Chapter 6 Practice Exam

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Chapter 6 Practice Exam
Problems and answers to the practice
exam as well as what section they
came from so you can study!
1. What is the sum of the angle
measures of a 35-gon?
5940
6660
6120
6300
38%
24%
63
00
61
20
19%
66
60
19%
59
40
A.
B.
C.
D.
1. What is the sum of the angle
measures of a 35-gon?
𝑛 − 2 180 = π‘‘π‘œπ‘‘π‘Žπ‘™ π‘‘π‘’π‘”π‘Ÿπ‘’π‘’π‘  𝑖𝑛 π‘Ž π‘π‘œπ‘™π‘¦π‘”π‘œπ‘›
35 − 2 180 =
33 180 =
5940
Correct Answer is A. 5940
6-1
2. Find the missing values of the
variables. The diagram is not to scale.
119°
52%
29%
=1
19
π‘₯
𝑦
=1
03
;
=7
4
𝑦
=9
3;
π‘₯
π‘₯
𝑦
=9
3
10%
=7
4;
=1
03
10%
=7
4;
π‘₯ = 74; 𝑦 = 103
π‘₯ = 74; 𝑦 = 93
π‘₯ = 93; 𝑦 = 74
π‘₯ = 103; 𝑦 = 119
𝑦
A.
B.
C.
D.
74°
π‘₯
106°
x°
y°
2. Find the missing values of the
variables. The diagram is not to scale.
119°
106°
x°
y°
74°
106 + 𝑦 = 180
𝑦 = 74
π‘₯ + 𝑦 + 119 + 74 = 360
π‘₯ + 74 + 74 + 119 = 360
π‘₯ + 267 = 360
π‘₯ = 93
Correct answer is C. π‘₯ = 93; 𝑦 = 74
6-1
3. How many sides does a regular
polygon have if each exterior
angle measures 30?
A.
B.
C.
D.
15 sides
12 sides
14 sides
11 sides
62%
24%
14%
0%
3. How many sides does a regular
polygon have if each exterior
angle measures 30?
360
= 𝑒π‘₯π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’ π‘šπ‘’π‘Žπ‘ π‘’π‘Ÿπ‘’ π‘œπ‘“ π‘Ž π‘Ÿπ‘’π‘”π‘’π‘™π‘Žπ‘Ÿ π‘π‘œπ‘™π‘¦π‘”π‘œπ‘›
𝑛
360
= 30
𝑛
360
=𝑛
30
𝑛 = 12
Correct answer is B. 12 sides
6-1
4. Find the values of the variables in the
parallelogram. The diagram is not to scale.
31
67%
96
19%
10%
5%
π‘₯
𝑧
=1
27
=9
6
𝑦
=3
1,
π‘₯
=3 𝑧
1,
=5
3,
𝑦
=5
3,
=1
27
𝑧
=5 𝑧
3,
=5
3,
𝑦
=3
1,
= 53, 𝑦 = 31, 𝑧 = 96
= 53, 𝑦 = 53, 𝑧 = 127
= 31, 𝑦 = 53, 𝑧 = 96
= 31, 𝑦 = 53, 𝑧 = 127
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
π‘₯
A.
B.
C.
D.
𝑦
x°
=5
3,
z°
=9
6
y°
4. Find the values of the variables in the
parallelogram. The diagram is not to scale.
31
𝑆𝑖𝑛𝑐𝑒 π‘₯ 𝑖𝑠 π‘‘β„Žπ‘’ π‘Žπ‘™π‘‘π‘’π‘Ÿπ‘›π‘Žπ‘‘π‘’ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ
π‘Žπ‘›π‘”π‘™π‘’ π‘‘π‘œ 31π‘œ
π‘₯ = 31
96
𝐴𝑛𝑑 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘Ž π‘π‘Žπ‘Ÿπ‘Žπ‘™π‘™π‘’π‘™π‘œπ‘”π‘Ÿπ‘Žπ‘š
π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘”π‘Ÿπ‘’π‘’π‘›π‘‘ π‘ π‘œ 𝑧 = 96π‘œ
y°
z°
x°
6-2
𝐴𝑛𝑑 π‘₯ + 𝑦 + 96 = 180
31 + 𝑦 + 96 = 180
𝑦 + 127 = 180
𝑦 = 53
Correct answer is
C. π‘₯ = 31; 𝑦 = 53; 𝑧 = 96
5. Find AM in the parallelogram if
PN =15 and AO = 6. The diagram is
not to scale.
