Download QBA 1 Review

QBA 1 Review
Short Answer
1. Find m
.
A
>>
3. Violin strings are parallel. Viewed from above, a
violin bow in two different positions forms two
transversals to the violin strings. Find x and y in the
diagram.
xº
100º
C
(3x - 70)º
>>
B
(4x + y)º
(8x + y)º
2. Find m
.
R
T
S (3x)º
>>
4. Use the information
,
and the theorems you have learned to show that
.
U
(4x – 24)º
>>
V
1
2
5. Write a two-column proof.
Given:
Prove:
3
1
l
2
m
Complete the proof.
Proof:
Statements
1.
2.
3.
4.
60º
Reasons
1. Given
2. [1]
3. Substitution (Steps 1 and 2)
4. [2]
l
m
6. In a swimming pool, two lanes are represented by
lines l and m. If a string of flags strung across the
lanes is represented by transversal t, and x = 10,
show that the lanes are parallel.
7. Write and solve an inequality for x.
D
2x + 4
8
A
B
C
(3x + 4)º l
(4x – 6)º
m
t
8. Write a two-column proof.
Given:
Prove:
t
1 2
m
l
Complete the proof.
Proof:
Statements
1. [1]
1. Given
2.
2. [2]
3. [3]
3.
9. Apply the transformation M to the triangle with the
given vertices.
Identify and describe the transformation.
M: (x, y)
(x – 6, y + 2)
E(3, 0), F(1, –2), G(5, –4)
10. Apply the transformation M to the polygon with the
given vertices.
Identify and describe the transformation.
M: (x, y)
(–x, –y)
A(–3, 6), B(–3, 1), C(1, 1), D(1, 6)
Reasons
y
6
G
–6
E
F
P
Q
6
R
11. Determine whether triangles
are congruent.
and
–6
x
12. Prove that the triangles with the given vertices are
congruent.
A(3, 1), B(4, 5), C(2, 3)
D(–1, –3), E(–5, –4), F(–3, –2)
13. Daphne folded a triangular sheet of paper into the
shape shown. Find
, given
,
, and m
.
N
F
D
E
P
M
E
17. Find m
m
, given
.
,
, and
D
C
22º
B
A
14. One of the acute angles in a right triangle has a
measure of
. What is the measure of the other
acute angle?
15. Find
E
D
C
42º
61º
A
B
F
18. Given:
Identify all pairs of congruent corresponding parts.
A
M
.
(
L
B
(4x + 7)º
C
O
19. Given that
m
.
(6x - 9)º
E
M
23º
N
A
16. Find
and m
118º
K
and
, and
N
, given
C
,
.
B
D
= 23°, find
20.Given:
,
,
. T is the midpoint of
.
R
S
T
U
Prove:
Complete the proof.
Proof:
Statements
Reasons
1.
2.
and
are right angles.
3.
4.
5.
6.
7. T is the midpoint of
.
8.
9.
10.
1. Given
2. [1]
3. Right Angle Congruence Theorem
4. Given
5. [2]
6. Given
7. Given
8. Definition of midpoint
9. [3]
10. Definition of congruent triangles
21. Tom is wearing his favorite bow tie to the school dance. The bow tie is in the shape of two triangles.
Given:
,
,
,
Prove:
B
C
A
D
E
Complete the proof.
Proof:
1.
2.
3.
4.
5. [3]
,
Statements
,
Reasons
1. Given
2. Given
3. [1]
4. [2]
5. Definition of congruent triangles
22. Given the lengths marked on the figure and that
bisects
, use SSS to explain why
.
R
4 cm
E
A
3 cm
3 cm
D
||
T
U
Complete the explanation.
4 cm
C
||
S
B
23. The figure shows part of the roof structure of a
house. Use SAS to explain why
.
24. Given: P is the midpoint of
and
.
Prove:
T
It is given that [1]. Since
and
are
right angles, [2] by the Right Angle Congruence
Theorem. By the Reflexive Property of
Congruence, [3]. Therefore,
by
SAS.
R
P
S
Q
Complete the proof.
Proof:
Statements
1. P is the midpoint of
and
Reasons
.
1. Given
2. [1]
2.
,
3. [2]
3. Vertical Angles Theorem
4. [3]
4.
25. What additional information do you need to prove
26. Determine if you can use ASA to prove
by the SAS Postulate?
. Explain.
C
B
E
D
A
A
B
||
C
||
D
27. Determine if you can use the HL Congruence Theorem to prove ACD
know.
P
A
DBA. If not, tell what else you need to
B
|
^
^
|
C
D
Q
28. For these triangles, select the triangle congruence
statement and the postulate or theorem that supports
it.
29. A pilot uses triangles to find the angle of elevation
from the ground to her plane. How can she find
m ?
L
D
C
40°
12 km
O
20 km
J
K
B
A
A
C
Find the value of x.
3x
+1
31.
5
2x
30.
12 km
B
20 km
31.
Given:
,
Prove:MLP is isosceles.
,
L
M
N
P
O
Complete the proof.
Proof:
Statements
,
Reasons
1.
2.
3.
