CBSE NCERT Solutions for Class 10 mathematics Chapter 6

CBSENCERTSolutionsforClass10mathematicsChapter6

Exercise6.1

Q.1. Allcirclesare_____.(congruent,similar)similar

Solution: Allcircleshavethesameshapei.e.theyareround.Butthesizeofacirclemayvary.

Thuscirclesaresimilar.Eachcirclehasadifferentradiussothesizeofthecirclemayvary.

Q.2. Allsquaresare_____.(similar,congruent)similar

Solution: Weknowthat,

Allthesidesofasquareareequal.

Sincetheratiosofthelengthsoftheircorrespondingsidesareequal.Hence,allsquaresaresimilarsincesizeofsquaresmaybedifferent,buttheshapewillbealwayssame.

Q.3. All_____trianglesaresimilar.(Isosceles,equilateral)equilateral

Solution: Weknowthat,allthesidesofanequilateraltriangleareequal.

Allequilateraltrianglesaresimilarbecauseoftheirsameshape.

Q.4. Twopolygonsofthesamenumberofsidesaresimilar,iftheircorrespondinganglesare_____andtheircorrespondingsidesareproportional.[proportional/equal]

Solution: Twopolygonsofsamenumberofsidesaresimilar,iftheircorrespondinganglesareequalandtheircorrespondingsidesareproportional.

Forexample,iftwotriangleswithangles30°,60°and90°aresimilar,thentheratioHypotenuseof1stcircleHypotenuseof2stcirclewillbethesame.

Q.5. Givetwodifferentexamplesofpairofsimilarfigures.

NCERTMathematics Chapter6Triangles

PracticemoreonTriangles Page1 www.embibe.com

Solution: Twofiguresaresaidtobesimilariftheratioofcorrespondingsidesareequal.

Twoequilateraltriangleswithsides1cmand2cm.

Ratioofthecorrespondingsidesare:ABDE=12ACDF=12BCEF=12Hereratiooftheequilateraltrianglesaresame.Therefore,theabovefiguresaresimilar.

Twosquareswithsides1cmand2cm.

Ratioofthecorrespondingsidesare:ABEF=12ACEG=12BDFH=12CDGH=12Herealsoratiosofthecorrespondingsidesareequal.Hence,theabovetwofiguresaresimilar.

Q.6. Givetwodifferentexamplesofapairofnon-similarfigures.

Solution: Twofiguresaresaidtobenon-similariftheratioofthecorrespondingsidesarenotequal.

ConsideraTrapeziumandasquare.

Ratioofthecorrespondingsidesare:PQAB=3cm3cm=1PSAD=43QRBC=6cm3cm=2SRDC=53Thus,theratioofthecorrespondingsidesarenotequal.Therefore,figuresarenotsimilar.

Consideratriangleandaparallelogram

Ratioofthecorrespondingsidesare:

ACPS=34BCSR=33=1

Hence,theabovetwofiguresarenon-similar.

Q.7. Statewhetherthefollowingquadrilateralsaresimilarornot.

Solution:

Tocheckwhetherthegivenquadrilateralsaresimilarornotweneedtochecktheratioofthecorrespondingsidesandangles.

Correspondingsidesoftwoquadrilateralsareproportionali.e.,1:2buttheircorrespondinganglesarenotequal.Hence,quadrilateralsPQRSandABCDarenotsimilar.

Exercise6.2

Q.1. Inthefiguregivenbelow,DE∥BC.FindEC.

Solution:

LetEC=xcmSinceDE∥BCSo,usingbasicproportionalitytheoremweget:ADDB=AEEC⇒1.53=1x⇒x=3×11.5⇒x=2Hence,EC=2cm.

Q.2. InFig.DE∥BC.FindAD

Solution:

LetAD=xcmSinceDE∥BCHence,usingBasicproportionalitytheorem,ADDB=AEEC⇒x7.2=1.85.4⇒x=2.4cmHence,AD=2.4cm

Q.3. ThediagonalsofaquadrilateralABCDintersecteachotheratthepointOsuchthatAOBO=CODO.ShowthatABCDisatrapezium.

Solution:

DrawalinesegmentOE∥ABIn△ABCSince,OE∥AB.Hence,AOOC=BEEC.

Butbythegivenrelation,wehave:AOBO=CODO⇒AOOC=OBODHence,OBOD=BEECSo,usingconverseofbasicproportionalitytheorem,EO∥DC.Therefore,AB∥OE∥DC⇒AB∥CDTherefore,ABCDisatrapezium.

Q.4. EandFarepointsonthesidesPQandPRrespectivelyofa△PQR.StatewhetherEF∥QRwherePE=3.9cm,EQ=3cm,PF=3.6cmandFR=2.4cm.

Solution:

Given:PE=3.9cm,EQ=3cm,PF=3.6cmandFR=2.4cmNow,PEEQ=3.93=1.3

PFFR=3.62.4=1.5Since,PEEQ≠PFFRHence,EFisnotparalleltoQR.

Q.5. EandFarepointsonthesidesPQandPRrespectivelyofa∆PQR.Foreachofthefollowingcases,statewhetherEF∥QR:PE=4cm,QE=4.5cm,PF=8cmandRF=9cm

Solution:

Given,PE=4cm,QE=4.5cm,PF=8cm,RF=9cmPEEQ=44.5=89

PFFR=89SincePEEQ=PFFRHence,EF∥QR(usingBasicproportionalitytheorem)

Q.6. EandFarepointsonthesidesPQandPRrespectivelyofa∆PQR.Foreachofthefollowingcases,statewhetherEF∥QR:

PQ=1.28cm,PR=2.56cm,PE=0.18cmandPF=0.36cm

Solution:

Given,PQ=1.28cm,PR=2.56cm,PE=0.18cmandPF=0.36cmEQ=PQ-PE=1.28-0.18=1.1cmandFR=PR-PF=2.56-0.36=2.2cmPEEQ=0.181.1=18110=955PFFR=0.362.2=955Since,PEEQ=PFFRHence,EF∥QR(usingbasicproportionalitythorem)

Q.7. Inthefigure,ifLM∥CBandLN∥CD,provethatAMAB=ANAD

Solution:

Inthegivenfigure,LM∥CB.

Hence,usingbasicproportionalitytheorem,AMMB=ALLC …(i)Since,LN∥CDHence,usingbasicproportionalitytheorem,ANND=ALLC …(ii)

Fromiand(ii)

AMMB=ANND

⇒MBAM=NDAN⇒MBAM+1=NDAN+1⇒MB+AMAM=ND+ANAN⇒ABAM=ADAN⇒AMAB=ANAD

Q.8. InfiguregivenbelowDE∥ACandDF∥AE.ProvethatBFFE=BEEC.

