Chapter6SquareRootsandCubeRootsSquareRootsandCubeRoots

6.0Introduction

Letusmakesquareshapesusingunitsquares.Observethenumberofunitsquaresused.Aunitsquareisasquarewhosesideis1unit

S.No.FigureLengthofthesideinunitsNo.ofunitsquaresused111224339Similarlymakenexttwosquares.Canyouguesshowmanytotalunitsquaresarerequiredformakingasquarewhosesideis6units?Fromtheaboveobservationswecouldmakesquareshapeswith1,4,9,16,25…unitsquares.Thenumbers1,4,9,16,25,…canbeexpressedas

1=1×1=12

4=2×2=22

9=3×3=32

16=4×4=42

25=……×……=……..36=…….×……=……..Observethepatternoffactorsineachcase………………………………..………………………………...m=n×n=n2wherem,nareintegers.

Youmighthaveobservedinthegivenpatternthatthenumbersareexpressedastheproductoftwoequalfactors.Suchnumbersarecalledperfectsquares.Observethefollowingperfectsquarenumbers

Ex:(i)9=3 3(ii)49=7 7(iii)1.44=1.2 1.2

(iv)2.25=1.5 1.5(v) (vi)In case (i) and (ii)wehave noticed the perfect square numbers 9 and49 are integers.Thegeneral formof suchperfect squarenumbersism=n n(wheremandnareintegers).Incase(iii),(iv)and(v),(vi)theperfectsquarenumbersarenotintegers.Hence,theyarenotsquarenumbers.Ifaninteger‘m’isexpressedasn2wherenisanintegerthen‘m’isasquarenumberor‘m’isasquareof‘n’.Perfectsquare:Arationalnumberthatisequaltothesquareofanotherrationalnumber.Squarenumber:Anintegerthatisasquareofanotherinteger.Thus“Allsquarenumbersareperfectsquares”butallperfectsquaresmaynotbesquarenumbers.Ex:2.25isaperfectsquarenumberbecauseitcanbeexpressedas2.25=(1.5)2=1.5 1.5,itisnotsquareofaninteger.Therefore,itisnotasquarenumber.Is42asquarenumber?Weknowthat62=36and72=49,if42isasquarenumberitmustbethesquareofaninteger.Whichshouldbebetween6and7.Butthereisnosuchintegerbetween6and7.Therefore42isnotasquarenumber.Observetheperfectsquaresinthegiventable

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100

Arethereanyothersquarenumbersthatexistotherthanthenumbersshowninthetable.

DoThis:1.Findtheperfectsquaresbetween(i)100and150(ii)150and2002.Is56aperfectsquare?Givereasons?

6.1Propertiesofsquarenumbers:Observeandfillthefollowingtable.NumberSquareNumberSquareNumberSquare

111112121441241214422.........3913........2352941614196........57652515225256256.........16..............................74917289....................86418324...........................8119361....................1010020400....................

Observethedigitsintheunitsplaceofthesquarenumbersintheabovetable.Doyouobserveallthesenumbersendwith0,1,4,5,6or9atunitsplace,noneoftheseendwith2,3,7or8atunitsplace.“Thatisthenumbersthathave2,3,7or8intheunitsplacearenotperfectsquares.”Canwesaythatallnumbersendwith0,1,4,5,6or9,atunitplacearesquarenumbers?Thinkaboutit.

TryThese:

1.Guessandgivereasonwhichofthefollowingnumbersareperfectsquares.Verifyfromtheabovetable.(i)84(ii)108(iii)271(iv)240(v)529

Writethesquaresof1,9,11,19,21Haveyounoticedanyrelationshipbetweentheunitsdigitofnumbersandtheirsquares?Itisobservedthatifanumberhas1or9intheunitsplace,thentheunitsdigitinitssquarenumberisonly1.Ifanumberhas4or6intheunitsplace,thentheunitsdigitinitssquareisalways6Similarly,exploretheunitsdigitofsquaresofnumbersendingwith0,2,3,5,7and8.