4.
5.
6.
and
7.
8. MLN PLO
9.
10. MLP is isosceles.
1. Given
2. Given
3. Definition of congruent line segments
4. Reflexive Property of Equality
5. Subtraction Property of Equality
6. Segment Addition Postulate
7. Substitution Property of Equality
8. [1]
9. [2]
10. Definition of isosceles triangle
32. Two Seyfert galaxies, BW Tauri and M77,
represented by points A and B, are equidistant from
Earth, represented by point C. What is m ?
33.
Find m
.
P
C
|
|
xº
115°
A
B
D
R
(2x + 15)º
Q
34. Find CA.
38. Point O is the centroid of
. Find
A
,
.
C
)
X
Y
s+ 2
O
B
)
)
C
2 s  10
39. In
B
A
Z
, show that midsegment
and that
.
is parallel to
y
35. Find the measure of each numbered angle.
|
L
–4

117
2
>
3 1
|
C (-4, 2)
2
–2
K
>
36. Find the measures
and
4
6
x
B (4, -2)
–2
A (-4, -4)
.
2
–4
B
40. Given
with
,
find the length of midsegment
6.4
, and
,
.
C
AC = 6
X
A
X
2.3
C
Y
A
37. Given that
.
Y
bisects
and
, find
3
B
X
41. Vanessa wants to measure the width of a reservoir.
She measures a triangle at one side of the reservoir
as shown in the diagram. What is the width of the
reservoir (BC across the base)?
120 m
B
X
Z
BC = 5
120 m
W
A
150 m
100 m
Y
100 m
C
42. Identify the pairs of congruent angles and
corresponding sides.
A
12
8
2.5 m
))
C
(
18
B
1.3 m
13.5
D
44. Maya is making a miniature dinner table for her
little sister. She wants the table top to be similar to
their real dinner table top. Find the width of the
miniature table top to the nearest tenth of a
centimeter.
((
)
9
Dining table top
E
9.5 cm
6
x
M iniature
F
43. Determine whether the rectangles are similar. If so,
write the similarity ratio and a similarity statement.
M
9
45. A video game designer is modeling a tower that is
320 ft high and 260 ft wide. She creates a model so
that the similarity ratio of the model to the tower is
. What is the height and the width of the model
in inches?
N
6
P
R
O
15
S
46. Apply the dilation D to the polygon with the given
vertices. Name the coordinates of the image points.
J(1, 4), K(6, 4), L(6, 1), M(1, 1)
y
10
6
5
U
T
4
J
K
M
L
3
2
1
1
2
3
4
5
6
7
47. Explain why
9 x
8
and then find BC.
B
28
D
A
32
(
E
(
48
C
48. Find
.
P
N
5
M
Q
8
2
R
49. An artist used perspective to draw guidelines in her
picture of a row of parallel buildings. How many
centimeters is it from Point B to Point C?
H
3 cm
F
4 cm
C
D
B
5 cm
A
50. Find
.
B
3x - 3
18
D
x+ 5
A
15
C
QBA 1 Review
Answer Section
SHORT ANSWER
1. ANS:
m
= 35°
Corresponding Angles Postulate
Subtract x from both sides.
Add 70 to both sides.
Divide both sides by 2.
m
m
PTS:
OBJ:
LOC:
KEY:
2. ANS:
m
Substitute 35 for x. Simplify.
1
DIF: 2
REF: 1a24483e-4683-11df-9c7d-001185f0d2ea
3-1.1 Using the Corresponding Angles Postulate
STA: MCC9-12.G.CO.9
MTH.C.11.01.03.03.01.005
TOP: 3-1 Angles Formed by Parallel Lines and Transversals
corresponding angles | parallel lines
DOK: DOK 2
=
Alternate Exterior Angles Theorem
Subtract
from both sides.
Divide both sides by .
Substitute 24 for .
m
PTS:
OBJ:
LOC:
TOP:
DOK:
3. ANS:
1
DIF: 2
REF: 1a26aa9a-4683-11df-9c7d-001185f0d2ea
3-1.2 Finding Angle Measures
STA: MCC9-12.G.CO.9
MTH.C.11.01.03.03.01.007 | MTH.C.11.02.01.01.007
3-1 Angles Formed by Parallel Lines and Transversals
KEY: alternate exterior angles | parallel lines
DOK 2
By the Corresponding Angles Postulate,
By the Alternate Interior Angle Postulate,
.
.
Subtract the first equation from the second.
Divide both sides by 4.
Substitute 10 for x.
Simplify.
PTS: 1
DIF: 3
REF: 1a26d1aa-4683-11df-9c7d-001185f0d2ea
OBJ: 3-1.3 Application
STA: MCC9-12.G.CO.9
LOC: MTH.C.11.01.03.03.01.005 | MTH.C.11.01.03.03.01.006
TOP: 3-1 Angles Formed by Parallel Lines and Transversals
KEY: corresponding angles | parallel lines | alternate interior angles
DOK: DOK 3
4. ANS:
By substitution,
and
.
By the Substitution Property of Equality,
.
By the Converse of the Alternate Interior Angles Theorem,
.
