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CHAPTER
For a complete
list of the
postulates and
theorems in
this chapter,
see p. S82.
Study Guide:
Review
6
Organizer
Vocabulary
kite . . . . . . . . . . . . . . . . . . . . . . . . 427
rhombus . . . . . . . . . . . . . . . . . . . 409
base angle of a trapezoid . . . . 429
leg of a trapezoid . . . . . . . . . . . 429
side of a polygon . . . . . . . . . . . . 382
organize and review key concepts
and skills presented in Chapter 6.
concave . . . . . . . . . . . . . . . . . . . . 383
midsegment of a trapezoid . . 431
square . . . . . . . . . . . . . . . . . . . . . 410
convex . . . . . . . . . . . . . . . . . . . . . 383
parallelogram . . . . . . . . . . . . . . 391
trapezoid. . . . . . . . . . . . . . . . . . . 429
diagonal . . . . . . . . . . . . . . . . . . . 382
rectangle . . . . . . . . . . . . . . . . . . . 408
vertex of a polygon . . . . . . . . . . 382
isosceles trapezoid . . . . . . . . . . 429
regular polygon. . . . . . . . . . . . . 382
GI
base of a trapezoid . . . . . . . . . . 429
Objective: Help students
<D
@<I
Online Edition
Multilingual Glossary
Complete the sentences below with vocabulary words from the list above.
1. The common endpoint of two sides of a polygon is a(n)
Resources
2. A polygon is
? .
−−−−
if no diagonal contains points in the exterior.
?
−−−−
? is a quadrilateral with four congruent sides.
−−−−
4. Each of the parallel sides of a trapezoid is called a(n) ? .
−−−−
3. A(n)
PuzzleView
Test & Practice Generator
6-1 Properties and Attributes of Polygons (pp. 382–388)
Multilingual Glossary Online
■
Tell whether the figure is a polygon. If it is a
polygon, name it by the number of its sides.
The figure is a closed plane figure
made of segments that intersect only
at their endpoints, so it is a polygon.
It has six sides, so it is a hexagon.
■
Tell whether the polygon is regular or
irregular. Tell whether it is concave or convex.
The polygon is equilateral, but it is
not equiangular. So it is not regular.
No diagonal contains points in the
exterior, so it is convex.
Lesson Tutorial Videos
CD-ROM
Answers
1. vertex of a polygon
2. convex
3. rhombus
4. base of a trapezoid
5. not a polygon
6. polygon; triangle
Find each measure.
the sum of the interior angle measures of
a convex 11-gon
(n - 2)180°
Polygon ∠ Sum Thm.
(11 - 2)180° = 1620°
Substitute 11 for n.
7. polygon; dodecagon
9. irregular; convex
10. reg.; convex
■
11. 1800°
12. 162°
the measure of each exterior angle of a
regular pentagon
sum of ext. = 360°
13. 90°
Chapter 6
Polygon Ext. ∠
SumThm.
360° = 72°
measure of one ext. ∠ = _
5
438
Tell whether each polygon is regular or irregular.
Tell whether it is concave or convex.
8.
9.
10.
Find each measure.
11. the sum of the interior angle measures
of a convex dodecagon
12. the measure of each
interior angle of a
regular 20-gon
13. the measure of each
exterior angle of a
regular quadrilateral
14. the measure of each
interior angle of
hexagon ABCDEF
ÇÃ
Â
xÃÂ
nÃÂ
nÃÂ
ÇÃÂ
xÃÂ
Chapter 6 Polygons and Quadrilaterals
ge07se_c06_0438_0447.indd 438
438
Tell whether each figure is a polygon. If it is a
polygon, name it by the number of its sides.
5.
6.
7.
■
8. irregular; concave
14. m∠A = m∠D = 144°; m∠B =
m∠E = 126°; m∠C = m∠F =
90°
EXERCISES
EXAMPLES
KEYWORD: MG7 Glossary
12/3/05 10:14:46 AM
Answers
6-2 Properties of Parallelograms (pp. 391–397)
In PQRS, m∠RSP = 99°,
PQ = 19.8, and RT = 12.3.
Find PT.
+
,
/
*
−− −−
PT RT
PT = RT
PT = 12.3
■
■
-
→ diags. bisect each other
Def. of segs.
Substitute 12.3 for RT.
JKLM is a
parallelogram. Find
each measure.
16. 62.4
EXERCISES
EXAMPLES
■
15. 37.5
­ÝÊÊ{®Â
ÓÞÊʙ
ÎÝÂ
LK
ÞÊÊÇ
−− −−
JM LK
→ opp. sides JM = LK
Def. of segs.
2y - 9 = y + 7
Substitute the given values.
y = 16
Solve for y.
LK = 16 + 7 = 23
m∠M
m∠J + m∠M = 180°
(x + 4) + 3x = 180
x = 44
m∠M = 3 (44) = 132°
→ cons. supp.
Substitute the given
values.
Solve for x.
In ABCD, m∠ABC = 79°,
BC = 62.4, and BD = 75.
Find each measure.
15. BE
16. AD
17. 37.5
18. 79°
19. 101°
20. 101°
17. ED
18. m∠CDA
21. 9.5
19. m∠BCD
20. m∠DAB
22. 9.5
23. 54°
WXYZ is a parallelogram.
Find each measure.
21. WX
22. YZ
23. m∠W
24. m∠X
25. m∠Y
26. m∠Z
8
LÊÊÈ
24. 126°
£{>Â
9
7 È>Â
25. 54°
xLÊÊn
26. 126°
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27. T(6, -5)
27. Three vertices of RSTV are R(-8, 1), S(2, 3), and
V(-4, -7). Find the coordinates of vertex T.
