chapter 6 study guide answers - allen independent school district · 2015. 2. 2. · chapter 6...

Chapter 6 Study Guide Answers

1

polygonthree

polygon sidecommon sides vertex

nonconsecutive diagonal

trianglequadrilateral

n-gondodecagon

decagonnonagonoctagon

heptagonhexagonpentagon

equilateralequiangular regular

irregular

concave diagonalexterior


2

diagonals exteriorconvex regular

interior convexdiagonals triangles

equals

convex twosides sum

exterior


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


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


5

four quadrilateral special quadrilaterals

two parallelparallelogram

sides side


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)


7

parallelogramproperties

definition conditions


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°


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.


10

rectangleright

parallelogramproperties

rhombus specialcongruent


11

rectangle rhombusproperties parallelograms

square right congruentparallelogram rectangle

rhombus

special


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


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.


14

parallelogramrectangle

rhombus

squarerectangle rhombus


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.


16

kite quadrilateral two

trapezoid onebase nonparallel legs angles

consecutive base


17

legs congruent isoscelesisosceles

midsegmentendpoints

midpointsTriangle

Trapezoid


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


19

chapter 6 study guide answers - allen independent school district · 2015. 2. 2. · chapter 6...

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