;
Substitute 20 for x.
Substitution Property of Equality
Converse of the Alternate Interior Angles Theorem
PTS: 1
DIF: 2
REF: 1a2b9662-4683-11df-9c7d-001185f0d2ea
OBJ: 3-2.2 Determining Whether Lines are Parallel
STA: MCC9-12.G.CO.9
LOC: MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010
TOP: 3-2 Proving Lines Parallel
KEY: parallel | alternate interior angles
DOK: DOK 2
5. ANS:
[1] Vertical Angle Theorem
[2] Converse of the Same-Side Interior Angles Theorem
Proof:
Statements
Reasons
1.
1. Given
2.
2. Vertical Angle Theorem
3.
3. Substitution (Steps 1 and 2)
4. Converse of the Same-Side Interior Angles
4.
Theorem
PTS:
OBJ:
LOC:
KEY:
6. ANS:
1
DIF: 1
REF: 1a2dd1ae-4683-11df-9c7d-001185f0d2ea
3-2.3 Proving Lines Parallel
STA: MCC9-12.G.CO.9
MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010
TOP: 3-2 Proving Lines Parallel
parallel lines | proof
DOK: DOK 1
;
The angles are alternate interior angles, and they are congruent, so the lanes are parallel by the Converse of the
Alternate Interior Angles Theorem.
Substitute 10 for x in each expression:
The angles are alternate interior angles, and they are congruent, so the lanes are parallel by the Converse of the
Alternate Interior Angles Theorem.
PTS: 1
DIF: 2
REF: 1a30340a-4683-11df-9c7d-001185f0d2ea
OBJ: 3-2.4 Application
STA: MCC9-12.G.CO.9
LOC: MTH.C.11.01.03.03.01.010 | MTH.C.11.01.03.03.010
TOP: 3-2 Proving Lines Parallel
KEY: parallel lines | proof
7. ANS:
DOK: DOK 2
is the shorter segment.
Substitute
for
and 8 for
Subtract 4 from both sides.
Divide both sides by 2 and simplify.
.
PTS: 1
DIF: 2
REF: 1a305b1a-4683-11df-9c7d-001185f0d2ea
OBJ: 3-3.1 Distance From a Point to a Line
STA: MCC9-12.A.REI.3
LOC: MTH.C.11.01.03.007
TOP: 3-3 Perpendicular Lines
KEY: distance from a point to a line | inequality | perpendicular DOK: DOK 2
8. ANS:
[1]
[2] 2 intersecting lines form linear pair of
s
lines .
[3] 2 lines to the same line
lines .
Proof:
Statements
Reasons
1. Given
1.
2.
2. If 2 intersecting lines form linear pair of
s
lines .
3.
3. If 2 lines to the same line
lines .
PTS:
OBJ:
LOC:
TOP:
DOK:
9. ANS:
1
DIF: 1
REF: 1a329666-4683-11df-9c7d-001185f0d2ea
3-3.2 Proving Properties of Lines
STA: MCC9-12.G.CO.9
MTH.P.08.02.03.01.002 | MTH.C.11.01.03.03.010 | MTH.C.11.01.03.04.007
3-3 Perpendicular Lines
KEY: perpendicular | parallel | proof
DOK 1
y
7
E'
E
–7
F'
7
G'
x
F
G
–7
This is a translation 6 units left and 2 units up.
PTS: 1
DIF: 2
REF: 9141cc4b-6ab2-11e0-9c90-001185f0d2ea
OBJ: 4-1.1 Drawing and Identifying Transformations
NAT: NT.CCSS.MTH.10.9-12.G.CO.5
STA: MCC9-12.G.CO.5
TOP: 4-1 Congruence and Transformations
KEY: transformation | coordinate geometry
DOK: DOK 2
10. ANS:
y
7
A
D
B
C
–7
Y
X
Z
7
x
W
–7
This is a rotation of 180° about the origin.
PTS: 1
DIF: 2
REF: 9141f35b-6ab2-11e0-9c90-001185f0d2ea
OBJ: 4-1.1 Drawing and Identifying Transformations
NAT: NT.CCSS.MTH.10.9-12.G.CO.5
STA: MCC9-12.G.CO.5
TOP: 4-1 Congruence and Transformations
KEY: transformation | coordinate geometry
DOK: DOK 2
11. ANS:
The triangles are congruent because
can be mapped to
by a reflection:
.
PTS: 1
DIF: 2
REF: 91442ea6-6ab2-11e0-9c90-001185f0d2ea
OBJ: 4-1.2 Determining Whether Figures are Congruent
NAT: NT.CCSS.MTH.10.9-12.G.CO.6
STA: MCC9-12.G.CO.6
TOP: 4-1 Congruence and Transformations
KEY: reflection | rotation | transformation | translation | coordinate geometry
DOK: DOK 2
12. ANS:
The triangles are congruent because
can be mapped onto
by a rotation:
, followed
by a reflection:
.
