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Chapter '1POLYGONS
Pages 298-301 (Section '1.1)
1 The A is isosceles so LB = LC.
40+2x = ISO
2x = 140
x = 70
mLB = 70
2 L2 issupp toLl so mL2 = 50
L5 and L7 are vert Ls so mL5 = 70
L2 +L3 +L5 = ISO so mL3 = 60
L4 is supp to L3 so mL4 = 120
L6 is supp to L5 so mL6 = 110
8 LF+LK+LH ISO
90+70+LH = ISO
160+LH = ISO
mLH = 20
OJ II FK by Midline Theorem
So mLHGJ = 90 (II lines ~ corr Ls ••) and mLHJG = 70
(II lines ~ corr Ls -).
9 aA bA eN dA eN
10 Let the m vertex L = 2L
x+x+2x = ISO
4:1:= 180
x = 45 m vertexL = 2(45) = 90
()) •
3 The sum ofintLs ofaA are ISO".
LCAD=LCBA+LC = ISO
SO+ 60 + LC = ISO
mLC = 40
AE and BD are alta so mLADB and mLAEB = 90
LAEB+LCBA+LEAD = ISO
90 + 60 + LEAD = ISO
mLEAD = 30
LADB + LCAD +LADD = ISO
90+S0+LADD = ISO
mLABD = 10
LEAD + LADD +LAFB = ISO
30 + 10 +LAFB = ISO
mLAFB = 140
E
D~F
2x+21+50 = ISO
2x+21 = 130
x+y=65
65 + mLM = ISO
mLM = 115
111 mLORP = 2mLR. LR = ISO- 90 - 10 = SO
1mLORP =2(SO)= 40
mLMOR= lS0-90-40=50
12 LD +LE +LF = ISO
110+LF = ISO
LF = 70
LD+ 150 = ISO
LD=30
:. LD + LF = 30 + 70 = 100
13 Given: AABC is a rt A-
LB isrtL.
Prove: LA and LC are compo
AABC is a rt A, LB a rt L and the sum of the measures of
the 3 Ls = 180. Then
LB+LA+LC = ISO
90+LA+LC = ISO
LA + LC = 90 soLA andLC are compo
()) •14 Given: AABC isos
5 Let the m of the Ls be 4:1:,5x, 6x.
4:I:+5x+6x = ISO 4:1: 4(12)= 4S
15x = ISO 5x = 5(12)= 60
x = 12 6x = 6(12) = 72
8 LORS=LP+LD
4x+6 = (x+24)+(2x+4)
4x+6 = 3x+2S
x = 22
mLO = (2x + 4) = 2(22) + 4 = 48
7 From diagram, A and Y are midpoints. By Midline1Theorem, AY = 2WX or 9.
88 Section 7.1
A~LB a rtL.
Prove: LA = LC = 450 B C
An isos A has 2 - Ls. If LB = 90 and
LA+LB+LC= ISO, LA+LC=90.
Since LA = LC, 2LA = 90, mLA = 45 so mLC = 45.
15 mLI = 2x, mL2 = 3x, mL3 = 3x +4
2x+3x+3x+4 = ISO
Sx = 176
x = 22mLl = 44, mL2 = 66, mL3 = 70
m extL atL3 = lS0-70 = 110
18 The measure of an ext L of aLI. = sum of the measures of
the remote intLs so
17 By Midline Theorem MO =~FH = 10
Ll.KFM iiI!i Ll.OGMby SAS (K, 0 are midpoints so KF •• OG,
Ls F and G are rtLS, FM '" MG because M is a midpoint)
so KM •• MO and KM = 10. JK and JO can be proved
iiI!i MO in the same way.
a P of JKMO is 4(10) = 40 b Rhombus
B
4 Midline Theorem
2 Two pta determine a seg.3 Midline Theorem
5 Transitive prop of IIlines
6 Division prop
7 If a quad has opp sides '"
and II,then itis aO.7 EFGH is aO. )
20 Given: ABCD is a quad.
FmdptAB
GmdptBC
Hmdpt,CD
EmdptDA 0
Prove: EFGH is a parallelogram.