Solution:

InΔABCSinceDE∥ACHence,BDDA=BEEC …(i)(usingbasicproportionalitytheorem)

In∆BAE,SinceDF∥AEHence,BDDA=BFFE…(ii)(usingbasicproportionalitytheorem)Fromiand(ii),weget:BEEC=BFFE

Q.9. Inthefigure,DE∥OQandDF∥OR.ShowthatEF∥QR.

Solution:

In∆POQ,sinceDE∥OQ,

PEEQ=PDDO …(i)[Usingbasicproportionalitytheorem]

In∆POR,sinceDF∥OR,

PFFR=PDDO …(ii)[Usingbasicproportionalitytheorem]

Fromiand(ii),weget,PEEQ=PFFRUsingconverseofbasicproportionalitytheoremEF∥QR

Q.10. Inthefigure,A,BandCarepointsonOP,OQandORrespectivelysuchthatAB∥PQandAC∥PR.ShowthatBC∥QR.

Solution:

In∆POQ

Since,AB∥PQ,Hence,OAAP=OBBQ …(i)[Usingbasicproportionalitytheorem]

In∆POR

Since,AC∥PRHence,OAAP=OCCR …(ii)[Usingbasicproportionalitytheorem]

Fromiand(ii)

OBBQ=OCCR

Hence,BC∥QR(Usingconverseofbasicproportionalitytheorem)

Q.11. UsingTheorem6.1,provethatalinedrawnthroughthemidpointofonesideofatriangleparalleltoanothersidebisectsthethirdside.(RecallthatyouhaveproveditinClassIX).

Solution:

LetinthegivenfigurePQisalinesegmentdrawnthroughmid-pointPoflineABsuchthatPQ∥BCHence,AP=PB

Now,usingbasicproportionalitytheoremAQQC=APPB⇒AQQC=APAP⇒AQQC=1⇒AQ=QCHence,Qisthemid-pointofAC.

Q.12. Provethatthelinejoiningthemid-pointsofanytwosidesofatriangleisparalleltothethirdside.

Solution:

LetinthegivenfigurePQisalinesegmentjoiningmid-pointsPandQoflineABandACrespectively.

Hence,AP=PBandAQ=QCNow,sinceAPPB=APAP=1andAQQC=AQAQ=1Hence,APPB=AQQCNow,usingconverseofbasicproportionalitytheorem,weget,PQ∥BC.

Q.13. ABCDisatrapeziuminwhichAB∥DCanditsdiagonalsintersecteachotheratthepointO.ShowthatAOBO=CODO

Solution:

LetalinesegmentEFisdrawnthroughpointOsuchthatEF∥CD

In∆ABCandin∆BDC,FO∥ABandFO∥CD∵EF∥CD,AB∥CD

So,usingbasicproportionalitytheoremBFFC=AOOC …(1)andBFFC=BOOD...(2)Now,fromequation1and(2),weget,AOOC=BOOD⇒AOBO=OCOD

Exercise6.3

Q.1. Statewhichpairsoftrianglesinthefigurearesimilar.Writethesimilaritycriterionusedbyyouforansweringthequestionandalsowritethepairsofsimilartrianglesinthesymbolicform:

Solution: Givenfiguresare

∠A=∠P=60°

∠B=∠Q=80°∠C=∠R=40°HencebyAAArule∆ABC~∆PQR.

Q.2. Arethepairsoftrianglesinthefiguresimilar?Writethesimilaritycriterionusedbyyouforansweringthequestionandalsowritethepairsofsimilartrianglesinthesymbolicform:

Solution:

ABQR=24=12

BCRP=2.55=12CAPQ=36=12Since,ABQR=BCRP=CAPQHence,bySSSrule.∆ABC~∆QRP.

Q.3. Statewhetherthefollowingpairoftrianglesissimilarornot.

Solution: Givenfigureis:

Weknowthattwotrianglesaresimilariftheyhave,alltheiranglesequalandcorrespondingsidesareinthesameratio.Here,thecorrespondingsidesarenotproportional.Hence,thegiventrianglesarenotsimilar.

Q.4. Stateifthefollowingpairsoftrianglesinthefigurearesimilarornot.Writethesimilaritycriterionusedbyyouforansweringthequestionandalsowritethepairsofsimilartrianglesinthesymbolicform:

Solution:

MNPQ=2.55=12MLQR=510=12∠M=∠Q=70oHence,bySASrule∆MNL~∆QPR

Q.5. Statewhetherthefollowingpairoftrianglesissimilarornot.

Solution: Givenfigureis:

Weknowthattwotrianglesaresaidtobesimilariftheircorrespondinganglesarecongruentandthecorrespondingsidesareinproportion.Inotherwords,similartrianglesarethesameshape,butnotnecessarilythesamesize.Here,asthecorrespondingsidesarenotinproportional.Hence,thegiventrianglesisnotsimilar.

Q.6. Statewhetherthepairtrianglesaresimilarornot.Writethesimilaritycriterionusedbyyouforansweringthequestionandalsowritethepairofsimilartrianglesinthesymbolicform.

Solution:

In∆DEF,∠D+∠E+∠F=180o(Sumofanglesofatriangleis180o.)⇒70o+80o+∠F=180o⇒∠F=30oSimilarlyin∆PQR,∠P+∠Q+∠R=180o(Sumofanglesofatriangleis180o.)⇒∠P+80o+30o=180o⇒∠P=70o

Now,In∆DEFand∆PQRsince,∠D=∠P=70o∠E=∠Q=80o∠F=∠R=30oHence,byAAArule∆DEF~∆PQR

Q.7. CDandGHarerespectivelythebisectorsof∠ACBand∠EGFsuchthatDandHlieonsidesABandFEof△ABCand△EFGrespectively.If△ABC~△FEG,showthatCDGH=ACFG.

Solution:

In△ACDand△FGH,∠A=∠F(∵△ABC~△EFG)∠ACD=∠FGH(anglebisector)∠ADC=∠FHG(remainingangle)Hence,byAAArulewehave:△ACD~△FGHSo,CDGH=ACFG(correspondingsidesareproportional)

Q.8. CDandGHarerespectivelythebisectorsof∠ACBand∠EGFsuchthatDandHlieonsidesABandFEof△ABCand△EFGrespectively.If△ABC~△FEG,showthat△DCB~△HGE.