PTS:
OBJ:
STA:
KEY:
13. ANS:
1
DIF: 2
REF: 91469101-6ab2-11e0-9c90-001185f0d2ea
4-1.3 Applying Transformations
NAT: NT.CCSS.MTH.10.9-12.G.CO.6
MCC9-12.G.CO.5
TOP: 4-1 Congruence and Transformations
transformation | coordinate geometry
DOK: DOK 2
=
Step 1 Find m
.
Triangle Sum Theorem
Substitute 61° for
and 22° for
Simplify.
Subtract 83° from both sides.
Step 2 Find
.
.
Linear Pair Theorem and Angle Addition Postulate
Substitute 97° for
and 42° for
.
Simplify.
Subtract 139° from both sides.
PTS:
OBJ:
LOC:
KEY:
14. ANS:
1
DIF: 2
REF: 1a696caa-4683-11df-9c7d-001185f0d2ea
4-2.1 Application
STA: MCC9-12.G.MG.1
MTH.C.11.02.01.01.005 | MTH.C.11.03.02.04.002
TOP: 4-2 Angle Relationships in Triangles
triangle sum theorem
DOK: DOK 2
Let the acute angles be
m
m
m
m
PTS:
OBJ:
LOC:
KEY:
15. ANS:
and
, with m
.
The acute angles of a right triangle are complementary.
Substitute
for m
.
Subtract
from both sides.
1
DIF: 1
REF: 1a6993ba-4683-11df-9c7d-001185f0d2ea
4-2.2 Finding Angle Measures in Right Triangles
STA: MCC9-12.A.CED.1
MTH.C.11.03.02.05.001
TOP: 4-2 Angle Relationships in Triangles
triangle sum theorem
DOK: DOK 1
=
Exterior Angle Theorem
Substitute
for
.
Simplify.
Add 2 to both sides.
Divide both sides by 10.
PTS:
OBJ:
LOC:
KEY:
16. ANS:
,
for
, and 118 for
1
DIF: 2
REF: 1a6bcf06-4683-11df-9c7d-001185f0d2ea
4-2.3 Applying the Exterior Angle Theorem
STA: MCC9-12.A.CED.1
MTH.C.11.03.02.04.004
TOP: 4-2 Angle Relationships in Triangles
exterior angle theorem
DOK: DOK 2
,
Third Angles Theorem
Definition of congruent angles
Substitute
for
and
Subtract
from both sides.
Divide both sides by –3.
So
Since
for
.
.
,
.
PTS: 1
DIF: 2
REF: 1a6e3162-4683-11df-9c7d-001185f0d2ea
OBJ: 4-2.4 Applying the Third Angles Theorem
NAT: NT.CCSS.MTH.10.9-12.G.SRT.5
STA: MCC9-12.A.CED.1
LOC: MTH.C.11.03.02.04.005
TOP: 4-2 Angle Relationships in Triangles
KEY: third angles theorem | triangle sum theorem
DOK: DOK 2
17. ANS:
m
The Third Angles Theorem states that if two angles of one triangle are congruent to two angles of another triangle,
then the third pair of angles are congruent.
It is given that
PTS:
NAT:
LOC:
KEY:
18. ANS:
and
. Therefore,
. So, m
.
1
DIF: 3
REF: 1a6e5872-4683-11df-9c7d-001185f0d2ea
NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.A.CED.1
MTH.C.11.03.02.04.005
TOP: 4-2 Angle Relationships in Triangles
third angles theorem | triangle sum theorem
DOK: DOK 2
,
,
,
,
,
Corresponding angles and corresponding sides are parts which lie in the same position in the triangles.
Corresponding angles:
,
,
Corresponding sides:
,
,
PTS:
OBJ:
STA:
TOP:
DOK:
19. ANS:
m
1
DIF: 1
REF: 1a7093be-4683-11df-9c7d-001185f0d2ea
4-3.1 Naming Congruent Corresponding Parts
NAT: NT.CCSS.MTH.10.9-12.G.SRT.5
MCC9-12.G.CO.7
LOC: MTH.C.11.01.02.03.002 | MTH.C.11.02.02.002
4-3 Congruent Triangles
KEY: correspondence | corresponding parts
DOK 1
= 67°
Triangle Sum Theorem
Substitution.
Simplify.
Subtract 113 from both sides.
Corresponding parts of congruent triangles are
congruent.
Definition of congruent angles
Corresponding parts of congruent triangles are
congruent.
PTS: 1
DIF: 2
REF: 1a72f61a-4683-11df-9c7d-001185f0d2ea
OBJ: 4-3.2 Using Corresponding Parts of Congruent Triangles NAT: NT.CCSS.MTH.10.9-12.G.SRT.5
STA: MCC9-12.A.CED.1
LOC: MTH.C.11.02.02.001 | MTH.C.11.03.02.04.002
TOP: 4-3 Congruent Triangles
KEY: triangle sum theorem | congruent triangles | corresponding parts
DOK: DOK 2
20. ANS:
[1] Definition of perpendicular lines
[2] Third Angles Theorem
[3] Reflexive Property of Congruence
Proof:
Statements
1.
2.
and
are right angles.
3.
4.
5.
6.
7. T is the midpoint of
.