28. Write a two-column proof.
Given: GHLM is a parallelogram.
∠L ∠JMG
Prove: GJM is isosceles.
28. 1. GHLM is a . ∠L ∠JMG
(Given)
2. ∠G ∠L ( → opp. )
3. ∠G ∠JMG (Trans. Prop.
of )
−− −−
4. GJ MJ (Conv. of Isosc. Thm.)
5. GJM is isosc. (Def. of
isosc. )
29. m∠ A = m∠E = 63°; m∠G =
117°; since 117° + 63° = 180°,
∠G is supp. to ∠ A and to ∠E. So
1 ∠ of ACEG is supp. to both of
its cons. . By Thm. 6-3-4, ACEG
is a .
−− −−
30. RS = QT = 25, so RS QT.
m∠R = 76°, m∠Q = 104°, and
m∠R + m∠Q = 180°, so ∠R is
supp. to ∠Q. Since ∠R and ∠Q
are a pair of same-side int. ,
−− −−
and they are supp., RS QT. So
1 pair of opp. sides of QRST are
and . By Thm. 6-3-1, QRST
is a .
6-3 Conditions for Parallelograms (pp. 398–405)
EXERCISES
EXAMPLES
■
Show that MNPQ is a
parallelogram for
a = 6 and b = 1.6.
Ó>ÊÊx
ÓLÊÊn
ÇL
+
{>ÊÊÇ
*
MN = 2a + 5
QP = 4a - 7
MN = 2 (6) + 5 = 17
QP = 4 (6) - 7 = 17
MQ = 7b
NP = 2b + 8
MQ = 7 (1.6) = 11.2
NP = 2 (1.6) + 8 = 11.2
Since its opposite sides are congruent,
MNPQ is a parallelogram.
■
Determine if the quadrilateral
must be a parallelogram.
Justify your answer.
No. One pair of opposite angles are congruent,
and one pair of consecutive sides are
congruent. None of the conditions for a
parallelogram are met.
Show that the quadrilateral is a parallelogram for the
given values of the variables.
29. m = 13, n = 27
30. x = 25, y = 7
­Î˜ÊÊ£n®Â
­Ó˜Êʙ®Â ™“Â
,
xÞÊÊ£ä
­ÎÝÊÊ£®Â
+
­{ÝÊÊ{®Â
ÓÞÊÊ££
/
Determine if the quadrilateral must be a
parallelogram. Justify your answer.
31.
32.
33. Show that the quadrilateral with vertices B(-4, 3),
D(6, 5), F(7, -1), and H(-3, -3) is
a parallelogram.
Study Guide: Review
ge07se_c06_0438_0447.indd 439
31. Yes; the diags. of the quad.
bisect each other. By Thm. 6-3-5,
the quad. is a .
439
32. No; a pair of alt. int. are , so
1 pair of opp. sides are . A different pair of opp. sides are .
None of the conditions for a are met.
−−
−− 1
33. slope of BD = slope of FH = __
;
−−
−− 5
slope of BH = slope of DF = -6;
both pairs of opp. sides have the
−− −−
−−
same slope, so BD FH and BH
−−
DF; by def., BDFH is a .
12/3/05 10:14:49 AM
Study Guide: Review
439
Answers
6-4 Properties of Special Parallelograms (pp. 408–415)
34. 18
35. 39.6
In rectangle JKLM,
KM = 52.8, and JM = 45.6.
Find each length.
37. 19.8
38. 25.5
■
39. 10.5
40. 25.5
41. 21
■
42. 41°
43. 49°
44. 82°
■
45. 98°
46. m∠1 = 57°; m∠2 = 66°; m∠3 =
33°; m∠4 = 114°; m∠5 = 57°
47. m∠1 = 37°; m∠2 = 53°; m∠3 =
90°; m∠4 = 37°; m∠5 = 53°
−− −−
, so RT SU.
48. RT = SU = 2 √10
−−
Slope of RT = -3, and slope
−− 1
−− −−
, so RT ⊥ SU. The
of SU = __
3
−−
coordinates of the mdpt. of RT
−−
−−
and SU are (-4, -3), so RT
−−
and SU bisect each other. So the
diags. of RSTU are ⊥ bisectors
of each other.
−− −−
, so EG FH.
49. EG = FH = 3 √2
−−
Slope of EG = -1, and slope
−−
−−
−−
of FH = 1, so EG ⊥ FH. The
−−
coordinates of the mdpt. of EG
−−
−−
7
__
1
and FH are 2 , -__
, so EG
2
−−
and FH bisect each other. So the
diags. of EFGH are ⊥ bisectors
of each other.
(
■
KL
JKLM is a .
KL = JM = 45.6
NL
JL = KM = 52.8
1 JL = 26.4
NL = _
2
)
Rect. → → opp. sides Rect. → diags. → diags. bisect
each other
PQRS is a rhombus.
+
*
Find m∠QPR, given that
/
m∠QTR = (6y + 6)° and
m∠SPR = 3y°.
,
m∠QTR = 90°
Rhombus → diags. ⊥
6y + 6 = 90
Substitute the given value.
y = 14
Solve for y.
m∠QPR = m∠SPR
Rhombus → each
m∠QPR = 3 (14) ° = 42°
diag. bisects opp. The vertices of square ABCD are A(5, 0),
B(2, 4), C(-2, 1), and D(1, -3). Show that
the diagonals of square ABCD are congruent
perpendicular bisectors of each other.
AC = BD = 5 √
2
Diags. are .