Solution:

Since△ABC~△FEGHence,∠A=∠F∠B=∠E∠ACB=∠FGE⇒∠ACB2=∠FGE2And∠DCB=∠HGE(anglebisector)∠BDC=∠EHG(remainingAngle)Hence,byAAArulewehave:△DCB~△HGE

Q.9. CDandGHarerespectivelythebisectorsof∠ACBand∠EGFsuchthatDandHlieonsidesABandFEof△ABCand△EFGrespectively.If△ABC~△FEG,showthat△DCA~△HGF

Solution:

Since△ABC~△FEGHence,∠A=∠F∠B=∠E∠ACB=∠FGE⇒∠ACB2=∠FGE2⇒∠ACD=∠FGH(anglebisector)∠CDA=∠GHF(remainingangle)Hence,byAAArulewehave:△DCA~△HGF

Q.10. Inthefiguregivenbelow,EisapointonsideCBproducedofanisoscelestriangleABCwithAB=AC.IfAD⊥BCandEF⊥AC,provethat△ABD~△ECF.

Solution:

In△ABDand△ECF,Since,AB=AC(isoscelestriangles)So,∠ABD=∠ECF(anglesoppositetoequalsides)∠ADB=∠EFC=90°∠BAD=∠CEF(remainingangle)Hence,byAAArulewehave:△ABD~△ECF

Q.11. SidesABandBCandmedianADofatriangleABCarerespectivelyproportionaltosidesPQandQRandmedianPMof△PQRinthegivenfigure.Showthat△ABC~△PQR.

Solution:

Mediandividesoppositeside.So,BD=BC2andQM=QR2Giventhat,ABPQ=BCQR=ADPMSo,ABPQ=BDQM=ADPM(∵BC=2×BD,QR=2×QM).Hence,△ABD~△PQM.So,∠ABD=∠PQM=∠PQR(correspondinganglesofsimilartriangles)AndABPQ=BCQRHence,△ABC~△PQR.

Q.12. DisapointonthesideBCofatriangleABCsuchthat∠ADC=∠BAC.ShowthatCA2=CB.CD.

Solution:

In△ACDand△BACItisgiventhat∠ADC=∠BAC∠ACD=∠BCA(commonangle)∠CAD=∠CBA(remainingangle)Hence,byAAArulewehave:△ADC~△BACSo,bycorrespondingsidesofsimilartriangleswillbeproportionaltoeachother.CACB=CDCAHence,CA2=CB×CD.

Q.13. SidesABandACandmedianADofatriangleABCarerespectivelyproportionaltosidesPQandPRandmedianPMofanothertrianglePQR.Showthat∆ABC~∆PQR.

Solution:

Giventhat,ABPQ=ACPR=ADPM

LetusextendADandPMuptopointEandLrespectivelysuchthatAD=DEandPM=ML.NowjoinBtoE,CtoE,QtoLandRtoL.Weknowthatmediansdivideoppositesides.So,BD=DCandQM=MRAlso,AD=DE(byconstruction)AndPM=ML(Byconstruction)So,inquadrilateralABEC,diagonalsAEandBCbisectseachotheratpointD.Also,inquadrilateralPQLR,diagonalsPLandQRbisectseachotheratpointM.So,quadrilateralsABEDandPQLRareparallelograms.AC=BEandAB=EC(Sinceitisaparallelogram,oppositesideswillbeequal)AlsoPR=QLandPQ=LR(Sinceitisaparallelogram,oppositesideswillbeequal)

In∆ABEand∆PQL,

ABPQ=BEQL=AEPL(ACPR=BEQLandADPM=2AD2PM=AEPL)

Hence,bySSSrule,∆ABE~∆PQLSimilarly,∆AEC~∆PLRHence,∠BAE=∠QPLand∠EAC=∠LPRHence,∠BAC=∠QPRNow,in∆ABCand∆PQR,ABPQ=ACPRand∠BAC=∠QPRHence,bySASrule,∆ABC~∆PQR

Q.14. Averticalpoleoflength6mcastsashadow4mlongonthegroundandatthesametimeatowercastsashadow28mlong.Findtheheightofthetower.

Solution:

LetABbeatowerandCDbeapole

ShadowofABisBE.ShadowofCDisDF.Thesunraywillfallontowerandpoleatsameangle.∴∠DCF=∠BAEand∠DFC=∠BEA∠CDF=∠ABE=90o(Towerandpoleareverticaltoground)Hence,byAAArule,∆ABE~∆CDFThereforeABCD=BEDF⇒AB6=284⇒AB=42Hence,theheightofthetower=42meters.

Q.15. IfADandPMaremediansoftrianglesABCandPQR,respectivelywhere∆ABC~∆PQR,provethatABPQ=ADPM

Solution:

Since∆ABC~∆PQR

Thus,theirrespectivesideswillbeinproportionOr,ABPQ=ACPR=BCQR …(1)Also,∠A=∠P,∠B=∠Q,∠C=∠R…(2)Since,ADandPMaremedians,theywilldividetheiroppositesidesequally.Hence,BD=BC2and QM=QR2…(3)

Fromequation1and(3)

ABPQ=BDQM

∠B=∠Q(Fromequation2)Hence,bySASrule,∆ABD~∆PQMHence,ABPQ=ADPM(Correspondingsidesareproportional)

Q.16. Inthefollowingfigure,∆ODC~∆OBA,∠BOC=125∘and∠CDO=70∘.Find∠DOC,∠DCOand∠OAB.

Solution:

SinceDOBisastraightline

Hence,∠DOC+∠COB=180°linearpair⇒∠DOC=180°-125°=55°

In∆DOC,

∠DCO+∠CDO+∠DOC=180°(Anglesumproperty)

⇒∠DCO+70°+55°=180°.⇒∠DCO=55°Since,∆ODC~∆OBA.Thus,∠OCD=∠OABCorrespondinganglesequalinsimilartrianglesHence,∠OAB=55°.

Q.17. DiagonalsACandBDofatrapeziumABCDwithAB∥DCintersecteachotheratthepointO.Usingasimilaritycriterionfortwotriangles,showthatOAOC=OBOD

Solution:

In∆DOCand∆BOA

AB||CD

Hence,∠CDO=∠ABO[Alternateinteriorangles]∠DCO=∠BAO[Alternateinteriorangles]∠DOC=∠BOA[Verticallyoppositeangles]Hence,∆DOC~∆BOAUsingAAArule⇒DOBO=OCOA[Correspondingsidesareproportional]⇒OAOC=OBOD

Q.18. InthefiguregivenbelowQRQS=QTPRand∠1=∠2.Showthat△PQS~△TQR.