8.
9.
10.
Reasons
1. Given
2. Definition of perpendicular lines
3. Right Angle Congruence Theorem
4. Given
5. Third Angles Theorem
6. Given
7. Given
8. Definition of midpoint
9. Reflexive Property of Congruence
10. Definition of congruent triangles
PTS: 1
DIF: 2
OBJ: 4-3.3 Proving Triangles Congruent
STA: MCC9-12.G.CO.10
TOP: 4-3 Congruent Triangles
DOK: DOK 1
21. ANS:
[1] Vertical Angles Theorem
[2] Third Angles Theorem
[3]
Proof:
Statements
1.
,
,
2.
3.
4.
5.
REF:
NAT:
LOC:
KEY:
PTS:
OBJ:
STA:
TOP:
DOK:
22. ANS:
REF:
NAT:
LOC:
KEY:
1
DIF: 2
4-3.4 Application
MCC9-12.G.CO.10
4-3 Congruent Triangles
DOK 1
1a731d2a-4683-11df-9c7d-001185f0d2ea
NT.CCSS.MTH.10.9-12.G.SRT.5
MTH.P.08.02.03.017 | MTH.C.11.08.02.02.002
proof | congruent triangles
Reasons
1. Given
2. Given
3. Vertical Angles Theorem
4. Third Angles Theorem
5. Definition of congruent triangles
1a755876-4683-11df-9c7d-001185f0d2ea
NT.CCSS.MTH.10.9-12.G.SRT.5
MTH.P.08.02.03.017 | MTH.C.11.08.02.02.002
proof | congruent triangles
It is given that
,
, and
bisects
. By the definition of segment bisector,
three pairs of corresponding sides of the triangles are congruent. Therefore, ABC DEC by SSS.
PTS:
OBJ:
LOC:
KEY:
23. ANS:
[1]
1
DIF: 1
REF: 1a77e1e2-4683-11df-9c7d-001185f0d2ea
5-1.1 Using SSS to Prove Triangle Congruence
STA: MCC9-12.G.SRT.5
MTH.C.11.08.02.02.02.002
TOP: 5-1 Triangle Congruence: SSS and SAS
SSS | congruent triangles
DOK: DOK 2
. All
[2]
[3]
It is given that
. Since
and
are right angles,
Congruence Theorem. By the Reflexive Property of Congruence,
PTS: 1
DIF: 2
OBJ: 5-1.2 Application
LOC: MTH.C.11.08.02.02.02.004
KEY: proof | SAS | congruent triangles
24. ANS:
[1]. Definition of midpoint
[2]
[3] SAS
Proof:
Statements
1. P is the midpoint of
and
.
2.
3.
4.
PTS:
OBJ:
STA:
LOC:
TOP:
DOK:
25. ANS:
,
REF:
STA:
TOP:
DOK:
by the Right Angle
. Therefore,
by SAS.
1a7a1d2e-4683-11df-9c7d-001185f0d2ea
MCC9-12.G.SRT.5
5-1 Triangle Congruence: SSS and SAS
DOK 2
Reasons
1. Given
2. Definition of midpoint
3. Vertical Angles Theorem
4. SAS
1
DIF: 2
REF: 1a7ca69a-4683-11df-9c7d-001185f0d2ea
5-1.4 Proving Triangles Congruent NAT: NT.CCSS.MTH.10.9-12.G.SRT.4
MCC9-12.G.SRT.5
MTH.C.11.01.02.02.001 | MTH.C.11.08.02.02.02.004 | MTH.C.11.08.02.02.002
5-1 Triangle Congruence: SSS and SAS
KEY: proof | congruent triangles
DOK 1
The SAS Postulate is used when two sides and an included angle of one triangle are congruent to the
corresponding sides and included angle of a second triangle.
From the given,
.
From the figure,
by the Reflexive Property of Congruence.
You have two pair of congruent sides, so you need information about the included angles.
Use these pairs of sides to determine the included angles.
The angle between sides
is
.
The angle between sides
is
.
You need to know
to prove
PTS:
STA:
TOP:
DOK:
26. ANS:
by the SAS Postulate.
1
DIF: 3
REF: 1a7ee1e6-4683-11df-9c7d-001185f0d2ea
MCC9-12.G.SRT.5
LOC: MTH.C.11.08.02.02.02.004
5-1 Triangle Congruence: SSS and SAS
KEY: proof | congruent triangles | SAS
DOK 2
is given.
. Therefore,
because both are right angles. By the Vertical Angles Theorem,
by ASA.
is given.
. Therefore,
PTS:
OBJ:
LOC:
KEY:
27. ANS:
Yes.
because both are right angles. By the Vertical Angles Theorem,
by ASA.
1
DIF: 1
5-2.2 Applying ASA Congruence
MTH.C.11.08.02.02.02.006
proof | congruent triangles | ASA
REF:
STA:
TOP:
DOK:
1a816b52-4683-11df-9c7d-001185f0d2ea
MCC9-12.G.SRT.5
5-2 Triangle Congruence: ASA, AAS, and HL
DOK 2
is given. In addition, by the Reflexive Property of Congruence,
. Since
and
, by the Perpendicular Transversal Theorem
. By the definition of right angle,
is a right
angle. Similarly,
is a right angle. Therefore, ABD DCA by the HL Congruence Theorem.
