Solutions for Foundations Of Geometry 3rd Us Edition by Venema
INSTRUCTOR’S SOLUTIONS MANUAL
F OUNDATION S OF G EO M ET RY
THIRD EDITION
Gerard A. Venema Department of Mathematics and Statistics Calvin University
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The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs.
Reproduced by Pearson from electronic files supplied by the author
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SolutionstoExercisesinChapter11
SolutionstoExercisesinChapter26
SolutionstoExercisesinChapter314
SolutionstoExercisesinChapter427
SolutionstoExercisesinChapter550
SolutionstoExercisesinChapter672
SolutionstoExercisesinChapter786
SolutionstoExercisesinChapter896
SolutionstoExercisesinChapter10106
SolutionstoExercisesinChapter11127
SolutionstoExercisesinChapter12137
Bibliography143
SolutionstoExercisesinChapter1
1.6.1 Checkthattheformula A D 1 4 a C c b C d worksforrectanglesbutnotfor parallelograms.
Forrectanglesandparallelograms, a D c and b D d and Area=base∗height Forarectangle,thebaseandtheheightwillbeequaltothelengthsoftwoadjacent sides.Therefore
1.6.2 Theareaofacircleisgivenbytheformula A D π d 2 2.AccordingtheEgyptians, A isalsoequaltotheareaofasquarewithsidesequalto
;thus
. Equatingandsolvingfor π gives
1.6.3 Thesumofthemeasuresofthetwoacuteanglesin ^ABC is90◦,sothefirstshaded regionisasquare.Wemustshowthattheareaoftheshadedregioninthefirst square c2 isequaltotheareaoftheshadedregioninthesecondsquare a2 C b2 . Thetwolargesquareshavethesameareabecausetheybothhavesidelength a C b.Alsoeachofthesesquarescontainsfourcopiesoftriangle ^ABC (in white).Therefore,bysubtraction,theshadesdregionsmusthaveequalareaandso
1.6.4(a) Suppose
.Wemustshowthat
(b) Let u and v beodd.Wewillshowthat a, b and c arealleven.Since u and v are bothodd,weknowthat u2 and v2 arealsoodd.Therefore a D u2 v2 iseven (thedifferencebetweentwooddnumbersiseven).Itisobviousthat
iseven,and c D u2 C v2 isalsoevensinceitisthesumoftwooddnumbers.
(c) Supposeoneof u and v isevenandtheotherisodd.Wewillshowthat a,b, and c donothaveanycommonprimefactors.Now a and c arebothodd,so2 isnotafactorof a or c.Suppose x Z 2isaprimefactorof b.Theneither x divides u or x divides v,butnotbothbecause u and v arerelativelyprime.If x divides u,thenitalsodivides u2 butnot v2.Thus x isnotafactorof a or c.
If x divides v,thenitdivides v2 butnot u2.Again x isnotafactorof a or c Therefore a,b,c isaprimitivePythagoreantriple.
1.6.5 Let h C x betheheightoftheentire(untruncated)pyramid.Weknowthat
(bytheSimilarTrianglesTheorem),so x D h b a b (algebra).Thevolumeofthe truncatedpyramidisthevolumeofthewholepyramidminusthevolumeofthetop pyramid.Therefore
(a) Theperpendicularbisectorofalinesegment AB. Usingthecompass,constructtwocircles,thefirstabout A through B,the secondabout B through A.Thenusethestraightedgetoconstructaline throughthetwopointscreatedbytheintersectionofthetwocircles.
(b) Alinethroughapoint P perpendiculartoaline UsethecompasstoconstructacircleaboutP,makingsurethecircleisbig enoughsothatitintersects attwopoints, A and B.Thenconstructthe perpendicularbisectorofsegment AB asinpart(a).
(c) Theanglebisectorof jBAC. Usingthecompass,constructacircleabout A thatintersects AB and AC Callthosepointsofintersection D and E respectively.Thenconstructthe perpendicularbisectorof DE.Thislineistheanglebisector.