Solution:

In△PQR,

∠PQR=∠PRQHence,PQ=PR…(i)QRQS=QTPR....(given)Using(i),weget:QRQS=QTPQ…(ii)Also,∠RQT=∠PQS=∠1.Hence,bySASrule∆PQS~∆TQR.

Q.19. SandTarepointsonsidesPRandQRof∆PQRsuchthat∠P=∠RTS.Showthat∆RPQ~∆RTS.

Solution:

In∆RPQand∆RTS

∠QPR=∠RTS[Given]∠R=∠R[Commonangle]∠RQP=∠RST[Remainingangle]Hence,∆RPQ~∆RTS[byAAArule]

Q.20. InFig.if∆ABE≅∆ACD,showthat∆ADE~∆ABC.

Solution:

Since∆ABE≅∆ACD

Therefore,AB=AC…(1)⇒AD=AE…(2)Now,in∆ADEand∆ABC,Dividingequation2by(1)ADAB=AEAC ∠A=∠A[Commonangle]Hence,∆ADE~∆ABC[bySASrule]

Q.21. InthefollowingfigurealtitudesADandCEof∆ABCintersecteachotheratthepointP.Showthat∆AEP~∆CDP.

Solution:

In∆AEPand∆CDP∠CDP=∠AEP=90°∠CPD=∠APE(verticallyoppositeangles)∠PCD=∠PAE(remainingangle)Hence,byAAArulewehave:∆AEP~∆CDP

Q.22. Inthefigure,altitudesADandCEof∆ABCintersecteachotheratthepointP.Showthat∆ABD~∆CBE.

Solution:

In∆ABDand∆CBE∠ADB=∠CEB=90o∠ABD=∠CBE(Commonangle)∠DAB=∠ECB(Remainingangle)Hence,byAAArule,∆ABD~∆CBE

Q.23. InFig.altitudesADandCEof∆ABCintersecteachotheratthepointP.Showthat∆AEP~∆ADB.

Solution:

In∆AEPand∆ADB∠AEP=∠ADB=90o∠PAE=∠DAB(Commonangle)∠APE=∠ABD(Remainingangle)Hence,byAAArule,∆AEP~∆ADB

Q.24. Inthefigure,altitudesADandCEof∆ABCintersecteachotheratthepointP.Showthat∆PDC~∆BEC.

Solution:

In∆PDCand∆BEC∠PDC=∠BEC=90o∠BCE=∠PCD(Commonangle)∠CPD=∠CBE(Remainingangle)Hence,byAAArule,∆PDC~∆BEC

Q.25. EisapointonthesideADproducedofaparallelogramABCDandBEintersectsCDatF.Showthat△ABE~△CFB.

Solution:

In△ABEand△CFB,∠A=∠C(Oppositeanglesofparallelogram)∠AEB=∠CBF(alternateinterioranglesasAE∥BC)∠ABE=∠CFB(alternateinterioranglesasAB∥DC)Hence,byAAArulewehave:△ABE~△CFB

Q.26. Inthefigure,ABCandAMParetworighttriangles,right-angledatBandMrespectively.Provethat∆ABC~∆AMP

Solution:

In∆ABCand∆AMP

∠ABC=∠AMP=90o∠A=∠A(Commonangle)∠ACB=∠APM(Remainingangle)Hence,byAAArule,∆ABC~∆AMP

Q.27. Inthefigure,ABCandAMParetworighttriangles,right-angledatBandMrespectively.Provethat:CAPA=BCMP

Solution: Givenfigureis,

In∆ABCand∆AMP∠ABC=∠AMP=90o∠A=∠A(Commonangle)∠ACB=∠APM(Remainingangle)Hence,byAAArule,∆ABC~∆AMPHence, CAPA=BCMP (Corresponding sides areproportional)

Exercise6.4

Q.1. Let△ABC~△DEFandtheirareasbe64cm2and121cm2respectively.IfEF=15.4cm,findBC.

Solution: Given,△ABC~△DEF

Wehave,

area(△ABC)area(△DEF)=ABDE2=BCEF2=ACDF2SinceEF=15.4,area△ABC=64,area△DEF=121.Hence,64121=BC215.42⇒BC15.4=811⇒BC=8×15.411=8×1.4=11.2cm.Thus,BC=11.2cm.

Q.2. DiagonalsofatrapeziumABCDwithAB∥DCintersecteachotheratthepointO.IfAB=2CD,findtheratiooftheareasoftrianglesAOBandCOD.

Solution:

SinceAB∥CD,

∠OAB=∠OCD(Alternateinteriorangles)∠OBA=∠ODC(Alternateinteriorangles)∠AOB=∠COD(Verticallyoppositeangles)Hence,byAAArule,∆AOB~∆COD⇒area(∆AOB)area(∆COD)=ABCD2SinceAB=2CD,area(∆AOB)area(∆COD)=41=4:1Hence,theratiooftheareasoftrianglesAOBandCODis4:1.

Q.3. Inthefigure,ABCandDBCaretwotrianglesonthesamebaseBC.IfADintersectsBCatO,showthatarABCarDBC=AODO.

Solution: Weknowthattheareaofatriangle=12×Base×heightSince∆ABCand∆DBCareonsamebase,

Hence,ratiooftheirareaswillbesameasratiooftheirheights.

LetusdrawtwoperpendicularsAPandDMonBC.

In∆APO and ∆DMO

∠APO=∠DMO=90o,∠AOP=∠DOM(Verticallyoppositeangles)

∠OAP=∠ODM(Remainingangle)

Hence,byAAArule

∆APO ~ ∆DMO

Hence,APDM=AODOHence,area(∆ABC)area(∆DBC)=APDM=AODO

Q.4. Iftheareasoftwosimilartrianglesareequal,provethattheyarecongruent.

Solution:

Letusassumethat△ABC ~△PQR.Now,area(△ABC)area(△PQR)=ABPQ2=BCQR2=ACPR2Since,area△ABC=area(△PQR)Hence,AB=PQBC=QRAC=PRSince,correspondingsidesoftwosimilartrianglesareofsamelength.Hence,△ABC≅△PQR(bySSSrule)

Q.5. D,EandFarerespectivelythemid-pointsofsidesAB,BCandCAof∆ABC.Findtheratiooftheareasof∆DEFand∆ABC.

Solution:

SinceDandEaremid-pointsofABandBCof∆ABCHence,DE∥ACandDE=12AC(bymid-pointtheorem)Similarly,EF=12ABandDF=12BCNowin∆ABCand∆EFDABEF=BCFD=CADE=2

Therefore,bySSSrule,∆ABC~∆EFD

Hence,area(∆ABC)area(∆DEF)=ACDE2=4⇒area(∆DEF)area(∆ABC)=14=1:4

Q.6. Provethattheratiooftheareasoftwosimilartrianglesisequaltothesquareoftheratiooftheircorrespondingmedians.