PTS:
OBJ:
TOP:
DOK:
28. ANS:
1
DIF: 2
REF: 1a8608fa-4683-11df-9c7d-001185f0d2ea
5-2.4 Applying HL Congruence
STA: MCC8.MP.3 LOC: MTH.C.11.08.02.02.02.011
5-2 Triangle Congruence: ASA, AAS, and HL
KEY: proof | congruent triangles | HL
DOK 2
, HL
Because
and
are right angles,
are right triangles.
You are given a pair of congruent legs
and a pair of congruent hypotenuses
.
So a hypotenuse and a leg of
are congruent to the corresponding hypotenuse and leg of
by HL.
PTS:
STA:
TOP:
DOK:
29. ANS:
.
1
DIF: 3
REF: 1a86300a-4683-11df-9c7d-001185f0d2ea
MCC8.MP.3 | MCC9-12.G.SRT.5 LOC: MTH.C.11.08.02.02.02.010 | MTH.C.11.08.02.02.02.011
5-2 Triangle Congruence: ASA, AAS, and HL
KEY: proof | congruent triangles | HL
DOK 3
by SAS and
From the figure,
by SAS and
by CPCTC, so m
.
by CPCTC. m
by substitution.
by the Vertical Angles Theorem. Therefore,
by substitution.
PTS: 1
DIF: 2
REF: 1a886b56-4683-11df-9c7d-001185f0d2ea
OBJ: 5-3.1 Application
STA: MCC9-12.G.CO.7
LOC: MTH.C.11.08.02.01.003 | MTH.C.11.08.02.02.02.004
TOP: 5-3 Triangle Congruence: CPCTC
KEY: congruent triangles | corresponding parts | CPCTC
DOK: DOK 2
30. ANS:
[1] AAS
[2] CPCTC
[1] Steps 1 and 7 state that two angles and a nonincluded side of MLN andPLO are congruent. By AAS, MLN
PLO.
[2] Since MLN PLO, by CPCTC,
.
PTS: 1
DIF: 2
REF: 1a8af4c2-4683-11df-9c7d-001185f0d2ea
OBJ: 5-3.3 Using CPCTC in a Proof
STA: MCC9-12.G.CO.10
LOC: MTH.P.08.02.03.017 | MTH.C.11.08.02.01.003
TOP: 5-3 Triangle Congruence: CPCTC
KEY: congruent triangles | corresponding parts | CPCTC
31. ANS:
x=6
The triangles can be proved congruent by the SAS Postulate.
By CPCTC,
.
DOK: DOK 2
Solve the equation for x.
PTS: 1
DIF: 3
REF: 1a8f926a-4683-11df-9c7d-001185f0d2ea
NAT: NT.CCSS.MTH.10.9-12.G.SRT.5 STA: MCC9-12.G.SRT.5
LOC: MTH.C.11.08.02.004 | MTH.C.11.08.02.02.02.004
TOP: 5-3 Triangle Congruence: CPCTC
KEY: congruent triangles | corresponding parts | CPCTC
DOK: DOK 2
32. ANS:
m
=
BW Tauri and M77 are equidistant from Earth, so
. By the Isosceles Triangle Theorem,
From the Angle Addition Postulate,
and m = .
PTS: 1
DIF: 2
REF: 1a96e08e-4683-11df-9c7d-001185f0d2ea
OBJ: 5-4.1 Application
STA: MCC9-12.G.MG.1
LOC: MTH.C.11.03.02.03.02.002 | MTH.C.11.03.02.06.01.003
TOP: 5-4 Isosceles and Equilateral Triangles
KEY: isosceles triangle
DOK: DOK 1
33. ANS:
m
= 75 °
Isosceles Triangle Theorem
m
m
Triangle Sum Theorem
m
m
m
Substitute for m
and substitute
for m
and m .
Simplify and subtract 30 from both sides.
Divide both sides by 5.
Thus m
.
PTS: 1
DIF: 2
REF: 1a991bda-4683-11df-9c7d-001185f0d2ea
OBJ: 5-4.2 Finding the Measure of an Angle
STA: MCC9-12.A.CED.1
LOC: MTH.C.11.02.01.01.007
TOP: 5-4 Isosceles and Equilateral Triangles
KEY: isosceles triangle theorem
DOK: DOK 2
34. ANS:
CA = 14
Equiangular triangles are equilateral.
ABC is equilateral.
Definition of equilateral triangle.
Subtract s and add 10 to both sides of the equation.
Substitute 12 for s in the equation for AB.
Simplify.
.
Definition of equilateral triangle.
Substitute 14 for AB.
