Geometry A Semester Exam Review Answers 2015-2016

GEOMETRY A
Semester Exam Review Answers
Geometry A
Semester Exam Review Answers
2015-2016
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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 DC and DE .
7.
Each corresponding pair of points are the same distance from each other.
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GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 2
8.
A  4, 2 
9.
y
A
A
C
B
E O C
x
F
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.
 3, 5
b.
 3,5
c.
 4,3
d.
 5,3
e.
 3, 5 
f.
 5, 3
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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 can performed by a single reflection. The image of point  x, y  is  x,  y  .
b.
No, it cannot be performed by a single 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 yaxis, OR reflect across the x-axis, then reflect across the y-axis, then reflect across
the y-axis again, OR reflect across the x-axis three consecutive times.
15.
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GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 3
16.
A  E
BC  FG
17.
congruent
18.
RS  TS , RW  TU
19.
 RSW   TSU , RS  TS
or
W   U , RW  TU
20.
 RSW   TSU , W  U
21.
 R   T , W  U
or
 R   T ,  RSW   TSU
22.
No, since SSA is not a congruence theorem.
23.
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 by SAS, and by CPCTC the other
corresponding sides are congruent, making the length of the power line 625 ft.
24.
SAS
25.
AAA cannot be used to prove two triangles congruent.
26.
ASA
27.
SSS
28.
SSA cannot be used to prove triangles congruent.
29.
AAS
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GEOMETRY A
Semester Exam Review Answers
Unit 1, Topic 4
30.
Opposite sides are parallel and congruent. Diagonals bisect each other. Opposite angles
are congruent. Consecutive angles are supplementary.
31.
Diagonals are congruent. All angles are right angles.
32.
All sides are congruent. Opposite angles are bisected by diagonals. Diagonals are
perpendicular.
33.
DF  EF
34.
x  12, y  11, z  10
35.
a.
AD  BD
b.
AE  EC
c.
DE is parallel to BC
d.
DE
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
36.
37.
There are many different proofs. Below is one example.
Statements
1. m  n, 1  16
2. 1   9
3.  9  12
4. 1  12
5. 16  12
6. 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.
GEOMETRY A
Semester Exam Review Answers
38.
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
39.
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
Reasons
1. Given
2. Definition of midpoint
3. Given
4. Given
5. Definition of perpendicular
6. SAS
7. CPCTC
Unit 2, Topic 1
40.
a.
Statements
1. DE  BC
 ADE   ABC
2.
 AED   ACB
3.  ADE ~  ABC
b.
15 x

9 12
x  20
c.
9
y

24 30
y  11.25
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Reasons
1. Given
2. If parallel lines are cut by a transversal,
then corresponding angles are congruent.
3. AA Similarity
GEOMETRY A
41.
a.
Semester Exam Review Answers
SAS similarity
b.
20 40

30 L
L  60
42.
43.
44.
45.
a.
Corresponding sides are proportional.
b.
Corresponding angles are congruent.
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 these corresponding angles are congruent, the
lines are parallel.
b.
QR; TU
a.
B
b.
parallel to
c.
5
a.
1
b.
3: 2
c.
they are equal
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1
2
GEOMETRY A
46.
Semester Exam Review Answers
a.
y
10
A  5, 6 
9
C   7, 6 
8
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
B
P
-5
-6
-7
-8
-9
-10
b.
1:3
c.
they are equal
47.
10 12

15 x
x  18
48.
301 m x cm

100 m 10 cm
x  30.1
49.
a.
b.
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5
h

9 144
h  80 m
1442  802  27136  164.73 ft
B  4, 3
9 10
x
GEOMETRY A
50.
Semester Exam Review Answers
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
9
12
b.
18
AA Similarity
42o
42o
88o
88o
c.
SAS Similarity
9
128o
7
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18
128o
14
GEOMETRY A
Semester Exam Review Answers
Unit 2, Topic 2
51.
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
52.
All statements are true.
53.
The triangles are congruent. However sin A 
54.
a.
In  SQU , cos S 
b.
sin U 
c.
Yes. cos U 
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12
16
.
, 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
GEOMETRY A
Semester Exam Review Answers
55.
sin 20o 
V
16
16sin 20o  V
V  5.47 ft
56.
a.
sin  
15
17 Yes, this is safe.
17
15
  61.9o

b.
sin 70o 
h
17
17
h
17 sin 70o  h
70o
h  15.97 ft
57.
a.
cos15o 
h
5280
5280 cos15o  h
h  5100 ft
5280
v
15o
b.
h
v
sin15o 
5280
5280sin15o  v
v  1367 ft
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GEOMETRY A
Semester Exam Review Answers
58.
a.
x 2  152  202
x  625
x  25
b.
252  y 2  652
625  y 2  4225
y 2  3600
y  60
59.
a.
1002  202  A2
10000  400  A2
10400  A2
A  10400  102.0 ft
b.
tan  
20
100
  11.3o
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