Solution:

Letusassumethat∆ABC~∆PQR.LetADandPSbethemediansofthesetriangles.

So,ABPQ=BCQR=ACPR…1

∠A=∠P,∠B=∠Q,∠C=∠R

Since,ADandPSaremedians

So,BD=DC=BC2andQS=SR=QR2So,equation1becomesABPQ=BDQS=ACPRNowin∆ABDand∆PQS∠B=∠Qand,ABPQ=BDQS

Hence,∆ABD~∆PQS

Hence,ABPQ=BDQS=ADPS…2

Since,area(∆ABC)area∆PQR=ABPQ2⇒area∆ABCarea∆PQR=ADPS2[fromequation(2)]

Q.7. Provethattheareaofanequilateraltriangledescribedononesideofasquareisequaltohalftheareaoftheequilateraltriangledescribedononeofitsdiagonals.

Solution:

LetABCDbeasquareofsidea.Therefore,it'sdiagonal=2a.

Let∆ABEand∆DBFaretwoequilateraltriangles.Hence,AB=AE=BE=aandDB=DF=BF=2a.Weknowthatallanglesofequilateraltrianglesare60o.

Hence,allequilateraltrianglesaresimilartoeachother.

Hence,ratioofareasofthesetriangleswillbeequaltothesquareoftheratiobetweensidesofthesetriangles.

areaof∆ABEareaof∆DBF=a2a2=12Hence,areaof∆ABE=12(areaof∆DBF).

Q.8. ABCandBDEaretwoequilateraltrianglessuchthatDisthemidpointofBC.RatiooftheareasoftrianglesABCandBDEis2:11:24:1

Solution:

Since,allangleofequilateraltrianglesare60o,allequilateraltrianglesaresimilartoeachother.

Therefore,ratiobetweenareasofthesetriangleswillbeequaltothesquareoftheratiobetweensidesofthesetriangles.

Letsideof∆ABC=aTherefore,sideof∆BDE=a2Hence,area∆ABCarea∆BDE=aa22=41=4:1

Q.9. Sidesoftwosimilartrianglesareintheratio4:9.Areasofthesetrianglesareintheratio2:34:981:1616:81

Solution: Weknowthat,

Iftwotrianglesaresimilartoeachother,theratiobetweenareasofthesetriangleswillbeequaltothesquareoftheratiobetweensidesofthesetriangles.

Giventhatsidesareintheratio4:9.Hence,ratiobetweenareasofthesetriangles=492=1681=16:81.

Exercise6.5

Q.1. Sidesofatrianglearegivenbelow.Determineifitisarighttriangle.Incaseofarighttriangle,writethelengthofitshypotenuse.7cm,24cm,25cm

Solution: Giventhatsidesare7cm,24cmand25cm.

Squaringthelengthsofthesesidesweget49,576and625.

Clearly,49+576=625or72+242=252.ThegiventrianglesatisfiesPythagorastheorem.So,itisarighttriangle.Weknowthatthelongestsideinarighttriangleisthehypotenuse.Hence,lengthofhypotenuse=25cm.

Q.2. Sidesoftrianglesaregivenbelow.Determinewhichofthemarerighttriangles.Incaseofarighttriangle,writethelengthofitshypotenuse.3cm,8cm,6cm

Here,64≠36+9

Clearly,sumofsquaresoflengthsoftwosidesisnotequaltosquareoflengthofthirdside.Hence,giventriangleisnotsatisfyingPythagorastheorem.So,itisnotarighttriangle.

Q.3. Sidesoftrianglesaregivenbelow.Determinewhichofthemarerighttriangles.Incaseofarighttriangle,writethelengthofitshypotenuse.50cm,80cm,100cm

Here,10000≠6400+2500

Clearly,sumofsquaresoflengthsoftwosidesisnotequaltosquareoflengthofthirdside.Therefore,giventriangleisnotsatisfyingPythagorastheorem.So,itisnotarighttriangle.

Q.4. Thesidesofatrianglearegivenbelow.Determinewhetheritisarighttriangle.Incaseofarighttriangle,writethelengthofitshypotenuse.13cm,12cm,5cm.

Squaringthelengthsofthesesideswemayget169,144and25.

Weknowthat,144+25=169or122+52=132.So,byconverseofPythagorastheorem,itisarighttriangle.Asweknowthatthelongestsideinarighttriangleisthehypotenuse.Hence,lengthofhypotenuse=13cm.

Q.5. Aguywireattachedtoaverticalpoleofheight18mis24mlongandhasastakeattachedtotheotherend.Howfarfromthebaseofthepoleshouldthestakebedrivensothatthewirewillbetaut?

Solution:

LetOBbethepoleandABbethewire.Therefore,byPythagorastheoremwehave:AB2=OB2+OA2⇒242=182+OA2⇒OA2=576-324⇒OA=252=6×6×7=67Therefore,distancefrombase=67m

Q.6. Anaeroplaneleavesanairportandfliesduenorthataspeedof1000kmperhour.Atthesametime,anotheraeroplaneleavesthesameairportandfliesduewestataspeedof1200kmperhour.Howfarapartwillbethetwoplanesafter112hours?

Solution:

Distancetraveledbytheplaneflyingtowardsnorthin112hrs

=1,000×112=1,500km

Distancetraveledbytheplaneflyingtowardswestin112hrs=1,200×112=1,800km

LetthesedistancesarerepresentedbyOAandOBrespectively.

NowapplyingPythagorastheorem

Distancebetweentheseplanesafter112hrs,AB=OA2+OB2=1,5002+1,8002=2250000+3240000=5490000=9×610000=30061So,distancebetweentheseplaneswillbe30061km.after112hrs.

Q.7. Twopolesofheights6mand11mstandonaplaneground.Ifthedistancebetweenthefeetofthepolesis12m,findthedistancebetweentheirtops.

Solution:

LetCDandABbethepolesofheight11mand6m.Therefore,CP=11-6=5mFromthefigurewemayobservethatAP=12mIn△APC,byapplyingPythagorastheoremweget:AP2+PC2=AC2⇒122+52=AC2⇒AC2=144+25=169⇒AC=13Therefore,thedistancebetweentheirtops=13m.

Q.8. DandEarepointsonthesidesCAandCBrespectivelyofatriangleABCright-angledatC.ProvethatAE2+BD2=AB2+DE2.