M
N
A.
B.
C.
D.
A
P
O
38%
38%
19%
.5
5
6
2
5%
12
6
15
7.5
5. Find AM in the parallelogram if
PN =15 and AO = 6. The diagram is
not to scale.
M
N
A
P
O
The diagonals of a parallelogram bisect each other
If AO = 6 then AM = 6.
Correct answer is B. 6
6-2
6. LMNO is a parallelogram.
If NM = x + 5 and OL = 2x + 3, find the
value of x and then find NM and OL.
O
A.
B.
C.
D.
π‘₯
π‘₯
π‘₯
π‘₯
N
L
= 4, 𝑁𝑀
= 2, 𝑁𝑀
= 2, 𝑁𝑀
= 4, 𝑁𝑀
= 9, 𝑂𝐿 = 9
= 7, 𝑂𝐿 =7
= 9, 𝑂𝐿 = 7
= 7, 𝑂𝐿 = 9
M
62%
19%
10%
10%
6. LMNO is a parallelogram.
If NM = x + 5 and OL = 2x + 3, find the value of x
O and then find NM and OL. N
L
M
Opposite sides of a parallelogram are congruent so NM = OL.
π‘₯ + 5 = 2π‘₯ + 3
2=π‘₯
𝑃𝑙𝑒𝑔𝑔𝑖𝑛𝑔 π‘π‘Žπ‘π‘˜ 𝑖𝑛 π‘₯:
2+5=7
2 2 +3=4+3= 7
6-2
Correct answer is B. π‘₯ = 2, 𝑁𝑀 = 7, 𝑂𝐿 =7
7. In the figure, the horizontal lines are
parallel and AB = 𝐡𝐢 = 𝐢𝐷. Find JM.
The diagram is not to scale.
M
A
9
D
48%
38%
10%
18
5%
9
J
C
27
36
9
18
36
K
B
27
L
A.
B.
C.
D.
7. In the figure, the horizontal lines are
parallel and AB = 𝐡𝐢 = 𝐢𝐷. Find JM.
The diagram is not to scale.
M
A
9
L
K
J
B
C
D
Since the lines are parallel and equidistant,
𝑀𝐿 = 𝐿𝐾 = 𝐾𝐽. The distance from 𝐽𝑀 =
9 + 9 + 9 = 27.
Correct answer is A. 27
6-2
8. Based on the information given, can you determine
that the quadrilateral must be a parallelogram? Explain.
𝐺𝑖𝑣𝑒𝑛: 𝑋𝑁 = 𝑁𝑍 and π‘π‘Œ = π‘π‘Š
X
Y
43%
29%
14%
14%
m
i..
l. .
.
al
a.
ca
n
ou
No
;y
go
na
ls
no
t
of
a
de
te
r
pa
r
sid
es
Ye
s
;d
ia
o
;t
w
Ye
s
Z
Ye
s
W
;o
pp
os
ite
sid
op
po
sit
e
es
ar
e
c..
.
.
N
A. Yes; opposite sides are congruent.
B. Yes; two opposite sides are both parallel and
congruent.
C. Yes; diagonals of a parallelogram bisect each
other.
D. No; you cannot determine that the quadrilateral
is a parallelogram.
8. Based on the information given, can you determine
that the quadrilateral must be a parallelogram? Explain.
𝐺𝑖𝑣𝑒𝑛: 𝑋𝑁 = 𝑁𝑍 and π‘π‘Œ = π‘π‘Š
X
Y
N
W
Z
Since you are given that the diagonals bisect
each other yes the quad is a parallelogram.
Correct answer is C. Yes; diagonals of a
parallelogram bisect each other.
6-3
9. Find values of x and y for which ABCD must be a
parallelogram. The diagram is not to scale.
A
B
3x – 14
y + 11
4y – 7
x+ 2
24%
19%
=1
0
=1
7
=8
,
𝑦
=8
π‘₯
,
𝑦
𝑦
=8
π‘₯
π‘₯
=6
,
=6
10%
,
= 8, 𝑦 = 6
= 6, 𝑦 = 8
= 8, 𝑦 = 17
= 8, 𝑦 = 10
=8
π‘₯
π‘₯
π‘₯
π‘₯
48%
𝑦
A.
B.
C.
D.