TryThese:

1.Whichofthefollowinghaveoneinitsunitsplace?(i)1262(ii)1792(iii)2812(iv)3632

2.Whichofthefollowinghave6intheunitsplace?(i)1162(ii)2282(iii)3242(iv)3632

Think,DiscussandWrite:

Vaishnaviclaimsthatthesquareofevennumbersareevenandthatofoddareodd.Doyouagreewithher?Justify.Observeandcompletethetable:NumbersNo.ofdigitsinitssquare

(Minimum)(Maximum)1-91210-99…….4100-9995.......1009-999978ndigit..............

TryThese:

1.Guess,Howmanydigitsarethereinthesquaresof(i)72(ii)103(iii)1000

2.

27liesbetween20and30272liesbetween202and302

Nowfindwhatwouldbe272fromthefollowingperfectsquares.(i)329(ii)525(iii)529(iv)729

6.2.Interestingpatternsinsquare:

1.Observethefollowingpatternandcomplete.

1=121+3=4=221+3+5=9=321+3+5+7=16=42

1=1=12

1+3=4=22

1+3+5=9=32

1+3+5+7=16=42

1+3+5+7+9=25=52

1+3+5+7+9+11=……….=()2

1+3+5+7+9+11+13=……….=()2

Fromthis,wecangeneralizethatthesumoffirst‘n’oddnaturalnumbersisequalto‘n2’.2.Observethefollowingpatternandsupplythemissingnumbers

(11)2=121(101)2=10201(1001)2=1002001(10001)2=…………..(1000001)2=……………

3.Observethepatternandcompleteit12=1112=1211112=1232111112=1234321111112=…………………..1111112=……………………ThesenumbersarecalledpalindromicnumbersornumericalpalindromeApalindromeisaword;phrase,asentenceoranumericalthatreadsthesameforwardorbackward.Ex.NOON,MALAYALAM,MADAM

4.Fromthefollowingpatternfindthemissingnumbers12+22+22=32

22+32+62=72

32+42+122=132

42+52+()2=212

52+()2+302=()2

62+72+()2=()2

Observethesumofthesquares:Doyoufindanyrelationbetweenthebasesofsquares?Howthebaseofthethirdnumberisrelatedtothebaseoffirstandsecondsquarenumbers?Howthebaseoftheresultantsquarenumberisrelatedtothebaseofthethirdsquarenumber?

5.Findthemissingnumbersusingthegivenpattern

32=9=4+5

52=25=12+1372=49=24+25(+)

112=121=…..+…..152=225=…..+…..(+)

Fromthis,wecanconcludethatthesquareofanyoddnumbersayncanbeexpressedasthesumoftwoconsecutivenumbersas

6.Numbersbetweensuccessivesquarenumbers:Observeandcompletethefollowingtable

SuccessivesquaresNumbersbetweenthesuccessiveRelationsquarenumbers

12=1;22=42,3(2numbersliesbetween1and4)2×Baseoffirstnumber1,(2×1=2)22=4;32=95,6,7,8(4numbersliesbetween2×Baseoffirstnumber2,(2×2=4)

.4and9)32=9;42=1610,11,12,13,14,15(6numberslies)2×Baseoffirstnumber3(2×3=6)

between9and16)42=16;52=25................................................2×Baseoffirstnumber4,(2×4=8)52=25;62=36.................................................................................................................................................................................................................................

Fromtheabovetablehaveyouobservedanyrelationbetweenthesuccessivesquarenumbersandnumbersbetweenthem?Withthehelpoftheabovetable,trytofindthenumberofnonsquarenumbersbetweenn2and(n+1)2.Thereare‘2n’nonsquarenumbersbetweenn2and(n+1)2.

DoThis:

1.Howmanynonperfectsquarenumbersaretherebetween92and102?2.Howmanynonperfectsquarenumbersaretherebetween152and162?