−−
1
Product of slopes is -1,
slope of AC = -_
7
−−
so diags. are ⊥.
slope of BD = 7
−−
mdpt. of AC
−−
3, _
1
= mdpt. of BD = _
Diags. bisect
2 2
each other.
In rectangle ABCD, CD = 18, and CE = 19.8.
Find each length.
34. AB
35. AC
36. BD
37. BE
In rhombus WXYZ, WX = 7a + 1,
WZ = 9a - 6, and VZ = 3a.
Find each measure.
7
38. WZ
39. XV
40. XY
8
6
<
41. XZ
In rhombus RSTV, m∠TZV = (8n + 18)°,
and m∠SRV = (9n + 1)°.
Find each measure.
<
42. m∠TRS
43. m∠RSV
44. m∠STV
9
45. m∠TVR
/
,
6
Find the measures of the numbered angles in
each figure.
46. rectangle MNPQ
ÎÎÂ
£
Ó
{
Î
47. rhombus CDGH
*
Ó
{
Î
x
+
£
xÎÂ
x
Show that the diagonals of the square with the given
vertices are congruent perpendicular bisectors of
each other.
48. R(-5, 0), S(-1, -2), T(-3, -6), and U(-7, -4)
49. E(2, 1), F(5, 1), G(5, -2), and H(2, -2)
6-5 Conditions for Special Parallelograms (pp. 418–425)
EXERCISES
EXAMPLES
■
51. valid
52. valid
440
Determine if the conclusion
is valid. If not, tell what
additional information is
needed to make it valid.
−− −−
Given: LP ⊥ KN
*
Conclusion: KLNP is a rhombus.
The conclusion is not valid.
If the diagonals of a parallelogram are
perpendicular, then the parallelogram is a
rhombus. To apply this theorem, you must first
know that KLNP is a parallelogram.
Determine if the conclusion is
valid. If not, tell what additional
information is needed to make
it valid.
−− −− −− −−
50. Given: ER ⊥ FS, ER FS
Conclusion: EFRS is a square.
−−
−−
51. Given: ER and FS bisect each other.
−− −−
ER FS
Conclusion: EFRS is a rectangle.
−− −− −− −− −− −−
52. Given: EF RS, FR ES, EF ES
Conclusion: EFRS is a rhombus.
,
-
Chapter 6 Polygons and Quadrilaterals
ge07se_c06_0438_0447.indd 440
Chapter 6
( )
50. Not valid; by Thm. 6-5-2, if the
diags. of a are , then the is a rect. By Thm. 6-5-4, if the
diags. of a are ⊥, then the is a rhombus. If a is both a
rect. and a rhombus, then the is a square. To apply this chain of
reasoning, you must first know
that EFRS is a .
440
EXERCISES
EXAMPLES
36. 39.6
12/3/05 10:14:52 AM
■
Use the diagonals to tell whether a
parallelogram with vertices P(-5, 3),
Q(0, 1), R(2, -4), and S(-3, -2) is a
rectangle, rhombus, or square. Give all
the names that apply.
PR = √
98 = 7 √
2
QS = √
18 = 3 √
2
Distance Formula
Distance Formula
Use the diagonals to tell whether a parallelogram
with the given vertices is a rectangle, rhombus,
or square. Give all the names that apply.
53. B(-3, 0), F(-2, 7), J(5, 8), N(4, 1)
Answers
54. D(-4, -3), H(5, 6), L(8, 3), P(-1, -6)
55. rect., rhombus, square
53. rhombus
54. rect.
56. 64°
55. Q(-8, -2), T(-6, 8), W(4, 6), Z(2, -4)
57. 25°
Since PR ≠ QS, PQRS is not a rectangle and
not a square.
−−
7 = -1
slope of PR = _
Slope Formula
-7
−− _
3
Slope Formula
slope of QS = = 1
3
Since the product of the slopes is -1,
the diagonals are perpendicular. PQRS is
a rhombus.
58. 65°
59. 123°
60. m∠R = 126°; m∠S = 54°
61. 51.6
62. 48.5
63. 3.5
64. n = 3 or n = -3
6-6 Properties of Kites and Trapezoids (pp. 427–435)
+
*
EXAMPLES
In kite PQRS, m∠SRT = 24°, and
m∠TSP = 53°. Find m∠SPT.
PTS is a right triangle. Kite → diags. ⊥
m∠SPT + m∠TSP = 90° Acute of rt. ,
are comp.
m∠SPT + 53 = 90
m∠SPT = 37°
■
Substitute 53 for m∠TSP.
Subtract 53 from both sides.
Find m∠D.
m∠C + m∠D = 180°
51 + m∠D = 180
m∠D = 129°
■
■
9
x£Â
<
57. m∠ZWV
58. m∠VZW
59. m∠WZY
Find each measure.
60. m∠R and m∠S
x{Â
8
{Ó
Çΰx
,
/
9
<
63. EQ
-
Îä
Ó°Ç
*
롣
/
+
64. Find the value of n so that PQXY is isosceles.
+
8
­È˜ÓÊÊÇ®Â
*
9
6
7
<
6
ÈÇ
8
61. BZ if ZH = 70
and EK = 121.6
62. MN
1 (XY + WZ)
AB = _
2
1 (42 + WZ)
73.5 = _
2
147 = 42 + WZ
105 = WZ
56. m∠XYZ
-
In trapezoid HJLN,
*
JP = 32.5, and HL = 50.
Find PN.
−− −−
JN HL
Isosc. trap. → diags. JN = HL = 50
Def. of segs.
JP + PN = JN
Seg. Add. Post.
32.5 + PN = 50
Substitute.
PN = 17.5 Subtract 32.5 from both sides.