1 ABCD is a quad. 1 Given
F mdpt AB, G mdpt BC,
H mdpt CD, E mdpt DA
2 DrawDB- 13 EF IIto andithe
length of DB- 14 HG IIto and ithe
length of DB
5 EFIIHG
6 EF-HG
22 Let mLEJF = x, mLEFJ = 180 - x -70180-x-70 110-x
a mLJFH = 2 --2-
180-x x+ 70LHJF = -2-' LJFH = -2- (ExtL Theorem)
mLH = 180_C8~-X+X~70)
x xmLH = 180 - (90 -2 +2 + 35)
mLH = 180-125 = 55
b mLH = 180-C8~-X + X+2LE)
mLH = 180 - (90 - ~ + ~ + L2l
mLH = 180-(90+ L2l
= 90- L2E
mLH = (90 - ~LE)
23 By Ext L Theorem, ~ = a + el and Cl = d + e2.
a + el +Cl = ~+ d + ~ ora + el + Cl = d + ~ +~.
21 LetmLA=x.IfA then AA-SOLEDA= x.
By Ext L Theorem, LBED == 2x. If A then AA
so LDBE = 2x. Then LBDE = 180 - (2x + 2x) or
LBDE = 180 - 4x. LBDC = 180 - (180 - 4x) - x.
LBDC = 180 - 180 + 4x - x = 3x. If A then AA so
LBCD = 3x. But AB - AC soLABC = 3x OfA then AA).By subtraction LCBD = x. In Ll.BCD,x + 3x + 3x = 180,
6 67x = 180, x = 257, mLA = 257,
180
180
92.5mLD =
A ---------7IZ/" I
M,." I" I
IIIB
mLD = 117.5 or
Mis mdpt of AB. C
Prove: M =dist from A, C, and B
Extend CM to Z so that CM •• MZ. Then ACBZ is a
parallelogram (the diagonals bis each other). ACBZ is a
rectangle because it has a rt L. The diagonals of a
rectangle are •• so AM •• CM - MB.
x+3y = 45+2y
x = 45-yand
45+2y+5x = 180
5x + 2y = 135 Substituting x = 45 - y,
5(45-y)+2y = 135
225-5y+2y = 135
-3y = -90 x = 45-y
y = 30 x=45-30= 15
mLPST = x + 3y = 15 + 3(30)
mLPST = 15 + 90 = 105
16 LABC "'LACB(IfA thenAA)
LA + LABC + LACB = 180
30 + x +x = 180
2x = 150
x = 75lID is one of the trisectors ofLABC soLDBC can belof
LABC(75)or~LABC(75)
LDBC = 25 or LDBC = 50CD bis LACB so LBCD =~LACB(75) = 37.5.
The sum oftheLs ofaLl. are 180.
LBCD + LOCB + LD = 180
37.5+25+LD = 1800r 37.5+50+LD =62.5 +LD = 180 or 87.5 +LD =
19 Given: .lI.ABCis a rt LI..
LC is rtL.
"
,
t.
Section 7.1 89
oillK
G~
2 In a rhombus all sidea are -.
3 Given
1 Given
1/ i
H
z
~X B Y
2 Given
3 In an isos trap base Ls are ••.
4 Reflexive prop
5 AAS
6 CPCTC
7 If a Ii. has 2 •• Ls, it is isos.
a If 11then l!s.9 CPCTC
10 An isos trap has 2 •••legs.
11 Subtraction prop
12 SAS
13 CPCTC
14 If2 pta are the same dist
from a 3rd, they are =dist
from that pt.
1 Given
2 Given
3 II lines => alt int Ls ••
4 Sameas3
5 Substitution
6 No Choice TheoremW
:~;':~o'J>",p
b HP"JK
c R is =dist from 0 and M.
1 OHJM is an isos trap, 1 Given
with bases HJ
and OM.
2 LHPJ •• LJKH
3 LPHJ"LKJH
4 HJ •• HJ
5 APHJ •• AKJH
6 LKHJ"LPJH
a 7.AHRJ is isos.
aHR •• RJ
b 9 HP••JK
10 OH •• MJ
11 LOHR •• LMJR
.12 AORH •• .AMRJ
13 RO •• RM
c 14 R is =mst from
OandM.
Conclusion: AWET •• li.RAG
1 TW" GR, WE •• AR 1 Given
2 LE, LA are rt Ls. 2 Given
3 Extend GA to P 3 Two pta determine a line.
soAP •• ET
15 Given:
14 Given: A<! nXV,AiJnlW,LZAC •• LXAB
Prove: LX •• LZ
1 A<! nXV, AiJ nlW2 LZAC"LXAB
3 LXBA •• LBAC
4 LBAC •• LACZ
5 LXBA"LACZ
6 LX"LZ
Prove: a AHRJ is isos.
13 Given: OHJM is an isos trap,
with bases HJ and OM.