TryThese:

Rehansaysthereare37nonsquarenumbersbetween92and112.Isheright?Giveyourreason.Exercise-6.1

1.Whatwillbetheunitsdigitofthesquareofthefollowingnumbers?(i)39(ii)297(iii)5125(iv)7286(v)8742

2.Whichofthefollowingnumbersareperfectsquares?(i)121(ii)136(iii)256(iv)321(v)600

3.Thefollowingnumbersarenotperfectsquares.Givereasons?(i)257(ii)4592(iii)2433(iv)5050(v)6098

4.Findwhetherthesquareofthefollowingnumbersareevenorodd?(i)431(ii)2826(iii)8204(iv)17779(v)99998

5.Howmanynumbersliebetweenthesquareofthefollowingnumbers(i)25;26(ii)56;57(iii)107;108

6.Withoutadding,findthesumofthefollowingnumbers(i)1+3+5+7+9=(ii)1+3+5+7+9+11+13+15+17=(iii)1+3+5+7+9+11+13+15+17+19+21+23+25=

6.3Pythagoreantriplets:Considerthefollowing

(i)32+42=9+16=25=52

(ii)52+122=25+144=169=132

Thenumbers(3,4,5)and(5,12,13)aresomeexamplesforPythagoreantriplets.Generallya,b,carethepositiveintegers.Ifa2+b2=c2then(a,b,c)aresaidtobepythagoreantriplet.Iftherearenocommonfactorsotherthan‘1’amonga,b,cthenthetriplet(a,b,c)iscalledprimitivetriplet.

DoThis1.CheckwhetherthefollowingnumbersformPythagoreantriplet

(i)2,3,4(ii)6,8,10(iii)9,10,11(iv)8,15,172.TakeaPythagoreantriplet.Writetheirmultiples.CheckwhetherthesemultiplesformaPythagoreantriplet.6.4SquareRoots

Observethefollowingsquaresandcompletethetable.A=4A=9A=16A=25Areaofthesquare(incm2)Sideofthesquare(incm)(A)(S)

4=2×229=3×3316=4×4_____25=5×5_____Thenumberofunitsquaresinarow/columnrepresentsthesideofasquare.

Doyoufindanyrelationbetweentheareaofthesquareanditsside?Weknowthattheareaofthesquare=side×side=side2

Iftheareaofasquareis169cm2.Whatcouldbethesideofthesquare?Letusassumethatthelengthofthesidebe‘x’cm.169=x2

Tofindthelengthoftheside,itisnecessarytofindanumberwhosesquareis169.Weknowthat169=132.Thenthelengthoftheside=13cm.Therefore,ifasquarenumberisexpressed,astheproductoftwoequalfactors,thenonethefactorsiscalledthesquarerootofthatsquarenumber.Thus,thesquarerootof169is13.Itcanbeexpressedas =13(symbolusedforsquarerootis ).Thusitistheinverseoperationofsquaring.Example1:32=9thereforesquarerootof9is3( =3)

42=16thereforesquarerootof16is4( =4)52=25thereforesquarerootof25is5( =5)

Ify2=xthensquarerootofxisy

(25isthesquareofboth5and−5.Therefore,thesquarerootof25is5or−5.Butinthischapterweareconfinedtothepositivesquarerootwhichisalsocalledprincipalsquareroot.Itiswrittenas∴ =5.)

Example2:1. =2because22=42. =4because42=163. =15because152=225etc.Completethefollowingtable:

SquareSquareroots12=1 =122=4 =232=9 =342=16 =452=25 =……..62=36 =……..72=…...... =……...82=…….. =………92=…….. =………102=…….. =………

6.5FindingtheSquarerootthroughsubtractionofsuccessiveoddnumbers:

Weknowthat,everysquarenumbercanbeexpressedasasumofsuccessiveoddnaturalnumbersstartingfrom1.