1.6.9 Let nABCD bearhombus(allfoursidesareequal),andlet E bethepointof intersectionbetween AC and DB 1 Wemustshowthat ^AEB ^CEB ^CED ^AED.Now jBAC jACB and jCAD jACD byProposition5.Byaddition wecanseethat jBAD jBCD andsimilarly, jADC jABC.Nowweknowthat ^ABC ^ADC byProposition4.Similarly, ^DBA ^DBC.Thisimpliesthat jBAC jDAC jBCA jDCA and jBDA jDBA jBDC jDBC
1Inthissolutionandthenext,theexistenceofthepoint E istakenforgranted.Itsexistence isobviousfromthediagram.Provingthat E existsisoneofthegapsthatmustbefilledinthese proofs.ThispointwillbeaddressedinChapter6.
1.6.10 Let nABCD bearectangle,andlet E bethepointofintersectionof AC and BD.We mustprovethat AC BD andthat AC and BD bisecteachother(i.e., AE EC and BE ED).ByProposition28, (' DA (' CB and (' DC (' AB .Therefore,by Proposition29, jCAB jACD and jDAC jACB.Hence ^ABC ^CDA and ^ADB ^CBD byProposition26(ASA).Sincethosetrianglesarecongruent weknowthatoppositesidesoftherectanglearecongruentand ^ABD ^BAC (byProposition4),andtherefore BD AC. Nowwemustprovethatthesegmentsbisecteachother.ByProposition29, jCAB jACD and jDBA jBDC.Hence ^ABE ^CDE (byProposition26) whichimpliesthat AE CE and DE BE.Thereforethediagonalsareequaland bisecteachother.
1.6.11 Theargumentworksforthefirstcase.Thisisthecaseinwhichthetriangleactually isisosceles.Thesecondcaseneveroccurs(D isneverinsidethetriangle).Theflaw liesinthethirdcase(D isoutsidethetriangle).Ifthetriangleisnotisoscelesthen either E willbeoutsidethetriangleand F willbeontheedge AC,or E willbeon theedge AB and F willbeoutside.Theycannotbothbeoutsideasshowninthe diagram.Thiscanbecheckedbydrawingacarefuldiagrambyhandorbydrawing thediagramusingGeoGebra(orsimilarsoftware).
(h) E ageometrystudentwhodoesnotknowhowtowriteproofs.
2.5.11(a) NegationofEuclideanParallelPostulate.Thereexistaline andapoint P not on suchthateitherthereisnoline m suchthat P lieson m and m isparallel to orthereare(atleast)twolines m and n suchthat P liesonboth m and n, m ,and n .
(b) NegationofEllipticParallelPostulate.Thereexistaline andapoint P that doesnotlieon suchthatthereisatleastoneline m suchthat P lieson m and m .
(c) NegationofHyperbolicParallelPostulate.Thereexistaline andapoint P thatdoesnotlieon suchthateitherthereisnoline m suchthat P lieson m and m orthereisexactlyoneline m withtheseproperties.
Note. Youcouldemphasizetheseparateexistenceof and P bystartingeachofthe statementsabovewith,‘‘Thereexistaline andthereexistsapoint P noton such that....’’
2.5.12 not (S and T ) K (not S) or (not T ).
S T S and T not (S and T ) not S not T (not C)or(not H)
S T S or T not (S or T ) not S not T (not C)and(not H)
True True True False False False False
True False True False False True False
False True True False True False False
False False False True True True True
2.5.13 H * C K (not H) or C.
H C H * C not H (not H)or C
True True True False True
True False False False False
False True True True True
False False True True True
ByDeMorgan, not (H and (not C))islogicallyequivalentto(not H) or C
2.5.14 not (H * C) K (H and (not C)).