Solution:

In△ACE,AC2+CE2=AE2…iIn△BCD,BC2+CD2=BD2…iiAddingiand(ii)weget:AC2+CE2+BC2+CD2=AE2+BD2…(iii)⇒CD2+CE2+AC2+BC2=AE2+BD2

In△CDE,DE2=CD2+CE2In△ABC,AB2=AC2+CB2Addingbothequationsandcomparingwithequation(iii),weget:DE2+AB2=AE2+BD2

Q.9. TheperpendicularfromAonsideBCofa△ABCintersectsBCatDsuchthatDB=3CD.Provethat2AB2=2AC2+BC2.

Solution: Giventhat,3DC=DB.DC=BC4DB:DC=3:1…1andDB=3BC4…2In△ACD,AC2=AD2+DC2AD2=AC2-DC2…3

In△ABD,AB2=AD2+DB2AD2=AB2-DB2…4Fromequation(3)and4AC2-DC2=AB2-DB2Since,giventhat3DC=DBAC2-BC42=AB2-3BC42(from1and2)⇒AC2-BC216=AB2-9BC216⇒16AC2-BC2=16AB2-9BC2⇒16AB2-16AC2=8BC2⇒2AB2=2AC2+BC2

Q.10. InanequilateraltriangleABC,DisapointonsideBCsuchthatBD=13BC.Provethat9AD2=7AB2.

Solution:

LetsideofequilateraltrianglebeaandAEbethealtitudeof∆ABC

So,BE=EC=BC2=a2

and,AE=a32GiventhatBD=13BC=a3So,DE=BE-BD=a2-a3=a6Now,in∆ADE,byapplyingPythagorastheoremAD2=AE2+DE2⇒AD2=a322+a62=3a24+a236=28a236or,9AD2=7AB2

Q.11. Inanequilateraltriangle,provethatthreetimesthesquareofonesideisequaltofourtimesthesquareofoneofitsaltitudes.

Solution:

Letsideofequilateraltrianglebea.AndAEbethealtitudeof∆ABC

So,BE=EC=BC2=a2Nowin∆ABEbyapplyingPythagorastheoremAB2=AE2+BE2⇒a2=AE2+a22⇒AE2=a2-a24⇒AE2=3a24⇒4AE2=3a2or,4AE2=3×squareofoneside.

Q.12. In∆ABC,AB=63cm,AC=12cmandBC=6cm.TheangleBis120o60o90o

Solution:

GiventhatAB=63cm,AC=12cmandBC=6cm

Wemayobservethat

AB2=108,AC2=144andBC2=36,AB2+BC2=AC2Thus,thegiven∆ABCissatisfyingPythagorastheorem.Therefore,thetriangleisarightangletriangleright-angledatBTherefore,∠B=90°.

Q.13. PQRisatriangleright-angledatPandMisapointonQRsuchthatPM⊥QR.ShowthatPM2=QM.MR.

Solution:

Let∠MPR=x

In∆MPR∠MRP=180o-90o-x⇒∠MRP=90o-xSimilarlyin∆MPQ∠MPQ=90o-∠MPR=90o-x∠MQP=180o-90o-90o-x⇒∠MQP=x

Nowin∆MPQand∆MRP,wemayobservethat

∠MPQ=∠MRP

∠PMQ=∠RMP∠MQP=∠MPRHence,byAAArule,∆MPQ~∆MRPHence,QMPM=MPMR⇒PM2=QM.MR

Q.14. Inthefiguregivenbelow,ABDisatriangleright-angledatAandAC⊥BD.ShowthatAB2=BC·BD.

Solution:

In△ABCand△ABD,∠CBA=∠DBA(commonangles)∠BCA=∠BAD=90°∠BAC=∠BDA(remainingangle)Therefore,△ABC~△ABD(byAAA)∴ABBD=BCAB⇒AB2=BC·BD

Q.15. Inthefigure,ABDisatriangleright-angledatAandAC⊥BD.ShowthatAC2=BC·DC

Solution: Let∠CAB=x

In∆CBA

∠CBA=180o-90o-x∠CBA=90o-xSimilarlyin∆CAD∠CAD=90o-∠CAB=90o-x∠CDA=180o-90o-(90o-x)∠CDA=x.

Nowin∆CBAand∆CAD,wemayobservethat

∠CBA=∠CAD

∠CAB=∠CDA∠ACB=∠DCA=90oTherefore∆CBA~∆CAD(byAAArule)Therefore,ACDC=BCAC⇒AC2=DC×BC.

Q.16. InFig.ABDisatriangleright-angledatAandAC⊥BD.ShowthatAD2=BD·CD

Solution: In∆DCA&∆DAB∠DCA=∠DAB=90°∠CDA=∠ADB(Commonangle)∠DAC=∠DBA(remainingangle)∆DCA~∆DAB(byAAAproperty)Therefore,DCDA=DADB⇒AD2=BD×CD

Q.17. ABCisanisoscelestriangleright-angledatC.ProvethatAB2=2AC2.

Solution:

Giventhat∆ABCisanisoscelestriangle.

Therefore,AC=CBApplyingPythagorastheoremin∆ABC(i.e.right-angledatpointC)AC2+CB2=AB2⇒2AC2=AB2(asAC=CB)

Q.18. ABCisanisoscelestrianglewithAC=BC.IfAB2=2AC2,provethatABCisarighttriangle.

Solution:

GiventhatAB2=2AC2

⇒AB2=AC2+AC2⇒AB2=AC2+BC2(asAC=BC)Therefore,byconverseofPythagorastheorem,giventriangleisaright-angledtriangle.

Q.19. ABCisanequilateraltriangleofside2a.Findeachofitsaltitudes.

Solution:

LetADbethealtitudeingivenequilateral∆ABC.

Weknowthataltitudebisectstheoppositeside. So,BD=DC=ain∆ADB∠ADB=90o

NowapplyingPythagorastheorem

AD2+BD2=AB2

⇒AD2+a2=2a2⇒AD2+a2=4a2⇒AD2=3a2⇒AD=a3Sinceinanequilateraltriangle,allthealtitudesareequalinlength.So,lengthofeachaltitudewillbe3a

Q.20. Provethatthesumofthesquaresofthesidesofarhombusisequaltothesumofthesquaresofitsdiagonals.