C
π‘₯
D
9. Find values of x and y for which ABCD must be a
parallelogram. The diagram is not to scale.
A
B
3x – 14
y + 11
4y – 7
x+ 2
D
3π‘₯ − 14 = π‘₯ + 2
2π‘₯ = 16
π‘₯=8
C
4𝑦 − 7 = 𝑦 + 11
3𝑦 = 18
𝑦=6
Correct answer is A. π‘₯ = 8, 𝑦 = 6
6-3
10. Based on the information in the
diagram, can you prove that the figure is a
parallelogram? Explain.
33%
29%
24%
c..
.
go
na
ls
ar
e
sid
es
;t
he
Ye
s
o
;t
w
Ye
s
di
a
et
op
po
sit
e
pr
ov
no
t
ca
n
ou
No
;y
a.
..
ha
.
..
ct
.
bi
se
ls
go
na
di
a
;t
he
Ye
s
.
14%
A. Yes; the diagonals bisect
each other.
B. No; you cannot prove
that the quadrilateral is a
parallelogram.
C. Yes; two opposite sides
are both parallel and
congruent.
D. Yes; the diagonals are
congruent.
10. Based on the information in the
diagram, can you prove that the figure is a
parallelogram? Explain.
Yes because the diagonals bisect each
other the figure will be a parallelogram.
Correct answer is A. Yes; the diagonals
bisect each other.
6-3
11. In the rhombus, π‘š∠1 = 140.
What are π‘š∠2 and π‘š∠3?
The diagram is not to scale.
|
67%
|
1
19%
3
10%
70
∠
π‘š
π‘š
∠
3=
20
2=
14
0,
3=
∠
π‘š
40
,
π‘š
∠
2=
π‘š
∠
3=
∠
π‘š
∠
2=
14
0,
π‘š
∠
π‘š∠2 = 40, π‘š∠3 = 70
π‘š∠2 = 140, π‘š∠3 = 20
π‘š∠2 = 40, π‘š∠3 = 20
π‘š∠2 = 140, π‘š∠3 = 70
π‘š
A.
B.
C.
D.
2=
40
,
3=
70
2
20
|
|
5%
|
1
|
11. In the rhombus, π‘š∠1 = 140. What are π‘š∠2 and π‘š∠3?
The diagram is not to scale.
|
|
3
2
Since it is a rhombus opposite angles are congruent so
π‘š∠2 = 140.
140 + 140 + π‘₯ + π‘₯ = 360
280 + 2π‘₯ = 360
2π‘₯ = 80
π‘₯ = 40
And a rhombus has diagonals that bisect the opposite
angles therefore π‘š∠3 = 20.
Correct answer is B. π‘š∠2 = 140, π‘š∠3 = 20.
6-4
12. Which description does NOT guarantee
that a quadrilateral is a square?
43%
24%
19%
...
nd
h.
.
d
. ..
ith
nt
a
..
14%
A. has all sides congruent and all
angles congruent
B. is a parallelogram with
perpendicular diagonals
C. has all right angles and has all
sides congruent
D. is both a rectangle and a
rhombus
12. Which description does NOT guarantee
that a quadrilateral is a square?
-has all sides congruent and all angles congruent
must be both rhombus and rectangle so square
-has all right angles and has all sides congruent
must be both rectangle and rhombus so square
-is both a rectangle and a rhombus
Therefore it is a square equal sides and all right angles.
BUT B. is a parallelogram with perpendicular diagonals
A rhombus has perpendicular diagonals but it does not
have to be square, there fore the correct answer is B.
6-4
13. DEFG is a rectangle. DF = 5x – 3 and
EG = x + 5. Find the value of x and the
length of each diagonal.
= 6, 𝐸𝐺
= 7, 𝐸𝐺
= 6, 𝐸𝐺
= 7, 𝐸𝐺
=6
= 12
=6
=7
57%
19% 19%
=7
𝐺
=7
,
=6
𝐺
=6
7,
,
=6
=1
2
5%
𝐺
= 1, 𝐷𝐹
= 2, 𝐷𝐹
= 2, 𝐷𝐹
= 2, 𝐷𝐹
,
π‘₯
π‘₯
π‘₯
π‘₯
=6
A.
B.
C.
D.
13. DEFG is a rectangle. DF = 5x – 3 and
EG = x + 5. Find the value of x and the
length of each diagonal.