Consider,1+3=4=22

1+3+5=9=32

1+3+5+7=16=42

1+3+5+7+9=25=52

Findingsquarerootisthereverseorderofthispattern.Forexample,find

Step1:49–1=48(Subtractingoffirstoddnumber)Step2:48–3=45(Subtractingof2ndoddnumber)Step3:45–5=40(Subtractingof3rdoddnumber)Step4:40–7=33Step5:33–9=24Step6:24–11=13Step7:13–13=0(Observeweknow1+3+5+7+9+11+13=72=4949–[1+3+5+7+9+11+13]=0Hence49isaperfectsquare.)From49,wehavesubtractedsevensuccessiveoddnumbersstartingfrom1andobtainedzero(0)at7thstep.

Note:Iftheresultofthisprocessisnotzerothenthegivennumberisnotaperfectsquare.DoThis:

(i)Bysubtractionofsuccessiveoddnumbersfindwhetherthefollowingnumbersareperfectsquaresornot?(i)55(ii)90(iii)121Itiseasytofindthesquarerootsofanysquarenumbersbytheabovesubtractionprocess.Butincaseofbiggernumberssuchas625,729..........itistimetakingprocess.So,Letustrytofindsimplewaystoobtainthesquareroots.Therearetwomethodsoffindingthesquarerootofthegivennumbers.Theyare(i)Primefactorizationmethod(ii)Divisionmethod6.6FindingtheSquareRootThroughPrimeFactorisationMethod:Letusfindthesquarerootof484byprimefactorizationmethod.Step1:Resolvethegivennumber484intoprimefactors

weget484=2×2×11×11

Step2:Makepairsofequalfactors,weget484=(2×2)×(11×11)

Step3:ChoosingonefactoroutofeverypairBydoingso,weget

=2×11=22Therefore,thesquarerootof484is22.

Or= (2×11)×(2×11)= (2×11)2

=

NowwewillseesomemoreexamplesExample3:Findthesquarerootof1296byPrimeFactorizationSolution:Resolving1296intoPrimefactors,weget

1296=(2×2)×(2×2)×(3×3)×(3×3)=2×2×3×3

∴ =36Example4:Findthesquarerootof2025Solution:Resolving2025intoPrimefactors,weget

2025=(3×3)×(3×3)×(5×5)=3×3×5

∴ =45Example5:Findthesmallestnumberbywhich720shouldbemultipliedtogetaperfectsquare.Solution:Resolving720intoPrimefactors,weget

720=(2×2)×(2×2)×(3×3)×5

Weseethat2,2,3existinpairs,while5isaloneSo,weshouldmultiplythegivennumberby5togetaperfectsquare.Therefore,theperfectsquaresoobtainedis720×5=3600

Example6:Findthesmallestnumberbywhich6000shouldbedividedtogetaperfectsquareandalsofindthesquarerootoftheresultingnumber.

Solution:Resolving6000intoPrimefactors,weget6000=2×2×2×2×3×5×5×5Wecanseethat,2,2,and5existsinpairswhile3and5donotexistsinpairsSo,wemustdividethegivennumberby3×5=15Thereforeperfectsquareobtained=6000÷15=400400=2×2×2×2×5×5Thesquarerootof400is

==2×2×5=20

Exercise-6.2

1.FindthesquarerootsofthefollowingnumbersbyPrimefactorizationmethod.(i)441(ii)784(iii)4096(iv)70562.Findthesmallestnumberbywhich3645mustbemultipliedtogetaperfectsquare.3.Findthesmallestnumberbywhich2400istobemultipliedtogetaperfectsquareandalsofindthesquarerootoftheresulting

number.4.Findthesmallestnumberbywhich7776istobedividedtogetaperfectsquare.5.1521treesareplantedinagardeninsuchawaythatthereareasmanytreesineachrowastherearerowsinthegarden.Findthe

numberofrowsandnumberoftreesineachrow.6.Aschoolcollected`2601asfeesfromitsstudents.Iffeepaidbyeachstudentandnumberstudentsintheschoolwereequal,

howmanystudentswerethereintheschool?7.Theproductoftwonumbersis1296.Ifonenumberis16timestheother,findthetwonumbers?8.7921soldierssatinanauditoriuminsuchawaythatthereareasmanysoldiersinarowastherearerowsintheauditorium.How