Find WZ.
67. isosc. trap.
In kite WXYZ, m∠VXY = 58°,
and m∠ZWX = 50°.
Find each measure.
Same-Side Int. Thm.
Substitute 51 for m∠C.
Subtract.
66. trap.
EXERCISES
/
■
65. kite
­n˜ÓÊÊ££®Â
9
Trap. Midsegment Thm.
Give the best name for a quadrilateral whose vertices
have the given coordinates.
65. (-4, 5), (-1, 8), (5, 5), (-1, 2)
Substitute.
66. (1, 4), (5, 4), (5, -4), (1, -1)
Multiply both sides by 2.
67. (-6, -1), (-4, 2), (0, 2), (2, -1)
7
<
Solve for WZ.
Study Guide: Review
ge07se_c06_0438_0447.indd 441
441
12/3/05 10:14:55 AM
Study Guide: Review
441
CHAPTER
6
Organizer
Tell whether each figure is a polygon. If it is a polygon, name it by the number of its sides.
GI
<
polygon; decagon
n˜Â
mastery of concepts and skills in
Chapter 6.
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not a polygon 2.
1.
Objective: Assess students’
3. The base of a fountain is in the shape of a quadrilateral, as shown.
Find the measure of each interior angle of the fountain.
£{˜
Â
3.
m∠A = 96°;
m∠B = 112°;
m∠C = 64°;
m∠D = 88°
1260°
Online Edition
4. Find the sum of the interior angle measures of a convex nonagon.
5. Find the measure of each exterior angle of a regular 15-gon. 24°
£Ó˜Â
££˜Â
6. In EFGH, EH = 28, HZ = 9, and
FH = 18; 7. JKLM is a parallelogram.
m∠EHG = 145°. Find FH and m∠FEH. m∠FEH
Find KL and m∠L. KL = 17; m∠L = 52°
Resources
Assessment Resources
Chapter 6 Tests
= 35°
<
ÞÊÊ££
{ÞÊÊÇ
­ÈÝÊÊ£®Â
­ÓÝÊʙ®Â
• Free Response
(Levels A, B, C)
8. Three vertices of PQRS are P(-2, -3), R(7, 5), and S(6, 1). Find the coordinates of Q.(-1, 1)
• Multiple Choice
(Levels A, B, C)
9. Show that WXYZ is
a parallelogram for
a = 4 and b = 3.
• Performance Assessment
8
Î>
9
xL
>Ê
Ê
ÊÎ
{LÊ
n
7
10. Determine if CDGH
must be a parallelogram.
Justify your answer.
<
11. Show that a quadrilateral with vertices K(-7, -3), L(2, 0), S(5, -4),
and T(-4, -7) is a parallelogram.
Test & Practice Generator
12. In rectangle PLCM,
LC = 19, and LM = 23.
Find PT and PM.
PT = 11.5; PM = 19
/
*
13. In rhombus EHKN,
m∠NQK = (7z + 6)°, and
m∠ENQ = (5z + 1)°.
Find m∠HEQ and m∠EHK.
+
m∠HEQ = 29°; m∠EHK = 122°
Determine if the conclusion is valid. If not, tell what additional information is
needed to make it valid.
−− −− −−− −−−
−− −−− −−− −− −−− −−−
15. Given: NP MQ, NM PQ, NQ MP
14. Given: NP PQ QM MN
Conclusion: MNPQ is a square.
Conclusion: MNPQ is a rectangle. valid
*
+
not valid
Use the diagonals to determine whether a parallelogram with the given vertices
is a rectangle, rhombus, or square. Give all the names that apply.
16. A(-5, 7), C(3, 6), E(7, -1), G(-1, 0) rhombus
18. m∠JFR = 43°, and m∠JNB = 68°.
Find m∠FBN. 103°
17. P(4, 1), Q(3, 4), R(-3, 2), S(-2, -1) rect.
19. PV = 61.1, and
YS = 24.7.
Find MY. 36.4
,
*
9
6
-
20. Find HR. 27 in.
442
Ó{ʈ˜°
−− −−
−−
9. XN = ZN = 12, so XN ZN. Thus WY
−−
−−−
bisects
XZ. WN = YN = 15, so WN
ge07se_c06_0438_0447.indd
442
−−
−−
−−
YN. Thus XZ bisects WY. The diags. of
WXYZ bisect each other. By Thm. 6-3-5,
WXYZ is a .
442
Chapter 6
9
-
Chapter 6 Polygons and Quadrilaterals
Answers
KEYWORD: MG7 Resources
,
ÓxÊÚÚ
Ê£ÓÊÊÊʈ˜°
8
10. No; 1 pair of opp. sides of the quad.
are . A pair of vert. formed by the
diags. are . None of the conditions for
a are met.
−−
11. Possible answer: slope of KL = slope
−− __
−−
1
of ST = 3 ; slope of KT = slope of
−−
4
LS = -__
; both pairs of opp. sides have
3
−− −−
−− −−
the same slope, so KL ST and KT LS;
by def., KLST is a .
14. Possible answer: MNPQ is a rhombus
by def. because its 4 sides are . To
show that MNPQ is a square, you need
to know that MNPQ is also a rect.
4/26/07 4:03:47 PM
CHAPTER
6
Organizer
FOCUS ON SAT
The scores for each SAT section range from 200 to 800.
Your score is calculated by subtracting a fraction for
each incorrect multiple-choice answer from the total
number of correct answers. No points are deducted for
incorrect grid-in answers or items you left blank.
Objective: Provide practice for
If you have time, go back through each
section of the test and check as many
of your answers as possible. Try to use a
different method of solving the problem
than you used the first time.
