Chapter 6 Study Guide Answers
1
polygonthree
polygon sidecommon sides vertex
nonconsecutive diagonal
trianglequadrilateral
n-gondodecagon
decagonnonagonoctagon
heptagonhexagonpentagon
equilateralequiangular regular
irregular
concave diagonalexterior
Chapter 6 Study Guide Answers
2
diagonals exteriorconvex regular
interior convexdiagonals triangles
equals
convex twosides sum
exterior
Chapter 6 Study Guide Answers
3
polygon polygon
polygon
not a polygon
not a polygon
not a polygon
hexagon heptagon
nonagon
irregularconvex
irregular
irregular irregular
regular
regular
convex
convex
concave
concave concave
Chapter 6 Study Guide Answers
4
(n–2)180°
(7–2)180°
(5)180°
900°
(n–2)180°
(10–2)180°
(8)180°
1440°
1440°10
=144°
(5–2)180°=540°
m∠A+m∠B+m∠C+m∠D+m∠E=540°35c+18c+32c+32c+18c=540
135c=540135 135
c=4m∠A=35(4)=140°m∠B=m∠E=18(4)=72°m∠C=m∠D=32(4)=128°
A dodecagon has 12 sides.
360°12
=30°
sum of ext. ∠s=360°15b+18b+33b+16b+10b+28b=360
120b=360
b=3120 120
(n–2)180°n
=135°
(n–2)180=135n
n( )
180n360=135n–180n 180n
360=45n45 45
n=8
Chapter 6 Study Guide Answers
5
four quadrilateral special quadrilaterals
two parallelparallelogram
sides side
Chapter 6 Study Guide Answers
6
42°
31
74
CF=74
m∠DEF=42°
DF=2(31)=62
YZ=XW8a4=6a+106a 6a 2a4=10 +4 +42a=142 2a=7
YZ=8a4=8(7)4=52
18b11+9b+2=180 m∠W+m∠Z=180°
27b9=180 +9 +9 27b=189 27 27b=7
m∠Z=9b+2=9(7)+2=65°
3w=w+8 w w 2w=82 2w=4
JG=w+8=4+8=12
2z=4z94z 4z 2z=9
z=4.5FH=2(2z)=2(2(4.5))=18
HJ=JFEJ=JG
2 2
J(3,4)
K(2,2)
L(2,4)Opposite sides are parallel
Find slope of KL by counting
Over 4 up 2
M(7,2) From point J, go over 4 and up 2, this is your point M.
M(7,2)
Chapter 6 Study Guide Answers
8
bothopposite parallel
JK=15a–11 =15(3)–11=34
LM=10a+4 =10(3)+4=34
KL=5b+6 =5(9)+6=51
JM=8b–21 =8(9)–21=51
Since both pairs of opposite sides are congruent, JKLM is a parallelogram.
JK=LM KL=JM
m∠Q=(6y+7)°=(6(6.5)+7)°=46°
m∠S=(8y–6)°=(8(6.5)–6)°=46°
m∠R=(15x–16)°=(15(10)–16)°=134°
Since one ∠ is supplementary to its consecutive ∠s, PQRS is a parallelogram.
46°+134°=180°
Chapter 6 Study Guide Answers
9
One angle is supplementary to both its consecutive angles. The quadrilateral is a parallelogram.
One pair of opposite angles are congruent. This is not enough information.
Both pairs of opposite angles are congruent. The quadrilateral is a parallelogram.
No. Two pairs of consecutive sides congruent does not form a parallelogram.
1(6) 5m= =
4(1) 3
JK LM05 5
m= = 74 3
5(1) 6 3m= = =
4(4) 8 4
KL JM0(6) 6 3
m= = = 7(1) 8 4
The quadrilateral has two pairs of parallel sides, therefore it is a parallelogram.
JK||LM
KL||JM
23 1m= =
62 4
AB CD10 1
m= = 15 4
d=√(62)2+(23)2=√17
d=√(15)2+(10)2=√17
AB||CD
AB=CD
The quadrilateral has one pair of parallel and congruent sides, therefore it is a parallelogram.
Chapter 6 Study Guide Answers
11
rectangle rhombusproperties parallelograms
square right congruentparallelogram rectangle
rhombus
special
Chapter 6 Study Guide Answers
12
35 TR=35
CE=1/2(58)=2958
50 HM=1/2(86)=4386
HJ=48 in.