π·π‘–π‘Žπ‘”π‘œπ‘›π‘Žπ‘™π‘  π‘œπ‘“ π‘Ž π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘”π‘Ÿπ‘’π‘’π‘›π‘‘ π‘ π‘œ 𝐷𝐹 = 𝐸𝐺
5π‘₯ − 3 = π‘₯ + 5
4π‘₯ = 8
π‘₯=2
𝑃𝑙𝑒𝑔𝑔𝑖𝑛𝑔 π‘₯ π‘π‘Žπ‘π‘˜ 𝑖𝑛 𝑀𝑒 𝑔𝑒𝑑:
5 2 −3=7
2+5=7
Correct answer is D. π‘₯ = 2, 𝐷𝐹 = 7, 𝐸𝐺 = 7
6-4
14. Which statement is true?
A. All squares are rectangles.
B. All quadrilaterals are rectangles.
C. All parallelograms are rectangles.
D. All rectangles are squares.
38%
29%
24%
es
.
ar
...
re
.. .
ec
ng
le
s.
10%
14. Which statement is true?
-All quadrilaterals are rectangles.
Not true could be a rhombus, kite, etc.
-All parallelograms are rectangles.
Not true could be a rhombus.
-All rectangles are squares.
Not true some do not have all sides congruent.
• All squares are rectangles.
This statement is always true to be a square it must
have 4 right angles which makes it a rectangle.
Correct answer is A. All squares are rectangles.
6-4
15. Parallelogram ABCD has the
angle measures shown. Can you
conclude that it is a rhombus, a
rectangle, or a square? Explain.
62%
19%
14%
in
f. .
.
h
a.
..
is
CD
a.
is
D
D
is
a.
..
..
5%
66º
66º
66º
66º
A. Parallelogram ABCD is a rectangle,
because the diagonal creates congruent
angles.
B. Parallelogram ABCD is a square, because
all four angles have the same measure.
C. Parallelogram ABCD is a rhombus,
because the diagonal bisects two angles.
D. There is not enough information.
15. Parallelogram ABCD has the
angle measures shown. Can you
conclude that it is a rhombus, a
rectangle, or a square? Explain.
66º
66º
66º
66º
-Parallelogram ABCD is a rectangle,
because the diagonal creates congruent angles.
Not true rectangles do not bisect the opposite angles.
-Parallelogram ABCD is a square, because all four
angles have the same measure.
Not square 66 + 66 is not a right angle.
-Parallelogram ABCD is a rhombus, because the
diagonal bisects two angles.
YES a rhombus does bisect the opposite angles.
Correct answer is C.
6-5
16. In quadrilateral ABCD, 𝐴𝐸 = π‘₯ + 6
and 𝐡𝐸 = 3π‘₯ − 18. For what value of x
is ABCD a rectangle?
A
B
A.
B.
C.
D.
E
D
C
71%
14%
14
16
5%
12
18
10%
18
12
16
14
16. In quadrilateral ABCD, 𝐴𝐸 = π‘₯ + 6 and
𝐡𝐸 = 3π‘₯ − 18. For what value of x is ABCD a
rectangle?
A
B
E
D
C
π·π‘–π‘Žπ‘”π‘œπ‘›π‘Žπ‘™π‘  𝑖𝑛 π‘Ž π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘Ÿπ‘’ π‘π‘œπ‘›π‘”π‘Ÿπ‘’π‘’π‘›π‘‘ π‘ π‘œ AE = BE.
π‘₯ + 6 = 3π‘₯ − 18
24 = 2π‘₯
12 = π‘₯
Correct answer is B. 12
6-5
17. In quadrilateral ABCD, π‘š∠𝐴𝐢𝐷 = 2π‘₯ + 4
and π‘š∠𝐴𝐢𝐡 = 5π‘₯ − 8. For what value of x is
ABCD a rhombus?
A
B
A.
B.
C.
D.
D
3
4
5
6
C
52%
29%
19%
0%
17. In quadrilateral ABCD, π‘š∠𝐴𝐢𝐷 = 2π‘₯ + 4 and
π‘š∠𝐴𝐢𝐡 = 5π‘₯ − 8. For what value of x is ABCD a
A
B
rhombus?
D
C
To be a rhombus the diagonals must bisect the opposite angles.
2π‘₯ + 4 = 5π‘₯ − 8
12 = 3π‘₯
4=π‘₯
The correct answer is B. 4.