GI
college entrance exams such as the
SAT.
<D
@<I
Online Edition
You may want to time yourself as you take this practice test. It should take you
about 6 minutes to complete.
Resources
1. Given the quadrilateral below, what value of x
would allow you to conclude that the figure is a
parallelogram?
ÓÝÊÊ£
ÝÊÊ£
3. Which of the following terms best describes the
figure below?
College Entrance Exam
Practice
(A) Rhombus
Questions on the SAT represent the
following math strands:
(B) Trapezoid
ÓÝÊÊ{
ÎÝÊÊÓ
(C) Quadrilateral
Number and Operation, 20–25%
(D) Square
(A) -2
Algebra and Functions, 35–40%
(E) Parallelogram
Geometry and Measurement,
25–30%
(B) 0
(C) 1
4. Three vertices of MNPQ are M(3, 1), N(0, 6),
and P(4, 7). Which of the following could be the
coordinates of vertex Q?
(D) 2
(E) 3
2. In the figure below, if ABCD is a rectangle, what
type of triangle must ABE be?
(A) (7, 0)
Items on this page focus on:
(B) (–1, 1)
• Geometry and Measurement
(C) (7, 2)
Text References:
(D) (11, 3)
Item
(E) (9, 4)
Data Analysis, Statistics, and
Probability, 10–15%
Lesson
1
2
3
4
5
6-3
6-4
6-6
6-2
6-1
(A) Equilateral
5. If ABCDE is a regular pentagon, what is the
measure of ∠C?
(B) Right
(A) 45°
(C) Equiangular
(B) 60°
(D) Isosceles
(C) 90°
(E) Scalene
(D) 108°
(E) 120°
College Entrance Exam Practice
1. Students who chose A may have set the
lengths of adjacent sides equal to each
other. Remind students that any quadrilateral in which both pairs of opposite
sides are congruent must be a parallelogram.
ge07se_c06_0438_0447.indd 443
2. Students who chose B may be thinking of the properties of a rhombus, in
which the diagonals are perpendicular.
Remind students that the diagonals of a
rectangle bisect each other, and ask students what that means about the sides
of ABE.
3. Students who did not choose B should
review the vocabulary from this chapter.
Ask students to review their graphic
organizer from page 431, which shows
the relationships between the quadrilaterals studied in this chapter.
443
5. Remind students that the sum of the
measures of the interior angles of a con12/3/05
AM
vex10:15:01
polygon
with n sides is (n - 2)180
and that a pentagon has 5 sides.
4. Students who chose A or E may have
found the correct slopes but applied
them incorrectly. Suggest that students
graph the points in the coordinate plane
to make sure they understand their relative locations.
College Entrance Exam Practice
443
CHAPTER
6
Organizer
Multiple Choice: Eliminate Answer Choices
Objective: Provide opportunities
For some multiple-choice test items, you can eliminate one or more of the answer
choices without having to do many calculations. Use estimation or logic to help you
decide which answer choices can be eliminated.
GI
to learn and practice common testtaking strategies.
<D
@<I
Online Edition
What is the value of x in the figure?
This Test Tackler
focuses on using
logic and estimation to eliminate answer choices to
multiple-choice test items. While
this strategy may not always give
students the specific answer, it may
save students time by eliminating
some of the choices. Encourage
students to read a multiple-choice
test item carefully. Then, before they
begin to solve the problem, have
them determine if any of the answer
choices can be eliminated.
3°
83°
63°
153°
™nÂ
ÎÓÂ
ÝÂ
ÎÓÂ
The sum of the exterior angle measures of a convex polygon is 360°.
By rounding, you can estimate the sum of the given angle measures.
£ÎxÂ
100° + 30° + 140° + 30° = 300°
If x = 153°, the sum of the angle measures would be far greater than 360°.
So eliminate D.
If x = 3°, the sum would be far less than 360°. So eliminate A.
From your estimate, it seems likely that the correct choice is B, 63°.
Confirm that this is correct by doing the actual calculation.
98° + 32° + 63° + 135° + 32° = 360°
The correct answer is B, 63°.
What is m∠B in the isosceles trapezoid?
216°
72°
108°
58°
Base angles of an isosceles trapezoid are congruent.
Since ∠D and ∠B are not a pair of base angles, their
measures are not equal. Eliminate G, 108°.
£änÂ
∠D and ∠C are base angles, so m∠C = 108°. ∠B and ∠C are same-side interior
angles formed by parallel lines. So they are supplementary angles. Therefore
the measure of angle B cannot be greater than 180°. You can eliminate F.
m∠B = 180° - 108° = 72°
The correct answer is H, 72°.
444
Chapter 6 Polygons and Quadrilaterals
ge07se_c06_0438_0447.indd 444
444
Chapter 6
12/3/05 10:15:03 AM
Try to eliminate unreasonable answer choices.
Some choices may be too large or too small or
may have incorrect units.
Item C
Answers
In isosecles trapezoid ABCD, AC = 18.2, and
DG = 6.3. What is GB?
Possible answers:
Read each test item and answer the questions
that follow.
*
+
2. Use the formula for the area of a
and substitute 4 for the height
1
bh =
and 2 for the base: A = __
__1 (4)(2) = 4. Eliminate C2
2
because it is much greater
than 4.
Item A
The diagonals of rectangle MNPQ intersect at
S. If MN = 4.1 meters, MS = 2.35 meters, and
MQ = 2.3 meters, what is the area of MPQ
to the nearest tenth?
1. Yes; D; the units are incorrect.
24.5
6.3
11.9
2.9
3. Yes; F; 180° is the sum of the
int. of a , and because a
hexagon has 3 more sides than a
, the sum of its int. must be
much greater than 180°.