=2(30.8)=61.6 in.HK=JG
RS=QR3a=4a144a 4a a=141 1a=14
QP=RS=3(14)=42
m∠QRP=1/2(18078)°=51°
WV=XT13b9=3b+43b 3b 10b9=4 +9 +910b=1310 10b=1.3
TV=XT=3(1.3)+4=7.9
m∠TZV=90°14a+20=90 20 20
14a=7014 14a=5
m∠VTZ=m∠XTZ=(5(5)5)°=20°
CG=GF5a=3a+173a 3a 2a=172 2a=8.5
CD=CG=5(8.5)=42.5
m∠GCD+m∠CDF=180°b+3+6b40=180 7b37=180 +37 +37
7b=217
m∠GCH=1/2m∠GCD=1/2(31+3)°=17°
7 7b=31
Chapter 6 Study Guide Answers
13
0(1) 1m= =
3(4) 7
EG FH43 7
m= = 0(1) 1
d=√(3(4))2+(0(1))2=√50 d=√(0(1))2+(43)2=√50
Therefore, EG⊥FH.
Therefore, EG=FH.
4+3 1+0 2 , 2( ) 1 1
2 , 2 ( )=1+0 3+(4)2 , 2( ) 1 1
2 , 2 ( )=
First we need to find the slopes of the diagonals.
Then we need to find the lengths of the diagonals.
Then we need to find the midpoints of the diagonals.
The diagonals are congruent, perpendicular, and have the same midpoint. They are congruent perpendicular bisectors of each other.
3(4) 1m= =
6(5) 11
SV TW92 11
m= = 10 1
d=√(6(5))2+(3(4))2=√122 d=√(10)2+(92)2=√122
Therefore, SV⊥TW.
Therefore, SV=TW.
5+6 4+3 2 , 2( ) 1 7
2 , 2 ( )= 0+1 2+92 , 2( ) 1 7
2 , 2 ( )=
First we need to find the slopes of the diagonals.
Then we need to find the lengths of the diagonals.
Then we need to find the midpoints of the diagonals.
The diagonals are congruent, perpendicular, and have the same midpoint. They are congruent perpendicular bisectors of each other.
Chapter 6 Study Guide Answers
15
P(1,4) Q(2,6)
R(4,3) S(1,1)
PR=√(4(1))2+(34)2=√26
SQ=√(21)2+(61)2=√26
The diagonals are congruent. Parallelogram PQRS is a rectangle.
34 1m= = 4(1) 5
61 5m= = 21 1
PR SQ
The diagonals are perpendicular. Parallelogram PQRS is a rhombus.
Since parallelogram PQRS is both a rhombus and a rectangle, it is a square.
Y(3,2) Z(1,3)
X(4,2) W(0,1)
WY=√(30)2+(21)2=√18
ZX=√(4(1))2+(2(3))2=√50
21 3m= = =1 30 3
2(3) 5m= = =14(1) 5
WY ZX
Since WXYZ is not a rectangle, it is not a square.
Find the lengths of the diagonals.
PR=SQ
Find the slopes of the diagonals.
PR⊥SQ
Find the lengths of the diagonals.
Find the slopes of the diagonals.
WY≠ZX
The diagonals are not congruent. Parallelogram WXYZ is not a rectangle.
WY⊥ZX
The diagonals are perpendicular. Parallelogram WXYZ is a rhombus.
Chapter 6 Study Guide Answers
16
kite quadrilateral two
trapezoid onebase nonparallel legs angles
consecutive base
Chapter 6 Study Guide Answers
17
legs congruent isoscelesisosceles
midsegmentendpoints
midpointsTriangle
Trapezoid
Chapter 6 Study Guide Answers
18
54° 52°
m∠BCD=2(9052)°=76°m∠ABC=(9027)°+52°=115°
52°
27°
CD=12
m∠QPS=(9039)°+59°=110°m∠PSR=2(9059)°=62°
78° 59°
QP=15
59°
39°20
12
22
15
m∠A=180°100° m∠F=180°49°
m∠S=m∠P2a254=a2+27a2 a2 a254=27 +54 +54
a2=81√a2=√81a=9
=80° =131°
AD=BC12x11=9x29x 9x 3x11=2 +11 +11
3x=93 3x=3
13.5+82
=10.758.5
8.5 EH=16.58.5=8