H C H * C not(H * C) not C H and (not C)
True True True False False False
True False False True True True
False True True False False False
False False True False True False
2.5.15 If ^ABC isrighttrianglewithrightangleatvertex C,and a, b,and c arethelengths ofthesidesoppositevertices A, B,and C respectively,then a2 C b2 D c2
2.5.16(a) Euclidean.If isalineand P isapointthatdoesnotlieon ,thenthereexists exactlyoneline m suchthat P lieson m and m
(b) Elliptic.If isalineand P isapointthatdoesnotlieon ,thentheredoesnot existaline m suchthat P lieson m and m . (c) Hyperbolic.If isalineand P isapointthatdoesnotlieon ,thenthereexist atleasttwodistinctlines m and n suchthat P liesonboth m and n and is paralleltoboth m and n.
2.6.1 ConversetoTheorem2.6.2.
Proof. Let and m betwolines(notation).Assumethereexistsexactlyonepoint thatliesonboth and m (hypothesis).Wemustshowthat Z m and ∦ m Suppose D m (RAAhypothesis).Thereexiststwodistinctpoints Q and R that lieon (IncidenceAxiom2).Since m D , Q and R lieonboth and m.This contradictsthehypothesisthat and m intersectinexactlyonepoint,sowecan rejecttheRAAhypothesisandconcludethat Z m Since and m haveapointincommon, ∦ m (definitionofparallel).
Note. Inthenextfewproofsitisconvenienttointroducethenotation (' AB forthe linedeterminedby A and B (IncidenceAxiom1).Thisnotationisnotdefinedinthe textbookuntilChapter3,butitfitswithIncidenceAxiom1andallowstheproofs belowtobewrittenmoresuccinctly.
2.6.2 Theorem2.6.3.
Proof. Let bealine(hypothesis).Wemustprovethatthereexistsapoint P such that P doesnotlieon .Thereexistthreenoncollinearpoints A, B,and C (Axiom3). Thethreepoints A, B and C cannotalllieon becauseiftheydidthentheywould becollinear(definitionofcollinear).Henceatleastoneofthemdoesnotlieon andtheproofiscomplete.
Note. Many(ormost)studentswillgetthelastproofwrong.Thereasonisthat theywanttoassumesomerelationshipbetweenthelinegiveninthehypothesisof thetheoremandthepointsgivenbyAxiom3.Inparticular,manystudentswill startwiththefirstthreesentencesoftheproofabove,butwilltheneitherassertor assumethat D (' AB
Proof. Let P beapoint(hypothesis).Wemustshowthattherearetwodistinct lines and m suchthat P liesonboth and m.Thereexistthreenoncollinearpoints A, B,and C (Axiom3).Therearetwocasestoconsider:either P isequaltooneof thethreepoints A, B,and C oritisnot.
Suppose,first,that P D A.Define D (' AB and m D (' AC (Axiom1).Obviously P D A liesonboththeselines.Itcannotbethat D m becauseinthatcase A, B, and C wouldbecollinear.Sotheproofofthiscaseiscomplete.Theproofsofthe casesinwhich P D B and P D C aresimilar.
Nowsupposethat P isdistinctfromallthreeofthepoints A, B,and C.Inthat casewecandefinethreelines D (' PA , m D (' PB ,and n D (' PC (Axiom1).These threelinescannotallbethesamebecauseiftheywerethen A, B,and C wouldbe collinear.Thereforeatleasttwoofthemaredistinctandtheproofiscomplete.
Note. Manystudentswillassertthatthethreelinesinthelastparagrapharedistinct. Butthatisnotnecessarilythecase,asFig.S2.5shows. Outlineofanalternativeproof: Findoneline suchthat P lieson .Thenusethe previoustheoremtofindapoint R thatdoesnotlieon .Thelinethrough P and R isthesecondline.Thisproofissimplerandbetterthantheonegivenabove,but moststudentsdonotthinkofit.Youmightwanttoleadtheminthatdirection bysuggestingthattheyprovetheexistenceofonelinefirstratherthanprovingthe existenceofbothatthesametime.