−−
6. Will the measure of GB be more than, less
−−
than, or equal to the measure of AC? What
answer choices can you eliminate and why?
4. The sum of the int. of a convex polygon must be a multiple
of 180°. Because 500 is not a
multiple of 180, it can be eliminated.
7. Explain how to use estimation to answer
this problem.
4.7 square meters
5.4 meters
9.4 square meters
12.8 meters
5. The student did not subtract 2
from the number of sides before
multiplying by 180.
Item D
1. Are there any answer choices you can
eliminate immediately? If so, which choices
and why?
2. Describe how to use estimation to eliminate
at least one more answer choice.
In trapezoid LMNP, XY = 25 feet. What are two
−−
−−
possible lengths for LM and PN ?
6. Less than; A can be eliminated
because 24.5 > 18.2.
ÓxÊvÌ
8
7. Use compatible numbers and
subtract: 18 - 6 = 12. GB is
about 12, so select B as the
answer.
9
*
Item B
What is the sum of the interior angles of a
convex hexagon?
8. J; the units are incorrect.
18 feet and 32 feet
9. Yes; the midsegment of a trap.
has a measure less than that
of the longest base and greater
than that of the shortest base.
Because 10 and 15 are both less
than 25, it cannot be the correct
answer.
49 feet and 2 feet
10 feet and 15 feet
7 inches and 43 inches
180°
720°
500°
1080°
3. Can any of the answer choices be eliminated
immediately? If so, which choices and why?
4. How can you use the fact that 500 is not a
multiple of 180 to eliminate choice G?
5. A student answered this problem with J.
Explain the mistake the student made.
8. Which answer choice can you eliminate
immediately? Why?
10. The student used numbers com1
patible with 50 and 0: __
(50 + 0)
2
= 25. However, the student
should have checked the answer,
1
because __
(49 + 2) = 25.5 ≠ 25.
2
9. A student used logic to eliminate choice H.
Do you agree with the student’s decision?
Explain.
10. A student used estimation and answered
this problem with G. Explain the mistake
the student made.
Test Tackler
445
Answers to Test Items
A. A
B. H
ge07se_c06_0438_0447.indd 445
12/3/05 10:15:05 AM
C. B
D. F
KEYWORD: MG7 Resources
Test Tackler
445
CHAPTER
6
KEYWORD: MG7 TestPrep
Organizer
CUMULATIVE ASSESSMENT, CHAPTERS 1–6
GI
Objective: Provide review and
practice for Chapters 1–6 and
standardized tests.
@<I
<D
Multiple Choice
Use the figure below for Items 6 and 7.
J
1. The exterior angles of a triangle have measures
of (x + 10)°, (2x + 20)°, and 3x°. What is the
measure of the smallest interior angle of the
triangle?
Online Edition
15°
Resources
Assessment Resources
N
M
55°
35°
65°
2. If a plant is a monocot, then its leaves have
−
KEYWORD: MG7 TestPrep
you need to prove that quadrilateral JKLM is
a parallelogram?
−
All orchids have leaves with parallel veins.
−
JM KL
−
−
MN LN
∠MLK and ∠LKJ are right angles.
∠JML and ∠KLM are supplementary.
The leaves of a Mexican vanilla plant
have parallel veins.
A Mexican vanilla plant is a monocot.
L
−
6. If JK ML, what additional information do
parallel veins. If a plant is an orchid, then it is a
monocot. A Mexican vanilla plant is an orchid.
Based on this information, which conclusion is
NOT valid?
Chapter 6 Cumulative Test
K
7. Given that JKLM is a parallelogram and that
m∠KLN = 25°, m∠JMN = 65°, and m∠JML = 130°,
which term best describes quadrilateral JKLM?
All monocots are orchids.
Rectangle
Rhombus
3. If ABC PQR and RPQ XYZ, which of
Square
the following angles is congruent to ∠CAB?
∠QRP
∠YXZ
∠XZY
∠XYZ
4. Which line coincides with the line 2y + 3x = 4?
3y + 2x = 4
2x + 2
y=_
3
a line through (-1, 1) and (2, 3)
Trapezoid
8. For two lines and a transversal, ∠1 and ∠2 are
same-side interior angles, ∠2 and ∠3 are vertical
angles, and ∠3 and ∠4 are alternate exterior
angles. Which classification best describes the
angle pair ∠2 and ∠4?
Adjacent angles
a line through (0, 2) and (4, -4)
Alternate interior angles
Corresponding angles
5. What is the value of x in polygon ABCDEF?
C
D
6x˚
B (4x + 1)˚
(5x + 8)˚
A
446
−
Which of the following would allow you
to conclude that these triangles are
congruent by AAS?
∠ACB ∠EDF
5.25x˚
∠BAC ∠FDE
F
18
36
∠CBA ∠FED
Chapter 6 Polygons and Quadrilaterals
10. H
1. A
11. C
2. J
12. H
3. D
13. B
4. J
14. 36
5. C
15. 15
6. J
16. 13.5
7. B
8. H
9. A
Chapter 6
−
9. For ABC and DEF, ∠A ∠F, and AC EF.
∠ABC ∠EDF
24
GENL11S_c06_0446_0447.indd 446
446
(4.5x + 5)˚ E
12
Answers
KEYWORD: MG7 Resources
(5x - 8)˚
Vertical angles
7/13/09 11:10:58 PM
10. The vertices of ABCD are A(1, 4), B(4, y),
C(3, -2), and D(0, -3). What is the value of y?