PTS: 1
DIF: 1
REF: 1a9b7e36-4683-11df-9c7d-001185f0d2ea
OBJ: 5-4.3 Using Properties of Equilateral Triangles
STA: MCC9-12.A.CED.1
LOC: MTH.C.11.03.02.03.03.001 | MTH.C.11.03.02.03.05.005
TOP: 5-4 Isosceles and Equilateral Triangles
KEY: equilateral triangle | side length
DOK: DOK 1
35. ANS:
m
=
,m
=
,m
=
Step 1:
is supplementary to the angle that is
.
. So
.
Step 2: By the Alternate Interior Angles Theorem,
So
.
.
Step 3: By the Isosceles Triangle Theorem,
and the angle opposite the other side of the isosceles triangle are
congruent. Let
be that unknown angle.
Then,
and
.
by the Triangle Sum Theorem.
. So
.
PTS:
STA:
TOP:
DOK:
36. ANS:
1
DIF: 3
REF: 1a9de092-4683-11df-9c7d-001185f0d2ea
MCC9-12.A.CED.1
LOC: MTH.C.11.03.02.04.002
5-4 Isosceles and Equilateral Triangles
KEY: multi-step | isosceles triangle theorem
DOK 2
Perpendicular Bisector Theorem
Substitute 6.4 for
.
Given
Substitute.
Segment Addition Postulate
Substitute.
Simplify.
PTS:
OBJ:
STA:
TOP:
DOK:
37. ANS:
1
DIF: 2
REF: 1acdb6d2-4683-11df-9c7d-001185f0d2ea
6-1.1 Applying the Perpendicular Bisector Theorem and Its Converse
MCC9-12.A.CED.1
LOC: MTH.C.11.01.02.04.01.003 | MTH.C.11.01.02.04.01.004
6-1 Perpendicular and Angle Bisectors
KEY: perpendicular bisector
DOK 2
Angle Bisector Theorem
Substitute.
PTS: 1
DIF: 1
REF: 1acff21e-4683-11df-9c7d-001185f0d2ea
OBJ: 6-1.2 Applying the Angle Bisector Theorems
STA: MCC9-12.A.CED.1
LOC: MTH.C.11.02.03.005
TOP: 6-1 Perpendicular and Angle Bisectors
KEY: angle bisector
DOK: DOK 1
38. ANS:
= 2.2
Centroid Theorem
Substitute 3.3 for
PTS:
OBJ:
LOC:
KEY:
39. ANS:
and simplify.
1
DIF: 2
REF: 1adbddea-4683-11df-9c7d-001185f0d2ea
6-3.1 Using the Centroid to Find Segment Lengths
STA: MCC9-12.G.CO.10
MTH.C.11.03.02.02.02.006
TOP: 6-3 Medians and Altitudes of Triangles
centroid
DOK: DOK 2
.
The slope of
.
. The slope of
The length of
. The slopes are equal so
. The length of
.
.
.
Find K by finding the midpoint of
Find L by finding the midpoint of
.
Find the slope of
Find the slope of
The slopes are equal so
Find the length of
Find the length of
.
.
.
.
PTS: 1
DIF: 2
REF: 1ae304fe-4683-11df-9c7d-001185f0d2ea
OBJ: 6-4.1 Examining Midsegments in the Coordinate Plane
STA: MCC9-12.G.GPE.4
LOC: MTH.C.11.03.02.02.05.005
TOP: 6-4 The Triangle Midsegment Theorem
KEY: midsegment | coordinate geometry DOK: DOK 2
40. ANS:
XY = 1.5
The length of a midsegment is half the length of the corresponding base.
PTS:
OBJ:
LOC:
KEY:
41. ANS:
1
DIF: 2
REF: 1ae32c0e-4683-11df-9c7d-001185f0d2ea
6-4.2 Using the Triangle Midsegment Theorem
STA: MCC9-12.G.CO.10
MTH.C.11.03.02.02.05.005
TOP: 6-4 The Triangle Midsegment Theorem
midsegment DOK: DOK 1
300 m
Midsegment Theorem
Substitute 150 for XY.
Multiply both sides by 2 and simplify.
The width of the reservoir BC is about 300 meters.
PTS:
OBJ:
LOC:
KEY:
42. ANS:
1
DIF: 2
6-4.3 Application
MTH.C.11.03.02.02.05.005
midsegment DOK: DOK 1
,
,
and
PTS:
OBJ:
LOC:
TOP:
DOK:
43. ANS:
REF: 1ae5675a-4683-11df-9c7d-001185f0d2ea
STA: MCC9-12.G.MG.1
TOP: 6-4 The Triangle Midsegment Theorem
,
. By the Third Angles Theorem
.
1
DIF: 1
REF: 1b7bc30e-4683-11df-9c7d-001185f0d2ea
8-1.1 Describing Similar Polygons STA: MCC9-12.G.SRT.2
MTH.C.11.08.03.01.001 | MTH.C.11.08.03.01.002 | MTH.C.11.08.03.005
8-1 Ratios in Similar Polygons
KEY: similar polygons |corresponding sides
DOK 1
The similarity ratio is
and rectangle MNOP ~ rectangle RSTU.
Step 1 Identify pairs of congruent angles.
,
,
All angles of a rectangle are right angles and are
and
congruent.
Step 2 Compare corresponding sides.