LHPJ"LJKH
c
1 Given
2 Given
3 1. lines form rt Ls.
4 Same as3
5 RtLsare".
6 Sides of an isos A are ••.
7 If l!s. then 11a Given
9 A mdpt divides a seg into
2 •• segs.
10 AAS
11 CPCTC
1 6ABC isos, base BC
2 DE 1. BA, DF 1. AC
3 LBEDrtL
4 LDFCrtL
5 LBED "LDFC6 AB •• AC
7 LB "LCa DmdptofBC
9 BD •• DC
10 ADEB •• ADFC
11 DE •• DF
~
H 6MJK •• OP,
HK~MP
Prove: LH ~LM K 0 P
Either LH ••LM or it isn't. Assume that it is.
Since LJ •• LO and JK •• OP are given, and LH ••LM is
assumed, AHJK •• AMOP by AAS. By CPCTC, HK •• MP,
which contradicts given information that HK ;. MP.
Therefore, LH ;. LM.
10 Given: LJ •• LO,
11 Given: 6ABC is isos.
ABbase
CD alt A BProve: CD median AB
1 6ABC is isos. 1 Given
2 AB base 2 Given
3 CD alt 3 Given
4 LADC, LBDC rt Ls 4 The aIt to the base of an
isos A forms rt Ls.5 CD •• CD 5 Reflexive prop
(i AC •• BC 6 Legs of an isos A are 55.
7 AADC •• ABDC 7 HL
a AD •• DB a CPCTC
9 CD median AB 9 If a seg divides a seg into
,t 2 •• segs, then it is the
median.
A12 Given: 6ABC is isos.
BaseBC
DmdptofBC
DE 1. BA
DF LAC B C
Prove: DE •• DF
t
4 LRAP supp to LRAG 4 If 2 Ls form a st L, they are
supp.
Section 7.2
18 Given: EF median to AC.
LCBD-LADB;
CD base Ofi80St:.FDC.
17 a Rectangle b Rectangle c Square d Rhombus
e Parallelogram f Parallelogram IfRhombus
I
N
1 Given
2 Given
3 Given
4 IfA then A5 IfA then A6 Reflexive prop
7 SSS
20 mLHGK=x
mLGKH=x
mLGHK= 180-2x
a + (180 - 2x) + KHT = 180
:. mLKHT=2x-a
mLSGH = 180 - x - b
LS, LR, and LT = 60°
In t:.GSH a + 60 + 180 - x - b = 180
x+b-a = 60
2x + 2b - 2a = 120
Int:.KHTc+60+2x-a = 180
c+2x-a = 120
2x+2b-2a = c+2x-a
2b-c = a
a = 2b-c
19 Given: P, T and R lie
in planef.
LTNR-LTSR,
NS J. f,
LTNP-LTSP
Conel: t:.NPR - t:.SPR
1 LTNR-LTSR
2 NS J. f3 LTNP-LTSP
4 NR-SR
5 SP.NP
6 PR-PR
7 t:.NPR - t:.SPR
5 Subtraction prop
6 RtLsare-.
7 SAS
8 CPCTC
9 Transitive prop
10 Ifat:. has 2 sides-,
it is i80B.
11 IfA then A12 RtLs are-.
13 AAS
14 Transitive prop
11 LG-LP
12 LRAG - LRAP
13 £:.RAG- t:.RAP
14 t:.WE:r - £:.RAG
5 LRAP is a rtL.
6 LWET -LRAP
7 t:.WET - t:.RAP
8 TW-RP
9 GR-RP
10 t:.RGP is i80S.
Prove: ABCD is rectangle.
1 LCBD - LADB 1 Given
2 AD n Be 2 Alt intLs - =>II lines
18 a G, H, J are the midpoints of FD , FE , and DE. 1-- 1- - 1-
respectively. :. GH = 2DE; HJ = 2FD and GJ =2 FE
since a line that joins the midpoints of 2 sides of a t:. isII to and = to ~ the third side. :. Perimeter t:.GHJ = ~
Perimeter t:.DEF = ~(145)= 7~
b The triangle determined by the midpoints of the sides
of a triangle has a perimeter that is half the perimeter
of the original triangle.
d S = (10)180
S = 1800
b S = (5)180S=9oo
Pages 309-313 (Section 7.3)
1 Sj = (n - 2)180
a S = (4-2)180
S = (2)180
S=360
c S = (6)180
S= 1080
e S = (91)180
S= 16,380
2 Sj = (n - 2)180
Sj = (5 - 2)180
Sj=3.180=540
mLE = 540-(LA+LB +LC +LD)
mLE = 540 - (160 + 50 + 140 + 150)
mLE = 540 - 500 = 40
3 II lines =>alt int Ls ••
4 Given
5 A median divides a seg
into'2 '" segs.
