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HONORS GEOMETRY A
Semester Exam Review Answers
Honors Geometry A
Semester Exam Review Answers
2015-2016
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HONORS GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 1
1.
point, line, and plane
2.
angle bisector construction
3.
Construct segment BC  , then construct the perpendicular bisector of CC  .
C
B
C
4.
Draw a line through point H, then copy the angle formed so that its vertex is at point H.
5.
Each point on the perpendicular bisector is equidistant from points A and B.
6.
Each point on the angle bisector is equidistant from the sides of the angle.
7.
Each corresponding pair of points are the same distance from each other.
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HONORS GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 2
8.
A  4, 2 
9.
y
A
A
C
B
E O C
F
x
B
D
10.
a.
A translation five units to the right and three units down.
b.
 x, y    x  5, y  3
c.
Triangles that undergo rigid transformations preserve both distance and angles,
therefore the triangles are congruent.
d.
See graph above for the transformed triangle.  x, y    x,  y 
a.
 x,  y 
b.
  x, y 
c.
 x  6, y  2 
d.
 y,  x 
e.
  x,  y 
f.
  y, x 
g.
 x, 2  y 
h.
 2  x, y 
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HONORS GEOMETRY A
11.
Semester Exam Review Answers
y D
a.
A
C
B
y 1
O
A
D
B
x
C
b.
Yes, reflection is a rigid transformation, therefore lengths and angle
measurements are preserved.
c.
Yes, it would be the same.
12.
translations, rotations, and reflections
13.
congruent
14.
a.
x-axis or the line x  2
b.
180 degrees about the point  2, 0 
a.
Yes, it is a reflection. The image of point  x, y  is  x,  y  .
b.
No, it is not a translation. Points A, B, and C do not translate to points
A, B and C  .
c.
Reflect across the y-axis, then reflect across the x-axis, then reflect across the y-axis
OR reflect across the x-axis three consecutive times OR reflect across the x-axis,
reflect across the y-axis, then reflect across the y-axis again.
a.
y  3 x  2
b.
y  3 x  2
c.
y  3x  10
15.
16.
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HONORS GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 3
17.
A  E
BC  FG
18.
congruent
19.
RS  TS , RW  TU
20.
 RSW   TSU , RS  TS
or
W   U , RU  TU
21.
 RSW   TSU ,  W  U
22.
 R   T , W  U
or
 R   T ,  RSW   TSU
23.
No, since SSA is not a congruence theorem.
24.
The corresponding 500 ft. sides are congruent. The corresponding 450 ft. sides are
congruent. The vertical angles included between the 500 ft. and 450 ft. sides are
congruent. Therefore the two triangles are congruent, and by CPCTC the other
corresponding sides are congruent, making the length of the power line 625 ft.
25.
SAS
26.
AAA cannot be used to prove two triangles congruent.
27.
ASA
28.
SSS
29.
SSA cannot be used to prove triangles congruent.
30.
AAS
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HONORS GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 4
31.
Opposite sides are parallel and congruent. Diagonals bisect each other. Opposite angles
are congruent. Consecutive angles are supplementary.
32.
Diagonals are congruent. All angles are right angles.
33.
All sides are congruent. Opposite angles are bisected by diagonals. Diagonals are
perpendicular.
34.
DF  EF
35.
EC  12, DE  11
36.
a.
AD  BD
b.
AE  EC
c.
DE is parallel to BC
d.
BC
e.
The ratio AB : AD is 2 :1
f.
The ratio AE : AC is 1: 2
a.
Alternate interior, alternate exterior, corresponding, and vertical
b.
linear pairs, same side interior, same side exterior
37.
38.
There are many different proofs. Below is one example.
Statements
1. m  n, 1  16
2. 1   9
3.
4.
5.
6.
 9  12
1  12
16  12
pq
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Reasons
1. Given
2. If two parallel lines are cut by a transversal, then corresponding
angles are congruent.
3. Vertical angles are congruent
4. Substitution (statement 2 into statement 3) or transitive property
5. Substitution (statement 1 into statement 4)
6. If two lines are cut by a transversal so that corresponding angles
are congruent, then the lines are parallel.
HONORS GEOMETRY A
Semester Exam Review Answers
39.
Statements
1.  AC  D, AB  DE
2.  ACB   DCE
3.  ACB   DCE
4. CE  CB
Reasons
1. Given
2. Vertical angles are congruent
3. AAS
4. CPCTC
40.
Statements
1. C is the midpoint of BE
2. BC  CE
3. AB  DC
4. AB  BE , DC  BE
5.
m B  mDCE  90o
 B   DCE
6.  ABC   DCE
7.  A   D
41.
5. Definition of perpendicular
6. SAS
7. CPCTC
Given: m  n
Prove:
m 2  m 4  m 5  180o
Statements
1. m  n
2. m1  m 2  m 3  180o
3.
Reasons
1. Given
2. Definition of midpoint
3. Given
4. Given
m1  m 4
m 3  m 5
4. m 4  m 2  m 5  180o
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Reasons
1. Given
2. The sum of three adjacent angles whose
vertices lie on a line have a sum of 180o
3.If two parallel lines are cut by a transversal,
then the alternate interior angles have the
same measure
4. Substitution (Statements 3 into statement 2)
HONORS GEOMETRY A
Semester Exam Review Answers
Unit 2, Topic 1
42.
a.
Statements
1. DE  BC
 ADE   ABC
2.
 AED   ACB
3.  ADE ~  ABC
b.
AD
8