Thus the similarity ratio is
, and rectangle MNOP ~ rectangle RSTU.
PTS: 1
DIF: 1
REF: 1b7dfe5a-4683-11df-9c7d-001185f0d2ea
OBJ: 8-1.2 Identifying Similar Polygons STA: MCC9-12.G.SRT.2
LOC: MTH.C.11.08.03.002 | MTH.C.11.08.03.004
TOP: 8-1 Ratios in Similar Polygons
KEY: similar polygons | similarity statement
DOK: DOK 1
44. ANS:
4.9 cm
x is the width of the miniature in centimeters. The rectangular miniature is similar to the rectangular table top, so
the corresponding lengths are proportional.
Cross Product Property
Simplify.
Divide both sides by 2.5.
The width of the miniature table top is about 4.9 cm.
PTS: 1
DIF: 2
REF: 1b8060b6-4683-11df-9c7d-001185f0d2ea
OBJ: 8-1.3 Application
STA: MCC9-12.G.MG.1
LOC: MTH.C.09.03.01.001 | MTH.C.11.08.03.006
TOP: 8-1 Ratios in Similar Polygons
KEY: similar figures | side length
DOK: DOK 2
45. ANS:
height = 7.68 in.; width = 6.24 in.
Step 1 Convert measurements to inches.
tower’s length
ft
in.
tower’s width
ft
in.
Step 2 Apply the scale factor formula.
new dimension = (scale factor)(original dimension)
model’s length
in.
model’s width
in.
PTS: 1
DIF: 3
STA: MCC9-12.G.MG.1
TOP: 8-1 Ratios in Similar Polygons
DOK: DOK 2
46. ANS:
J´(3, 12), K´(18, 12),
L´(18, 3), M´(3, 3)
PTS:
OBJ:
STA:
KEY:
47. ANS:
REF: 1b8087c6-4683-11df-9c7d-001185f0d2ea
LOC: MTH.C.11.08.03.005 | MTH.C.11.08.03.006
KEY: application | similarity ratio | scale model
1
DIF: 2
REF: 914db812-6ab2-11e0-9c90-001185f0d2ea
8-2.1 Drawing and Describing Dilations
NAT: NT.CCSS.MTH.10.9-12.G.CO.2
MCC9-12.G.SRT.1
TOP: 8-2 Similarity and Transformations
coordinate plane | dilation
DOK: DOK 2
by the Converse of the Corresponding Angles Postulate.
by the Corresponding Angles Postulate.
by AA Similarity.
Corresponding sides are proportional, so
.
Step 1 Prove triangles are similar.
As shown
, so
by the Converse of the Corresponding Angles Postulate.
by the Corresponding Angles Postulate.
Therefore
by AA Similarity.
Step 2 Find BC.
Corresponding sides are proportional.
Substitute 32 for DE, 48 for AC, and 28 for BE.
Cross Products Property
Simplify.
Divide both sides by 32.
PTS: 1
DIF: 2
REF: 1b85256e-4683-11df-9c7d-001185f0d2ea
OBJ: 8-3.3 Finding Lengths in Similar Triangles
NAT: NT.CCSS.MTH.10.9-12.G.SRT.5
STA: MCC9-12.G.SRT.5
LOC: MTH.C.11.08.03.03.002
TOP: 8-3 Triangle Similarity: AA, SSS, and SAS
KEY: similar triangles | side length | AA similarity DOK:
DOK 2
48. ANS:
= 1.25
It is given that
, so
by the Triangle Proportionality Theorem.
Substitute 2 for RQ, 8 for QM, and 5 for NM.
Cross Products Property
Divide by 8.
PTS: 1
DIF: 2
REF: 1b89ea26-4683-11df-9c7d-001185f0d2ea
OBJ: 8-4.1 Finding the Length of a Segment
STA: MCC9-12.G.SRT.5
LOC: MTH.C.11.01.02.01.001 | MTH.C.11.08.03.04.002
TOP: 8-4 Applying Properties of Similar Triangles
KEY: similar triangles | segment length
DOK: DOK 2
49. ANS:
3.75 cm
Two-Transversal Proportionality Corollary
Substitute the given values.
Cross Products Property
Simplify.
PTS: 1
DIF: 2
REF: 1b8c7392-4683-11df-9c7d-001185f0d2ea
OBJ: 8-4.3 Application
STA: MCC9-12.G.CO.9
TOP: 8-4 Applying Properties of Similar Triangles
KEY: perspective drawing
DOK: DOK 2
50. ANS:
= 12
Since
bisects
,
by the Triangle Angle Bisector Theorem.
Substitute the given values.
Cross Products Property
Distributive Property
Simplify.
Divide.
Substitute for x.
Simplify.
PTS:
OBJ:
LOC:
KEY:
1
DIF: 2
REF: 1b8eaede-4683-11df-9c7d-001185f0d2ea
8-4.4 Using the Triangle Angle Bisector Theorem
STA: MCC9-12.A.CED.1
MTH.C.11.08.03.04.008
TOP: 8-4 Applying Properties of Similar Triangles
triangle angle bisector theorem
DOK: DOK 2