6 Given
7 Legs ofi80~ are "".8 Transitive prop
9 IfA then l:i.10 Transitive prop (steps 1, 3, 9)
11 If l:i. then A12 Substitution
13 If the diagonals of a quad
are'" and bis each other,
the quad is aD.
3 LDAF-LFCB
4 EF median to AC
5AF-FC
6 CD base of i80St:.FDC
7 DF-FC
8 AF-DF
9 LFAD -LFDA
10 LFCB - LFBC
11 FB-FC
12 AF-FD-FC •••FB
13 ABCD is aD.
92 Section 7.3
4 ~ = (4 - 2)lS0 = 2(lS0) or 360
LF+LG+LH+LE = 360
110+S0+74+LE = 360
264 +LE = 360
LE = 96
L lis BUppto LE so
96 +L1 = ISO
mLl = 84
3 = n-2
5 = n
10 a 900 = (n-2)lS0900180 = n-2
5 = n-2
7 = n
c 2SS0 = (n - 2)lS02880180 = n-2
b 1440 = (n - 2)lS01440180 = n-2
S = (n -2)
10 = n
d lSOx - 720 = (n - 2)1S0lSO(x - 4) _ _ 2
ISO -n
x-4 = n-2
x-2 = n
16 = n-2
IS = n
436 = (n - 2)lS0436180 = n-2
2.42 = n - 2, Impossible, because 4.42 can't be the
4.42 = n number of sides of a polygon
540 = (n - 2)lS0640180 = n-2
f
e
b d= 6(6;3)
d-!.:.!- 2d=9
d d=3(3;3)
d-!:!- 2d=O
3 Use d = n(n2-3)
a d = &<5;3)
d-!.:!- 2d=5
c d = 4<4;3)
d-!..:..!- 2d=2
8 a Sj = (n - 2)lS0 b Sum of exterior Ls is 360°.
Sj = (12 - 2)lS0
Sj = 10 . ISO = lSoo
5 By Midline Theorem, KP IIMO so LJKP - LM
a mLJKP=70
y + 15 = 70, y = 55 so
b LJPK=55-10=45
In LWKP,LJ = ISO -70 - 45
c mLJ = lS0-115 = 65
Se= 360 so
a triangle = 360 b heptagon = 360
c nonagon =,360 d 1984-gon = 360
360 = (n - 2)1S0360180 = n-2
2 = n-2
a n = 4 quadrilateral
2(360) = (n - 2)1S0
720 = (n - 2)lS0
4 = n-2
b n = 6 hexagon
11
12 Sj = (n - 2)1S0, let x = the number of sides
lSO(n - 2) + 900 = IS0[(n - 2) + xl
lSOn - 360 + 900 = lSOn - 360 + lSOx
900 = lSOx
5 = x, the number of sides
13 d = n(n2-3)
a 14 = n(n2-3)
(2)(14) = n(n2- 3) x 2
2S = n(n-3)
0= n2-3n-2S
o = (n - 7)(n + 4)
n = -4 Or0(-4 cannot be a solution)Heptagon
b 2(35) = n(n2- 3) x 2
70 = n2-3n
n2-3n-70=0
(n + 7)(n - 10)
n = -7 or@ Decagon (-7 cannot be a solution)
7 3
2 CPCTC
3 CPCTC
1 Given
4 Given
5 An alt forms rt Ls with the
opp side of a 6. .
6 Given
7 Same as5
S RtLsare-.
9 AAS
10 CPCTC
6 XV alt of .6XYZ
7 LXVZ is a rt L.
S LADC-LXVZ
9 ~C.AXVZ
10 AD -XV
9 Given: 6ABC -.6XYZ
AD is alt
of6ABC.
XV is alt
of.6XYZ.
Prove: AD - XV
1 6ABC -.6XYZ
2 AC-XZ
3 LC -LZ
4 AD alt of 6ABC
5 LADCisartL.
tSection 7.3 93
•
It
c 2(209) = n(n - 3) x 22
418 = n2-3n
n2 -3n -418 = 0
(n - 22)(n + 19)
n = @ or -18 ltll.pn
14 8 A b N 08 d A 115 18 < x < 36
22 a Divide EFGH,J into :16s. The Ls of each 6 will have a
sum of 180 •. This will be 3(180.) = 540 •.
b Use the same procedure, 6 6s will have 6(180.) or
1080 •.
c Yes. The procedure seems the same for convex and non-
convex polygons.
d No, because in a non-convex polygon, the non-convex L
does not have the same exterior L as a convex L.