Solution:

In∆AOB,∆BOC,∆COD,∆AODApplyingPythagorastheoremAB2=AO2+OB2BC2=BO2+OC2CD2=CO2+OD2AD2=AO2+OD2

Addingalltheseequations,AB2+BC2+CD2+AD2=2AO2+OB2+OC2+OD2=2AC22+BD22+AC22+BD22(diagonalsbisecteachother.)=2AC22+BD22=AC2+BD2

Q.21. InFig.6.54,OisapointintheinteriorofatriangleABC,OD⊥BC,OE⊥ACandOF⊥AB.ShowthatOA2+OB2+OC2-OD2-OE2-OF2=AF2+BD2+CE2,

Solution:

In∆AOFApplyingPythagorastheoremOA2=OF2+AF2Similarlyin∆BODOB2=OD2+BD2similarlyin∆COEOC2=OE2+EC2AddingtheseequationsOA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2⇒OA2+OB2+OC2-OD2-OE2-OF2=AF2+BD2+EC2

Q.22. InFig.OisapointintheinteriorofatriangleABC,OD⊥BC,OE⊥ACandOF⊥AB.Showthat

AF2+BD2+CE2=AE2+CD2+BF2.

Solution:

In∆AOFApplyingPythagorastheoremOA2=OF2+AF2Similarlyin∆BODOB2=OD2+BD2similarlyin∆COEOC2=OE2+EC2AddingtheseequationsOA2+OB2+OC2=OF2+AF2+OD2+BD2+OE2+EC2⇒OA2+OB2+OC2-OD2-OE2-OF2=AF2+BD2+EC2

FromaboveresultAF2+BD2+EC2=OA2-OE2+OC2-OD2+OB2-OF2Therefore,AF2+BD2+EC2=AE2+CD2+BF2

Q.23. Aladder10mlongreachesawindow8mabovetheground.Findthedistanceofthefootoftheladderfrombaseofthewall.

Solution:

LetOAbethewallandABbetheladder.

ThereforebyPythagorastheorem,AB2=OA2+BO2⇒102=82+OB2⇒100=64+OB2⇒OB2=36⇒OB=6

Therefore,distanceoffootofladderfromofthewall=6m

Exercise6.6

Q.1. Inthegivenfigure,PSisthebisectorof∠QPRof∆PQR.ProvethatQSSR=PQPR.

Solution:

Giventhat,PSisanglebisectorof∠QPR.

ConstructalineRTparalleltoSPwhichmeetsQPproducedatT.∠QPS=∠SPR......(1)∠SPR=∠PRT(AsPS||TR,alternateinteriorangles)....(2)∠QPS=∠QTR(AsPS||TR,correspondingangles).....(3)Usingtheseequations,wemayfind∠PRT=∠QTRfrom(2)and(3)So,PT=PR(SinceΔPTRisisoscelestriangle)

NowinΔQPSandΔQTR,∠QSP=∠QRT(AsPS||TR)

∠QPS=∠QTR(AsPS||TR)

∠Qiscommon.ΔQPS~ΔQTR(byAAAproperty)So,QRQS=QTQP⇒QRQS-1=QTQP-1⇒SRQS=PTQP⇒QSSR=QPPT⇒QSSR=PQPR

Q.2. Nazimaisflyfishinginastream.Thetipofherfishingrodis1.8mabovethesurfaceofthewaterandtheflyattheendofthestringrestsonthewater3.6mawayand2.4mfromapointdirectlyunderthetipoftherod.Assumingthatherstring(fromthetipofherrodtothefly)istaut,howmuchstringdoesshehaveout(seethegivenfigure)?Ifshepullsinthestringattherateof5cmpersecond,whatwillbethehorizontaldistanceoftheflyfromherafter12seconds?

Solution:

LetABbetheheightoftipoffishingrodfromwatersurfaceandBCbethehorizontaldistanceofflyfromthetipoffishingrod.

Then,ACisthelengthofstring.ACcanbefoundbyapplyingPythagorastheoreminΔABC.AC2=AB2+BC2AC2=1.82+2.42AC2=3.24+5.76AC2=9.00Thus,lengthofstringoutis3m.

Now,shepullsstringatrateof5cmpersecond.

So,stringpulledin12second=12×5=60cm=0.6m.

After12seconds,letusassumetheflytobeatpointD.

Lengthofstringoutafter12secondisAD.AD=AC-stringpulledbyNazimain12second=3.00-0.6=2.4mInΔADB,AB2+BD2=AD2⇒1.82+BD2=2.42⇒BD2=5.76-3.24=2.52⇒BD=1.587mHorizontaldistanceoffly=BD+1.2=1.587+1.2=2.787=2.79m

Q.3. Inthefigure,DisapointonhypotenuseACof∆ABC,suchthatBD⊥AC,DM⊥BCandDN⊥AB.Provethat:DM2=DN.MC

Solution: LetusjoinDB.

DN||CB,DM||AB

Therefore,DNBMisaparallelogram.

Since,∠Bis90°,therefore,DNBMisarectangle.Hence,DN=MB,DM=NBand∠CDB=∠ADB=90°∠2+∠3=90o…(1)InΔCDM∠1+∠2+∠DMC=180°∠1+∠2=90°…(2)InΔDMB∠3+∠DMB+∠4=180°∠3+∠4=90°…(3)

Fromequation(1)and(2)

∠1=∠3

Fromequation(1)and(3)∠2=∠4So,ΔBDM~ΔDCMBMDM=DMMC⇒DNDM=DMMC⇒DM2=DN.MCHence,proved.

Q.4. Inthefigure,DisapointonhypotenuseACof∆ABC,suchthatBD⊥AC,DM⊥BCandDN⊥AB.Provethat:DN2=DM·AN

Solution: LetusjoinDB.

DN||CB,DM||AB

Therefore,DNBMisaparallelogram.

Since,∠Bis90°.Therefore,DNBMisarectangle.So,DN=MB,DM=NBand∠CDB=∠ADB=90°∠4+∠5=90o…(1)InΔADN∠5+∠6=90°…(2)

Fromequation(1)and(2)

∠4=∠6

and∠DNA=∠DNB=90°So,ΔADN~ΔBDNDNBN=ANDN⇒DNDM=ANDN(AsBN=DM)⇒DN2=DM×ANHence,proved.

Q.5. ABCisatriangleinwhich∠ABC>90oandAD⊥CBproduced.ProvethatAC2=AB2+BC2+2BC·BD.

Solution:

InΔADB,applyingPythagorastheoremAB2=AD2+DB2…(1)

InΔACD,applyingPythagorastheoremAC2=AD2+DC2

⇒AC2=AD2+DB+BC2

⇒AC2=AD2+DB2+BC2+2DB×BC

Nowusingequation(1)AC2=AB2+BC2+2BC.BD

Q.6. InthefiguregivenbelowABCisatriangleinwhich∠ABC<90oandAD⊥BC.ProvethatAC2=AB2+BC2-2BC.BD.