EC
9
Let AD  x, then EC  2 x
Substituting:
x 8

9 2x
2 x 2  72
x 2  36
x  6  AD
c.
6 DE

15 18
15 DE  108
DE  7.2
43.
44.
a.
SAS similarity
b.
20 40

30 L
L  60
Corresponding sides are proportional.
Corresponding angles are congruent.
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Reasons
1. Given
2. If parallel lines are cut by a transversal,
then corresponding angles are congruent.
3. AA Similarity
HONORS GEOMETRY A
45.
46.
47.
48.
Semester Exam Review Answers
a.
Since QRS ~ QTU , then  QRS   T because corresponding angles in
similar triangles are congruent.  QRS and T are corresponding angles of two
lines cut by a transversal. Since the corresponding angles are congruent, the lines
are parallel.
b.
QR; TU
a.
B
b.
parallel to
c.
5
a.
1
b.
2:3
c.
they are equal
a.
see graph below
1
2
b.
1:3
c.
no, rigid transformations
preserve lengths
y
10
9
8
A  5, 6 
C   7, 6 
7
6
5
4
3
2
1
-10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1O
A
C
-1
1
2
3
4
5
6
7
8
-2
-3
-4
-5
B
P
-6
-7
-8
-9
-10
49.
a.
The scale factor is 3
b.
The center of dilation is  5,8 
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B  4, 3
9 10
x
HONORS GEOMETRY A
50.
Semester Exam Review Answers
a.
10 12

15 x
x  18
b.
10
12

y y  1.5
10 y  15  12 y
15  2 y
7.5  y
51.
a.
b.
52.
5
h

9 144
h  80 m
1442  802  27136  164.73 ft
Answers below are examples. Your numbers will be different. As long as the angles are
congruent and sides proportional, then your answer is correct.
a.
SSS Similarity
6
8
12
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12
9
18
HONORS GEOMETRY A
b.
Semester Exam Review Answers
AA Similarity
42o
42o
88o
o
88
c.
SAS Similarity
9
128o
7
Unit 2, Topic 2
53.
a.
BC
b.
AB
c.
BC
AC
d.
AB
AC
e.
BC
AB
f.
AB
AC
g.
BC
AC
h.
AB
BC
i.
A
j.
C
k.
AC
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18
128o
14
HONORS GEOMETRY A
Semester Exam Review Answers
54.
All statements are true.
55.
The triangles are congruent. However sin A 
56.
a.
In  SQU , cos S 
b.
sin U 
c.
Yes. cos U 
16
12
.
, sin B 
20
20
20 4
8 4
 , In  SRT , cos S  
25 5
10 5
20 4

25 5
15 3
6 3
 , cos T  
25 5
10 5
57.
sin 20o 
V
16
16sin 20o  V
V  5.47 ft
58.
a.
sin  
15
17 Yes, this is safe.
  61.9o
15
17

b.
sin 70o 
24
L
L sin 70o  24
24
L
sin 70o
L  25.54 ft
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24
L
70o
HONORS GEOMETRY A
59.
Semester Exam Review Answers
a.
cos15o 
h
5280
5280 cos15o  h
h  5100 ft
5280
v
15o
b.
h
sin15o 
v
5280
5280sin15o  v
v  1367 ft
60.
a.
x 2  152  202
x  625
x  25
The area of the leftmost triangle is
1
 15  20  150 m 2 .
2
b.
252  y 2  652
The area of the middle triangle is
1
 25  60  750 m 2
2
625  y 2  4225
y 2  3600
y  60
Since tan  1 
c.
m 2  180  53.1  90  36.9o
65 m
h
x
20 m
1
y
2
m
15 m
Approximately 1764 square meters (see work in box).
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20
, m1  53.1o
15
h
, h  60sin 36.9o  36 m
60
m
cos 36.9o  , m  60 cos 36.9o  48 m
60
sin 36.9o 
Area of rightmost triangle is
1
 36  48  864 m 2
2
HONORS GEOMETRY A
Semester Exam Review Answers
61.
a.
1002  202  A2
10000  400  A2
10400  A2
A  10400  102.0 ft
b.
tan  
20
100
  11.3o
62.
a.
 
n 2  142  232  2  14  23cos 68o
n 2  483.753
n  21.99 miles
b.
63.
m B  58o
sin 58o sin 79o

BC
20
BC  23.15 ft
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 
1
A  14  23sin 68o  149.28 mi 2
2