.)
23 The 3 Ls are x, 90 - x, 180 - x. A decagon has 10 sides so
m of Ls = (10 - 2)180
m of Ls = (8)180
m of Ls = 1440
x + (90 - x) + (180 - x) = 1440 - 1220
-x + 270 = 220
-x = -50
mLx = 50, 90 - 50 = 40, 180 - 50 = 130
n(n - 3)24 2 > (n - 2)180
n2 - 3n > 2(180n - 360)
n2-3n > 360n -720
n2 - 363n + 720 > 0 By quadratic formula,
363:t -.J 131,769 - 2880n> 2
363 :t -.J 128,889n> 2
n > 363 ~ 359 = 0 or 2
A polygon can't have 2 sides so it must have 362 or more.
Pages 316-318 (Section 7.4)360 360
1 a E = '3 = 120 d E = 15 = 24360 360 15
b E =4= 90 e E ="23= 1523360
c E=s=45
x.axis
B (3,2)
C (3, 8)
y-axis
y-axisB (2b, 2c)
)
A (-5, 2)
D(-5,8)
17 A = bh
A = 8.6 = 48
16 .6 < x < 11 b 7 < y < 10
20 Given: LB '" LD,
LA"'LCProve: ABCD is a o.Let mLB and mLD = x,
mLA and mLC = y
The sum oftheLs ofa quad iil 360 so
18 Given: .6.0BA;
MismdptBO.
Nis mdpt BA.
Prove: MN II OA,MN = -210A A (2a, 0)
x-axis
midpt BA =ea; 2b, ¥) = (a + b, c) N(a + b, c)
. - (2b + 0 2c + 0)mldpt BD = -2 -, -2 - = (b, c) M(b, c)
OA= 2a1
MN=a :. MN=zOA- 0-0
slope OA = 2a _ 0 = 0
- c-cslope MN = a + b _ b = 0 :. MN II OA
19 34
2x + 2y 360
x + y 180
LB must be supp to LA so AD II BC
LB must be supp to LC so AB II DC
In a quad if both pairs of opp sides are II, the quad is a
parallelogram.
. n(n - 3)21 Theorem 57 d =--2-
2 Using the formula E = 360, and that an Int L = 180-n
IntL = 180-ExtL, 180 - E = I.'JOOa E='s=72 180-72 108
b 360E=6=60 180-60 120
360c E='9=40 180-40 140
d3(;0
E=12=30 180-30 1503(;0 1 1 162~e E=21= 177 180-177 =
7
In this formula the n represents. the number of vertices.
From each vertex you can draw n - 3 diagonals, so there
are n(n - 3) diagonals. Each diagonal will have been
drawn twice, so we divide by 2.
(94 Section 7.4
3604 From 180- IntL = ExtL and n = T
8 ANTE is isosceles because a stop sign is a reg octagon and
NE-NT.
1 PENTA is reg pentagon. 1 Given
2 PE - PA, EN - AT 2 In a reg polygon the sides
4 In a reg poly the sides are ••.
5 In a reg poly the Ls are -.
6SAS
7 CPCTC
8 A bis divides a seg into
360E=3i)=12
I = 180-E
I = 180 - 12 = 168
3 Given
4 .L lines form rt Ls.
5 RtLsare III.
6 Assumed from diagram
7 VertLsare III.
8 No Choice Theorem
9 Transitive prop
10 If a A has 2 - Ls, it is isos.
4 PE- PA,EN -AT
5 LE-LA
6 APEN-APAT
7 PN- PT
8 NR-TR
3 FC.L DD
4 LECD,LFCD rtLs
5 LECDIIILFCD
6 LAEF,LCED vertLs
7 LAEF-LCED
8 LF-LCED
9 LAEF-LF
10 AAEF is isos.
2 a segs.
9 PR 1.bis NT 9 Two pts =dist from endpts
of seg determine 1. bis.
W RS passes through P. W A seg has only 11.bis.
b The .Lbis ofa side ofa reg poly either bisects the L at
the vertex opposite that side or is the 1.bis of the side
opposite that side.
Conclusion: AAEF is iSfBceles.