6-5
18. 𝐽 π‘Žπ‘›π‘‘ ∠𝐾 π‘Žπ‘Ÿπ‘’ π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘Žπ‘›
π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘Žπ‘π‘’π‘§π‘œπ‘–π‘‘ 𝐽𝐾𝐿𝑀.
𝐼𝑓 π‘š∠𝐽 = 18π‘₯ + 8 , π‘Žπ‘›π‘‘
π‘š∠𝑀 = 11π‘₯ + 15, 𝑓𝑖𝑛𝑑 π‘š∠𝐾
A.1
B. 154
C. 77
D.26
67%
33%
26
77
0%
15
4
1
0%
18. 𝐽 π‘Žπ‘›π‘‘ ∠𝐾 π‘Žπ‘Ÿπ‘’ π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘Žπ‘› π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘Žπ‘π‘’π‘§π‘œπ‘–π‘‘ 𝐽𝐾𝐿𝑀.
𝐼𝑓 π‘š∠𝐽 = 18π‘₯ + 8 , π‘Žπ‘›π‘‘ π‘š∠𝑀 = 11π‘₯ + 15, 𝑓𝑖𝑛𝑑 π‘š∠𝐾
Base angles of an isosceles trapezoid are congruent.
18π‘₯ + 8 = 11π‘₯ + 15
7π‘₯ = 7
π‘₯=1
Plugging back into the equations we see the following:
π‘š∠𝐽 = 18 1 + 8 = 26
π‘š∠𝑀 = 11 1 + 15 = 26
However the question asked for angle K which is supplementary
180 − 26 = 154
Correct answer then is B. 154.
6-6
19. Find π‘š∠1 and π‘š∠3 in the kite.
The diagram is not to scale.
|
∠
∠
π‘š
73
,
1=
π‘š
∠
73
,
∠
1=
3=
73
0%
17
0%
3=
17
3=
17
,
∠
∠
π‘š
1=
π‘š
π‘š
∠
1=
17
,
3=
73
0%
π‘š
B
π‘š
2
∠
1
||
||
D
3
17°
100%
π‘š
|
A
C
A.
B.
C.
D.
π‘š∠1 = 17, π‘š∠3 = 73
π‘š∠1 = 17, π‘š∠3 = 17
π‘š∠1 = 73, π‘š∠3 = 17
π‘š∠1 = 73, π‘š∠3 = 73
19. Find π‘š∠1 and π‘š∠3 in the kite.
The diagram is not to scale.
|
A
2
Angle three can be found using the triangle.
B
17 + 90 + π‘š∠3 = 180
107 + π‘š∠3 = 180
||
||
D
3
17°
1
|
The measure of angle 1 is the same as 17 since
they are corresponding parts in congruent
triangles.
π‘š∠3 = 73
Correct answer is A. π‘š∠1 = 17, π‘š∠3 = 73
C
6-6
20. Find the values of the variables and
the lengths of the sides of this kite.
100%
||
A.
B.
C.
D.
π‘₯
π‘₯
π‘₯
π‘₯
= 9, 𝑦 = 14; 11, 20
= 14, 𝑦 = 9; 11, 11
= 14, 𝑦 = 9; 6, 16
= 9, 𝑦 = 14; 6, 16
16
=1
4;
6,
16
;6
,
π‘₯
𝑦
=9
,
=9
𝑦
=1
4,
=9
𝑦
=1
4,
𝑦
0%
π‘₯
π‘₯
=9
,
=1
4;
11
,2
0
x + 11
0%
;1
1,
11
0%
π‘₯
||
2x + 2
x+ 2
|
y–3
|
x+ 2
||
||
y–3
|
|
20. Find the values of
the variables and the
lengths of the sides of
this kite.
2x + 2
x + 11
The top two sides are congruent and the bottom two sides are
congruent so for ease solve the bottom two first they both have x.
2π‘₯ + 2 = π‘₯ + 11
π‘₯=9
Plugging into the top equations you get:
9 + 2 = 11
𝑦 − 3 = 11
𝑦 = 14
Plugging into the bottom equations:
2 9 + 2 = 20 π‘Žπ‘›π‘‘ 9 + 11 = 20
Correct answer is A. π‘₯ = 9, 𝑦 = 14; 11, 20
6-6
Chapter 6 Exam
• We will test tomorrow
Journals will be due at the
end of the test.
• Go over exams on Thursday.
• RETEST Friday
• Portfolios due on Monday!
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