VI. 1)
Right isosceles
Isosceles
2)
4)
4)
YZ
b)
12
c)
CAB
2)
Scalene
Equilateral and Equiangular
Not 
3)
Statements
1)
MC @ AC
2)
3)
4)
A and M are right angles
A  M
NCM  ACB
NMC  BAC
KD/MPH/KAL/CR 12/14
d)
Z
ZXY
FHG  EJF
by AAS
Reasons
1)
Given
2)
3)
4)
All right angles are congruent
Vertical angles are congruent
ASA
9
Geometry
2)
Statements
Reasons
1)
XY||WZ
1)
Given
2)
XZ ||WY
XZY  WYZ
YZW  XYZ
2)
|| lines  alt int s 
3)
4)
5)
YZ  YZ
XZY  WYZ
X  W
3)
4)
5)
Reflexive
ASA
CPCTC
3)
Statements
Reasons
1)
BD @ EC
1) Given
2)
3)
AC @ AD
1  2
BDA  ECA
2)
3)
Isosceles Triangle Theorem
SAS
4)
AB @ AE
4)
CPCTC
4)
Statements
1)
ÐK @ ÐM
Reasons
1)
Given
PL bisects KPM
2)
3)
4)
PL @ PL
KPL  MPL
KPL  MPL
2)
3)
4)
Reflexive
Def. of Angle Bisector
AAS
5)
KL @ LM
5)
CPCTC
6)
L is the midpoint of KM
6)
Def. of a Midpoint
KD/MPH/KAL/CR 12/14
10
Geometry
5)
Statements
Reasons
1)
XQ || TR
1)
Given
2)
XR bisects QT
Q  T
X  R
2)
|| lines  alt int s 
3)
Def. of Segment Bisector
4)
QS @ ST
XQS  RTS
4)
AAS
5)
XS @ SR
5)
CPCTC
6)
QT bisects XR
6)
Def. of Segment Bisector
3)
KD/MPH/KAL/CR 12/14
11
Geometry
MIDTERM EXAM REVIEW ANSWER KEY
UNIT 5 RELATIONSHIPS IN TRIANGLES
I.
1)
always
2)
never
II.
1)
D
2)
C
III.
1)
a)
b)
Yes
2)
13 and 63 units
3)
DE , RE , RD
4)
6; S, U, N
5)
a)
IV.
1)
3 6
V.
1)
obtuse
VI.
Isosceles right triangle
No
OS
KD/MPH/KAL/CR 12/14
b)
2)
RS
acute
c)
RO
2)
2 34
3)
right
4)
not a triangle
12