1 AB IIIAD 1 Given
2 LD IIILD 2 IfA thenAA
F
9 Given: ABIIIAD,
FC.L DD
10 S=(n-2)1805040 = (n - 2)180
28 = n-2
30 = n
P
'V5:J.AN T
are III.
are-.
3 In a reg polygon the Ls
4 BAS
5 CPCTC
6 If2 sides ofa A are III,the
A is isos.
d 180-162=18
360+18=20
20 sides4 1e 180-1726= 76
. 1360+7&=50
50 sides
a 180 - 1« = 36
360+36= 10
10 sidesb 180-120=60
360+60=6
6 sides
c 180-156=24
360+24= 15
15 sides
3 LE-LA
4 APEN- APAT
5 PN-PT
6 APNT is isos.
5 Given: PENTA is reg pentagon.
Prove: APNT is isos.
360 3603 FromE=o,n=T
a == 6 sides
b ~=9sides
c ~= 10 sides
d s:o = 180 sides
e ~ =48sides
The polygon is a pentagon.n = 5
9(3:0) = (n _ 2)180
540 = (n-2)180
3 = n-2
11
12 mIL = 7x, mEL = 2x
7x + 2x = 180
9x = 180
x = 20
Exterior L = 2(20) = 40Number of sides = ~ = 9
The polygon is an equiangular nonagon.
13 a A b SeA d S e S f N
AE
7 Ifx = m extL, 4I = m intL.
x+4x = 180 E=3:~n=3:0360
5x = 180 n=36
x = 36 n= 10
x=E = 36
The polygon is an equilateral decagon.
8 a Given: PENTA reg pentagon
RS.L bis ofNT.
Prove: RS passes through P.
1 PENTA reg pentagon 1 Given
2 RS.L bis of NT. 2 Given
3 Draw PN, PR, PT 3 Two pts determine a seg.
Section 7.4 95
14 Gi""", ABCDEF" •••• h........ •@,Prove: ACDF is a rectangle. F - - -- - - - 0
1 ABCDEF reg hexagon 1 Given . ' E2 AB IIIBC IIIDE •• EF 2 A reg hexagon has all
sides III.
)
)
vax yxY
Pages 320-323 Chapter 7 Review Problems
because these are the Ls between 105° and 145° that can
form equiangular polygons. By using the equation above,1
mLV can be (36, 60, 7'F,. 90, 100, 108).
17 A regular semioctagon means ~ a regular octagon soLE a
LD a LC. LA = ~LE. An angle of a regular octagon is
1= 180- 360n360
1= 180-8I = 180 - 45 = 135
1ThenLE= 135,LA=672
3x+3y+9=1351 1
2x+y-42=672Solving: 3x + 3y = 126
-3(2x + y) = 72
3x+3y = 126
-6x-3y = -216
-3x = -90
x = 30Then 3(30) + 3y + 9 = 135
90 + 3y = 126
3y=36
y = 12
16 LT+x+y= 180;
LV + 2x + 2y = 180
-2LT - 2x - 2y = -360
LV -2LT = -180
LV=2LT-180
The set of possibilities for LTis:4(108, 120, 128;;, 135, 140, 144)
3 A reg hexagon has aU Ls III.
4SAS
5 CPCTC
6 Sameas2
7 If opp sides of a quad are
III,it is a £:7.
8 Two pta determine a line.
9 Sameas3
10 IfA, then.().11 Subtraction
12 Reflexive prop
13 SAS
14 CPCTC15 Ifthe diagonals of a quad
are III,it is aD.
3 L1111L2
4 AABCIII6DEF
5 ACIIIDF
6 AFIIICD
7 ACDF is a £:7.
8 Draw AD and CF
9 LAFEIIILCDE
10 LDFE IIILDEF
11 LAFD IIILFAC
12 AF=.AF
136DFAIII6CAF
14 AD IIIFC
15 ACDFisaD.
15 Given: RO.l plane GHJ;0, M, and Kare coplanar. R
GHJKMP is a regular hexagon.
ffiSbisectaLGHJ.
RHIIIRJ
Prove: liliOO is regular.
1 RO.l plane GHJ 1 Given2 LROH, LROJ rt Ls 2 .1 lines intersect at rt Ls.
3 GHJKMP is a regular 3 Given
hexagon.4 LGHJ = 120" 4 Ls of a regular hexagon =
120".
5 HObisLGHJ
6 LOHJ = 60"
7 RHIIIRJ
8 ROIllRO
96ROHIII6ROO
10 OH 11100
11 60HJ is isos.