3
5
4
6
Short Response
Short-Response Rubric
17. In ABC, AE = 9x - 11.25, and AF = x + 4.
Items 17–19
2 Points = The student’s answer is
an accurate and complete execution of the task or tasks.
B
11. Quadrilateral RSTU is a kite. What is the length
−
of RV ?
D
1 Point = The student’s answer contains attributes of an appropriate
response but is flawed.
E
R
F
8 in.
S
U
V
A
10 in.
0 Points = The student’s answer
contains no attributes of an
appropriate response.
C
a. Find the value of x. Show your work and
T
explain how you found your answer.
−
4 inches
6 inches
5 inches
13 inches
12. What is the measure of each interior angle in a
−
b. If DF EF, show that AFD CFE.
Extended-Response
Rubric
State any theorems or postulates used.
18. Consider quadrilateral ABCD.
Item 20
y B
regular dodecagon?
30°
150°
144°
162°
4
4 Points = The student correctly
finds XZ and the range for AC.
Explanations are complete, and
work demonstrates a thorough
understanding of concepts related
to triangle inequality theorems.
C
A
x
13. The coordinates of the vertices of quadrilateral
-4
RSTU are R(1, 3), S(2, 7), T(10, 5), and U(9, 1).
Which term best describes quadrilateral RSTU?
Parallelogram
Rhombus
Rectangle
Trapezoid
-2
D
2
4
-3
3 Points = The student’s answers
are correct, but explanations may
contain minor flaws. Work demonstrates an understanding of
major concepts related to triangle
inequality theorems.
a. Show that ABCD is a trapezoid. Justify your
answer.
Mixed numbers cannot be entered into the grid
for gridded-response questions. For example,
if you get an answer of 7__14 , you must grid either
29
7.25 or __
.
4
b. What are the coordinates for the endpoints
of the midsegment of trapezoid ABCD?
19. Suppose that ∠M is complementary to ∠N and
2 Points = The student answers correctly, but explanations are missing
or incomplete. Work demonstrates
a limited understanding of triangle
inequality theorems.
Extended Response
1 Point = The student answers
incorrectly but makes a reasonable
attempt to show work.
∠N is complementary to ∠P. Explain why the
measurements of these three angles cannot be
the angle measurements of a triangle.
Gridded Response
14. If quadrilateral MNPQ is a parallelogram, what is
the value of x?
M
(3x + 3)˚
Q
N
−
−
a. If AB = 5, BC = 6, AC = 8, and m∠B < m∠Y,
P
b. If AB = 3, BC = 5, AC = 5, and m∠B > m∠Y,
−
find the length of XZ so that XYZ is a right
triangle. Justify your reasoning and state any
theorems or postulates used.
determined by six coplanar points when no three
are collinear?
c. If AB = 8 and BC = 4, find the range of possible
16. Quadrilateral RSTU is a rectangle with
−
−
diagonals RT and SU. If RT = 4a + 2 and
SU = 6a - 25, what is the value of a?
values for the length of AC. Justify your answer.
Cumulative Assessment, Chapters 1–6
Answers
2
(9x - 11.25)
17a. By the Centroid Thm., __
3
= x + 4. So 6x - 7.5 = x + 4, -11.5
= -5x, and x = 2.3.
GENL11S_c06_0446_0447.indd 447
b. Possible answer: By the Centroid Thm.,
−− −−
2
FC = __
DC. It is given that DF EF
3
, so DF = EF. From part a, EF = 3.15,
2
and AF = 6.3. So DF = 3.15. FC = __
3
2
__
(DF + FC) = 3 (3.15 + FC), so FC
−− −−
= 6.3. Thus AF = FC, and AF FC.
∠DFA ∠EFC by the Vert. Thm.
Thus AFD CFE by SAS.
−−
−− 3
18a. Slope of AB = __
, and slope of CD =
2
−−
−−
−−
__3 , so AB CD. Slope of BC = -2, and
2
−−
−−
slope of AD = 5. So BC is not to
−−
AD. ABCD is a trap. by def.
0 Points = The student answers
incorrectly and does not attempt
all parts of the problem.
explain why XYZ is obtuse. Justify your
reasoning and state any theorems or
postulates used.
15. What is the greatest number of line segments
−
−
and BC YZ.
(2y - 5)˚
3y ˚
−
20. Given ABC and XYZ, suppose that AB XY
1
1
1
b. -1__
, -__
, 1__
,4
2
2
2
(
)(
)
19. ∠M ∠P by the Comps. Thm, and
thus m∠M = m∠P. Also, m∠N = 90°
- m∠P by the def. of comp. . If these
were of a , m∠M + m∠N +
m∠P = 180° by the Sum Thm. By
subst., m∠P + (90° - m∠P) + m∠P
= 180°. Thus, m∠P = 90°. But m∠P
< 90° because ∠N is comp. to ∠P. So
there is a contradiction, and therefore,
these cannot be the of a .
−−
−− −−
20a. Since m∠B < m∠Y, AB XY, and BC
−−
YZ, it follows by the Hinge Thm.
that XZ > AC = 8. Since 64 > 25
+ 36, ABC is obtuse by the Pyth.
Inequals. Thm. Since the longer side
lies opp. the greater ∠, ∠B is the
447
largest ∠ in ABC. Since ABC is
obtuse, ∠B must be obtuse. Since
m∠B < m∠Y, m∠Y must also be
7/13/09 11:11:10 PM
obtuse, so XYZ is obtuse.
−−
−− −−
b. Since m∠B > m∠Y, AB XY, and BC
−−
YZ, it follows by the Hinge Thm.
that XZ < AC = 5. So the hyp. of
−−
XYZ would be YZ, the longest side.