FIGURES2.5: Twoofthelinescouldbethesame
2.6.4 Theorem2.6.5.
Proof. Let bealine(hypothesis).Wemustshowthatthereexisttwolines m and n suchthat , m and n aredistinctandboth m and n intersect Thereexisttwodistinctpoints P and Q suchthat P and Q lieon (Axiom2).There existsapoint R suchthat R doesnotlieon (Theorem2.6.3).Let m betheline determinedby P and R andlet n bethelinedeterminedby Q and R (Axiom1). Since P liesonboth amd m, m intersects .Since Q liesonboth amd n, n intersects .Since R doesnotlieon ,neither m nor n isequalto .Tocompletethe proofwemustshowthat m Z n
Suppose m D n (RAAhypothesis).Thenboth P and Q lieon m aswellas ,so m D (theuniquenesspartofAxiom1).Butthiscontradictsanearlierstatement intheproof,sowemustrejecttheRAAhypothesisandconcludethat m Z n.
2.6.5 Theorem2.6.6.
Proof. Let P beapoint(hypothesis).Wemustprovethatthereisatleastoneline suchthat P doesnotlieonthatline.
Let A,B and C bethreenon-collinearpoints(Axiom3).Define D (' AB , m D (' AC and n D (' BC (Axiom1).Therearetwocasestoconsider:eitherPisequal tooneofthethreepoints A,B or C oritisnot.
Suppose,first,that P D A.Then P D A doesnotlieon n becauseifitdid, A,B and C wouldbecollinear.Sotheproofforthefirstcaseiscomplete.Thereare similarproofsfor P D B and P D C
Nowassume P isdistinctfrom A,B and C.Wewillshowthat P cannotlieon allthreeofthelines , m,and n.Suppose P liesonallthreeof ,m,and n (RAA hypothesis).Thefactthat P and A lieonboth m and impliesthat m D (Axiom1). Thefactthat P and B lieonboth n and impliesthat n D (Axiom1).Thus D m D n.Butthiscannotbebecause A,B and C arenon-collinear.Thereforewe mustrejecttheRAAhypothesisandtheproofiscomplete.
2.6.6 Theorem2.6.7.
Proof. Let A,B and C bethreenon-collinearpoints(Axiom3).Define D (' AB , m D (' AC and n D (' BC (Axiom1).Inordertocompletetheproofwemustshow that , m,and n aredistinctlinesandthattheredoesnotexistapoint P suchthat P liesonallthreeofthelines.
C
Notwoofthelinescanbeequalbecause A, B,and C arenoncollinear.The argumentthatthereisnopoint P thatliesonallthreeofthelinesisthesameasthat inthelasttwoparagraphsoftheprecedingproof.
2.6.7 Theorem2.6.8.
Proof. Let P beapoint(hypothesis).Wemustprovethatthereexistpoints Q and R suchthat P, Q,and R arenoncollinear.
Thereexistsapoint Q suchthat Q Z P (byAxiom3thereareatleastthree points,so P isnottheonlypoint).Thereexistsapoint R suchthat R doesnotlieon (' PQ (Theorem2.6.3).Thereisnolineonwhichallthreeof P, Q,and R liebecause theonlylineonwhichthefirsttwolieis (' PQ and R doesnotlieonthatline.
2.6.8 Theorem2.6.9.
Proof. Let P and Q betwopointssuchthat P Z Q (hypothesis).Wemustprove thatthereexistsapoint R suchthat P, Q,and R arenoncollinear.
Thereexistsauniqueline suchthatboth P and Q lieon (IncidenceAxiom1). Thereisapoint R suchthat R doesnotlieon (Theorem2.6.3).Wewillcomplete theproofbyshowingthat P, Q,and R arenoncollinear.Supposethereexistsaline m suchthatallthreeofthepoints P, Q,and R lieon m (RAAhypothesis).Then m D bytheuniquenesspartofIncidenceAxiom1.Butthisisimpossiblesince R doesnotlieon butdoeslieon m.HencewemustrejecttheRAAhypothesisand concludethat P, Q,and R arenoncollinear.