12 LOHJ IIILOJH
13 LOJH = 60°
14 LHOO = 60°
15 6HOO is equiangular.
16 HO IIIHJ aOO
17 6HOO is equilateral.
18 6HOJ is regular.
5 Given
6 Division prop
7 Given
8 Reflexive prop
9 HL
10 CPCTC11 An isos 6 has 2 a sides.
12 If A then.().
13 Substitution prop
14 SumofLsof6= 180°.
15 Ifthe Ls of a 6 are ii, itis
equiangular.
16 If.(). then A17 If the sides of a 6 are "', it
is equilateral.
18 A regular polygon is
equilateral and equiangular.
1 Given: LDBC aLE E~
Concl: LAaLBDC ABC
1 LDBC aLE 1 Given
2 LC +LDBC + 2 The sum of the Ls of a 6
LBDC= 180 is 180°.
3 LC + LA + LE = 180 3 Same as2
4 LC+LDBC+LBDC= 4 Transitive prop
LC+LA+LE
5 LBDC aLA 5 Subtraction prop
96 Chapter 7 Review
The polygon has 11 Bides.
8 The measure of the Ls of a quadrilateral is 360.
Let4thL=x
40 + 70 + 130 + x = 360
240+ x = 360
x = 120
7 Let the m Ls be J:, 2x, 3L
X + 2x + 3x = ISO
6x = ISO
x = 303x = 90
(~)3X = 45
S mLl = 50 because 90 + 40 + x = ISO
mL2 = 50 because 90 + 40 + x = ISO
9 The ext L of a.to is equaJ to the sum of the remote
intLs 80,
x+50+40 = 2x+50
x+90 = 2x+50
x+40 = 2x
4O=x
2x+50+L'YZA = 180
2(40)+ 50 + L'YZA = ISO
SO+50+L'YZA = 180
130+L'YZA = ISO
mL'YZA = 50
11 Use ifA then IJ. and L3 = 180 -70 - 90 80L3 = 20.
10 CE = 6 by Midline Theorem
mLBCE = mLD = SO
mLBEC = ISO - 60 - SO
mLBEC=40
13 a St=(n-2)lS0
St = (33 - 2)lS0
St = (31)lS0
St= 5580
b 8,,=360
14 St = (n-2) ISO
1620 = (n - 2) ISO1620180 = (n-2)
9 = n-2
n = 11
1 Given
2 Given
3 Given
4 Given
5 If a line is .l to a plane, it
is .l toa line that passes
through its foot.
6 .llines form rt Ls.
7 RtLsare-.
S Reflexive prop
9 AAS
10 CPCTC
6 LTVS, LTVX rtLs
7 LTVS-LTVX
STY-TV
9 .t.STV - .t.rrV
10 TS-TX
I) Given: SV lies in plane m.
VX lies in plane m.
LS-LX,
TV .lplanem
Prove: TS - TX
1 SV lies in plane m.
2 VX lies in plane m.
3 LS-a
4 TV.l planem
5 TV .l SV and VX
2 Given: 00, .@J,LW-LV,
) LORW-LOSV
Prove: PR-"ST W V100 1 Given
2 LW-LV, 2 Given
LORW-LOSV
3 OW-OV 3 Radii of0 are-.
4 .toOWR- .toOVS 4 AAS
5 OR-OS 5 CPCTC6 op-(jif 6 Radii of0 are-.7 PR-"ST 7 SubtractionCr>3 Given: AC-AE, o A
LCBD-LEFD
Prove: LBDC-LFDE E F1 AC-A'E 1 Given
2 LCBD-LEFD 2 Given
3 LC-LE 3 IfA then IJ.4 LBDC-LFDE 4 No Choice Theorem
R
4 Given: LS-LRQP, s<2JLROS-LP
Prove: LSRO-LPRO o P
t 1 LS-LRQP 1 Given
2 LROS-LP 2 Given
3 LSRO-LPRO 3 No Choice Theorem
No sides
given -. T
Chapter7Re~ew 97
15 A pentadecagon has 15 sides.
d- n(n-3)- 2
d_15(l5-3)- 2
d_ 15(12)- 2
180d=z=90
20 LADC - LACB (IfA then.o.)
LABC + LACB = 180 - 50
LADC + LACB = 130 ( -JLADC = LACB = 65
LDBC - LDCA so LADD - LDCB
LD = 180 - (LDBC + LDCB)
LD = 180 - (LDBC + LABD)
LD = ISO-LADC
LD= 180-65
mLD= 115
21aAbScS dN
22 ~=(n-2)lS0
~ = (6)lS0 = 10S0
1080 - 540 = 540
The octagon must be non convex.