Solution:

In△ADB,applyingPythagorastheoremweget:AD2+DB2=AB2⇒AD2=AB2-DB2…(1)In△ADC,applyingPythagorastheoremweget:AD2+DC2=AC2....(2)

Nowusingequation(1),weget:AB2-BD2+DC2=AC2⇒AB2-BD2+BC-BD2=AC2⇒AC2=AB2-BD2+BC2+BD2-2BC.BDHence,AC2=AB2+BC2-2BC.BD.

Q.7. Inthefigure,ADisamedianofatriangleABCandAM⊥BC.Provethat:AC2=AD2+BC·DM+BC22

Solution:

InΔAMD,byusingPythagorastheorem,

AM2+MD2=AD2…(1)InΔAMCAM2+MC2=AC2…(2)⇒AM2+MD+DC2=AC2⇒AM2+MD2+DC2+2MD.DC=AC2Usingequation(1)weget,AD2+DC2+2MD.DC=AC2

Nowusingtheresult,DC=BC2

AD2+BC22+2MD.BC2=AC2

⇒AD2+BC22+MD×BC=AC2Hence,AC2=AD2+BC·DM+BC22

Q.8. Inthegivenfigure,ADisamedianofatriangleABCandAM⊥BC.ProvethatAB2=AD2-BC·DM+BC22

Solution:

InΔABM,applyingPythagorastheoremAB2=AM2+MB2

=AD2-DM2+MB2=AD2-DM2+BD-MD2=AD2-DM2+BD2+MD2-2BD.MD=AD2+BD2-2BD.MD=AD2+BC22-2BC2×MDAB2=AD2+BC22-BC×MDHence,AB2=AD2-BC·DM+BC22

Q.9. Inthefigure,ADisthemedianoftriangleABCandAM⊥BC.Provethat:AC2+AB2=2AD2+12BC2

Solution: InΔAMB,byPythagorastheorem,

AM2+MB2=AB2…(1)

InΔAMCAM2+MC2=AC2…(2)

Addingequations(1)and(2)2AM2+MB2+MC2=AB2+AC2

⇒2AM2+BD-DM2+MD+DC2=AB2+AC2

⇒2AM2+BD2+DM2-2BD.DM+MD2+DC2+2MD.DC=AB2+AC2⇒2AM2+2MD2+BD2+DC2+2MD-BD+DC=AB2+AC2⇒2AM2+MD2+BC22+BC22+2MD-BC2+BC2=AB2+AC2⇒2AD2+BC22=AB2+AC2Hence,AC2+AB2=2AD2+12BC2

Q.10. Provethatthesumofthesquaresofthediagonalsofparallelogramisequaltothesumofthesquaresofitssides.

Solution:

LetABCDbeaparallelogram

LetusdrawperpendicularDEonextendedsideBAandAFonsideDC.InΔDEADE2+EA2=DA2…iInΔDEBDE2+EB2=DB2⇒DE2+EA+AB2=DB2⇒DE2+EA2+AB2+2EA.AB=DB2⇒DA2+AB2+2EA.AB=DB2…(ii)

InΔADF

AD2=AF2+FD2

InΔAFCAC2=AF2+FC2=AF2+DC-FD2=AF2+DC2+FD2-2DC.FD=AF2+FD2+DC2-2DC.FD⇒AC2=AD2+DC2-2DC.FD…(iii)

SinceABCDisaparallelogram

AB=CDandBC=AD

InΔDEAandΔADF∠DEA=∠AFD∠EAD=∠FDAEA∥DF∠EDA=∠FADAF∥EDADiscommoninbothtriangles.Since,respectiveanglesaresameandrespectivesidesaresameΔDEA≅ΔAFDSo,EA=DF

Addingequation(ii)and(iii)

DA2+AB2+2EA.AB+AD2+DC2-2DC.FD=DB2+AC2

⇒DA2+AB2+AD2+DC2+2EA.AB-2DC.FD=DB2+AC2⇒BC2+AB2+AD2+DC2+2EA.AB-2AB.EA=DB2+AC2⇒AB2+BC2+CD2+DA2=AC2+BD2

Q.11. InFig.6.61,twochordsABandCDintersecteachotheratthepointP.Provethat△APC~△DPB

Solution:

In△APCand△DPB

∠A=∠Dand∠C=∠B(Angleonsamesegment)

Therefore,△APC~△DPB(AAcriteria)

Q.12. InfiguretwochordsABandCDintersecteachotheratthepointP.ProvethatAP·PB=CP·DP.

Solution:

In△APCand△DPB

∠A=∠Dand∠C=∠B(Angleonsamesegment)

Therefore,△APC~△DPB(AAcriteria)Weknowthatcorrespondingsidesofsimilartrianglesareproportional∴APDP=PCPB=CABD⇒APDP=PCPB∴AP.PB=PC.DP

Q.13. Inthefigure,twochordsABandCDofacircleintersecteachotheratthepointP(whenproduced)outsidethecircle.Provethat△PAC~△PDB.

Solution:

In△PACand△PDB

∠APC=∠DPB(Commonangle)∠ACP=∠DBP(Exteriorangleofacyclicquadrilateralisequaltotheoppositeinteriorangle.)Therefore,△PAC~△PDB(AAcriteria)

Q.14. Inthefigure,twochordsABandCDofacircleintersecteachotheratthepointP(whenproduced)outsidethecircle.ProvethatPA.PB=PC.PD

Solution:

In△APCand△DPB

∠APC=∠DPB(Commonangle)∠ACP=∠DBP(Exterioranglesofcyclicquadrilateral)Therefore,△APC~△DPB(AAcriteria)Weknowthatcorrespondingsidesofsimilartrianglesareproportional.PAPD=ACDB=PCPB⇒PAPD=PCPB∴PA.PB=PC.PD

Q.15. DisapointonsideBCof∆ABCsuchthatBDCD=ABAC.ProvethatADisthebisectorof∠BAC.

Solution:

ConstructalineCEparalleltoDAwhichmeetsBAproducedatE.

Therefore,∠BAD=∠BEC(Correspondingangles)......(1)∠DAC=∠ACE(Alternateangles)......(2)InΔDBAandΔCBE,BDCD=ABAC(Given)......(3)BDCD=BAAE(Basicproportionalitytheorem)......(4)From(3)and(4),AE=ACTherefore,∠ACE=∠BEC......(5)So,from(1),(2)and(5)⇒∠BAD=∠DACTherefore,ADisanglebisectorof∠BAC.

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