According to the Pyth. Thm., XY 2 + XZ 2
= YZ 2 , 3 2 + XZ 2 = 5 2, XZ 2 = 25 - 9
= 16, and XZ = 4.
c. 4 < AC < 12. By the Inequal. Thm.,
AB + AC > BC. By subst. 8 + AC > 4.
This is true when AC > 0. AB + BC >
AC, so 8 + 4 = 12 > AC. Also, AC +
BC > AB, so AC + 4 > 8, and AC > 4.
Standardized Test Prep
447
Organizer
Ohio
Sandusky
Objective: Choose appropriate
GI
problem-solving strategies
and use them with skills from
Chapters 5 and 6 to solve realworld problems.
<D
@<I
Handmade Tiles
Online Edition
During the nineteenth century, an important
industry developed in east central Ohio thanks
to an “earthy” discovery—clay! The region’s rich
soil and easy access to river transportation helped
establish Ohio as the pottery and ceramic capital
of the United States. Today the majority of the
earthenware clay used in handmade tiles is still
mined in Ohio.
Handmade Tiles
Reading Strategies
Have students write down the
important information as they read
Problem 1. Then have them summarize the problem in their own words.
Choose one or more strategies to solve each
problem.
Using Data Ask students what else
they need to know about the tile to
solve Problem 1. the height of the
Ask them to describe how they
can find this information. Find the
shorter leg of the 30°-60°-90° formed by the height of the .
(2 in.)
1. In tile making, soft clay is pressed into long
rectangular wooden trays. After the clay
has dried, tiles are cut from the rectangular
slab. A tile manufacturer wants to make
parallelogram-shaped tiles with the
dimensions shown. What is the maximum
number of such tiles that can be cut
from a 12 in. by 40 in. slab of clay? 36 tiles
{ʈ˜°
ÎäÂ
Èʈ˜°
2. An interior designer is buying tiles that are in
the shape of isosceles trapezoids. Each tile has
bases that are 1 in. and 3 in. long, and the tiles can
be arranged as shown to form a rectangle. How
many tiles should the designer buy in
order to frame a 25 in. by 49 in. window? 76 tiles
6.26 cm, 3. A tile manufacturer wants to make a tile in
the shape of a rhombus where one diagonal
12.52 cm
is twice the length of the other diagonal.
What should the lengths of the diagonals be
in order to make a tile with sides 7 cm long? Round
to the nearest hundredth.
448
For Problem 2, ask students which
problem-solving strategies might be
helpful. Possible answer: Find a pattern.
For each side, the corner tile gives 1 in.
of length. After that, every 2 tiles provide
another 4 in. of length. So it takes
-1
1+
tiles to make a side that
2
is inches long.
GENL11S_c06_0448_0449.indd 448
_
448
Chapter 6
Îʈ˜°
Chapter 6 Polygons and Quadrilaterals
Problem-Solving Focus
KEYWORD: MG7 Resources
£Êˆ˜°
Discuss different strategies that could be
used to solve Problem 2. Some students
might make a table that shows the number
of tiles and the corresponding distance.
Others might start by solving a simpler
problem, such as the number of tiles
needed for a 5 in. by 9 in. window.
8/7/09 GENL
12:52
Problem
Solving
Strategies
The Millennium
Force Roller Coaster
The Millennium Force
Roller Coaster
When it opened in May 2000, the Millennium Force roller
coaster broke all previous records and became
the tallest and fastest roller coaster in the world.
One of 16 roller coasters at Cedar Point in
Sandusky, Ohio, the Millennium Force takes
riders on a wild journey that features
1.25 miles of track, a top speed of 93 miles
per hour, and a breathtaking 310-foot drop!
Reading
Strategies
Make sure that students understand
what the words ascent and descent
mean. If students are unfamiliar
with the terms, ask them if they can
guess their meanings from context
clues in the problems.
Choose one or more strategies to solve
each problem.
Using Data Have students
identify the segments in the figure
that represent the ascent and
−− −−
descent. AB; CD Ask students to
identify a 45°-45°-90° right triangle
−−
in the figure. the rt. with AB as
its hyp.
1. The first hill of the Millennium Force
is 310 ft tall. The ascent to the top of
the hill is at a 45° angle. What is the
length of the ascent to the nearest
tenth of a foot? 438.4 ft
2. The Millennium Force was the first
coaster in which an elevator lift
system was used to pull the trains to the
top of the first hill. The system pulls the
trains at a speed of 20 ft/s. How long does
it take a train to reach the top of the hill? 22 s
The figure shows the support structure for
the first hill of the Millennium Force.
For 3 and 4, use the figure.
−−
3. The length of the first descent CD
is 314.8 ft. To the nearest foot, what
is the total horizontal distance AD
that the train covers as it goes over
the first hill? 425 ft
−−
4. Engineers designed the support beam XY
−−
so that X is the midpoint of the ascent AB
−−
and Y is the midpoint of the descent CD.
What is the length of the beam to the
nearest foot? 242 ft
2:13 PM
L11S_c06_0448_0449.indd
449
ENGLISH
LANGUAGE
LEARNERS
ÈäÊvÌ
8
Problem-Solving Focus
9
For Problem 2, focus on the final
step of the problem-solving process:
(4) Look Back. In particular, ask
students if their answer seems
reasonable. Based on their
experiences with roller coasters,
does the amount of time for
the ascent seem realistic? If not,
encourage students to check their
work for errors.
ΣäÊvÌ
{xÂ
ÈäÊvÌ
Real-World Connections
449
7/24/09 9:37:35 PM
Real-World Connections
449