23 A triangle has !l diagonals.
A quadrilateral has 2. diagonals.A pentagon has :idiagonals.A hexagon has IIdiagonals.It is not an arithmetic progression.
24 mIntL=x
mextL=yx = 4(y) +30 ( 'Ix-4y = 30
x+y = 180
Solve (x + y = lS0) mutt by (-1)
-x-y = -ISO
x-4y = 30
-5y = -150
y = 30lID Equiangular dodecagon"3)=12
A
25 Given: BC - FE,
~LC -LE
Prove: 6.ABF is isosceles. C E
IDe-FE 1 Given
2 LC -LE 2 Given
3 LBDC,LFDEvertLs 3 Assumed from diagram
4 LBDC - LFDE 4 VertLs are-.
5 6.BDC - 6,FDE 5 AAS
6 BD - FD, CD - DE 6 CPCTC
7 CFEIIEB 7 Addition
SLA-LA 8 Reflexive prop
9 6.AFC - 6.ABE 9 AAS
10 AB - AF 10 CPCTC( J11 6.ABF is isos. 11 If2legs of 6. are -, then it
is isos.
I
5x = 90
x = 18
mL2 = 72
4 Ls ofa reg poly are-.
5 SAS
6 CPCTC
~E
1 Given U2 Sides of a reg poly are a.
3 Same as2
6x=90
x=15
mL2 = 75
Chapter 7 Review
This is true for regular pentagons and squares.
4 L1-L2
5 6.PAT - 6,PEN
6 PT-PN
n = 3The polygon is an equilateral triangle.
18
17 Given: PENTA is a reg pentagon.
Prove: PT - PN1 PENTA is a reg pentagon.
2 PA-PE
3 AT-EN
19 a 6.ABC is isos so LABC - LACB.
LA + LABC + LACB = ISO
50+x+x = lS0
50+2x = lS0
2x = 130
LABC=x = 6512x = LABF
12(65) = LABF1
mLABF=322
b LACB + LACD = 18065 + LACD = ISO By Substitution
LACD = 1151 1
LACE =2LACD = 572
LBCE = LACB + LACE1 1
mLBCE = 65 + 572= 1222
C LEBC + LBCE + LE = lS03~+ 122~+LE = 180
155+LE = ISO
mLE = 25
18 E = 360, ~ = (n-2)180n n
3:0 = 2[(n-:)18~]
360 = (2n-4)180
360 = 360n - 720
360n = 1080
n
26 h=b+e
LAFG = h by the No Choice Theorem.
LEFD = h by vertical Ls =.d = h + e and e = d - h
Substituting for e:
h = b + (d - h) = b + d - h
2h = b + d1h = "2(b + d)
30 Assume there is a polygon.Then n(n - 3) - 3602 -
n(n -3) = 720
n2-3n-720 = 0
Using the quadratic formula-(-3) 1:../9 + 2880
2- (-3) 1:..J2889
2
Prove:
27 Given: 2889 is not a perfect square, so there is no solution
possible.
1 Given
2 Given
3 Given
4 If8 then 1115 "2C180-4x)
6 ~(LPRS)
7 The m of an extL ofab. is
equal to the sum of the
remote int Ls.
6 LWRS=45-x
7 LPST = 90 + 2x
PR= PS;
RYbisLPRS.
SVbisLPST.1mLV="2mLP
(mLP= 4x)
1 PR= PS
2 RYbisLPRS
3 SVbisLPST
4 LPRS ""LPSR5 LPSR = 90 - 2x
8 LPSV = LVST = 45 + x 8 A bis divides an L into 2 ""
Ls.
9 mLWRS + mLRSV = 9 Addition
180-2x
10 LWRS +LRSV +
LV = 180
11 180 - 2x + LV = 180
12 LV = 2x113 2x = "2(4x)114 mLV =2(LP)
10 Sum ofLs of a b. equals
180°.
11 Substitution prop
12 Algebraic Properties13 Multiplication
14 Substitution prop
28 Possibilities are:
@@@@@@@ 108,108,108,108,
108,@@@@@@
Probability = ~
29 y-axis
B(- 4. 4) f (4.4)C (0.-4)1
II x-axis
A(-4.0) 0(0.0) G (4, 0)
The rotation sends A to C, B to F, and C to G.
Area recto ABFG = bh = 8(4) = 32
Chapter 7 Review 99