manyrowsarethereintheauditorium?9.Theareaofasquarefieldis5184m2.Findtheareaofarectangularfield,whoseperimeterisequaltotheperimeterofthesquare

fieldandwhoselengthistwiceofitsbreadth.6.7Findingsquarerootbydivisionmethod:Wehavealreadydiscussedthemethodoffindingsquarerootbyprimefactorisationmethod.Forlargenumbers,itbecomeslengthyanddifficult.So,toovercomethisproblemweusedivisionmethod.Letusfindthesquarerootof784bydivisionmethod.Step1:Pairthedigitsofthegivennumber,startingfromunitsplacetotheleft.Placeabaroneachpair.

Step2:Findthelargestnumberwhosesquareislessthanorequaltothefirstpairorsingledigitfromleft(i.e.2).Takethisnumberasthedivisorandthequotient.

Step3:Subtracttheproductofthedivisorandquotient(2×2=4)fromfirstpairorsingledigit(i.e.7–4=3)

Step4:Bringdownthesecondpair(i.e.84)totherightoftheRemainder(i.e.3).Thisbecomesthenewdividend(i.e.384).

Step5:Forthenextpossibledivisordoublethequotient(i.e2×2=4)andwriteaboxonitsright.

Step6:Guessthelargestpossibledigittofilltheboxinsuchawaythattheproductofthenewdivisorandthisdigitisequaltoorlessthanthenewdividend(i.e.48×8=384).

Step7:Bysubtracting,wegettheremainderzero.Thefinalquotient28,isthesquareroot

=28

Think,DiscussandWriteObservethefollowingdivisions,givereasonswhy8inthedivisor48isconsideredintheaboveexample?

Now,wewillseesomemoreexamples.Example7:Findthesquarerootof1296

Solution:Step1

Step2

Step3

Step4

Step5Observe

Example8:Findthesquarerootof8281

Solution:ThereforeObserve

Example9:FindthegreatestfourdigitnumberwhichisaperfectsquareSolution:Greatestfourdigitnumberis9999

Wefindsquarerootof9999bydivisionmethod.

Theremainder198showsthatitislessthan9999by198Thismeansifwesubtract198from9999,wegetaperfectsquare.∴9999–198=9801istherequiredperfectsquare.

Example10:Findtheleastnumberwhichmustbesubtractedfrom4215tomakeitaperfectsquare?Solution:Wefindbydivisionmethodthat

Theremainderis119Thismeans,ifwesubtract119from4215.Wegetaperfectsquare.Hence,therequiredleastnumberis119.

6.8Squarerootsofdecimalsusingdivisionmethod:

LetusbeginwiththeexampleStep1:Placethebarsontheintegralpartofthenumberi.e.17intheusualmanner.Placethebarsoneverypairof

decimalpartfromlefttoright

Step2:Findthelargestnumber(i.e.4)whosesquareislessthanorequaltothefirstpairofintegralpart(i.e.17).Takethisnumber4asadivisorandthefirstpair17asthedividend.Gettheremainderas1.

Divideandgettheremainderi.e.1

Step3:Writethenextpair(i.e.64)totherightoftheremaindertoget164,whichbecomesthenewdividend.

Step4:Doublethequotient(2 4=8)andwriteitas8intheboxonitsright.Since64isthedecimalpartso,putadecimalpointinthequotient(i.e.4)

Step5:Guessthedigittofilltheboxinsuchawaythattheproductofthenewdivisorandthedigitisequaltoor

lessthanthenewdividend164.Inthiscasethedigitis2.Divideandgettheremainder.

Step6:Sincetheremainderiszeroandnopairsleft.

Now,letusseesomemoreexamples.Example11:Findthesquarerootof42.25usingdivisionmethod.

Solution:Step1:

Step2:

Step3:

.Example12:FindSolution:

Therefore =9.8

6.9Estimatingsquarerootsofnonperfectsquarenumbers:

Sofarwehavelearntthemethodforfindingthesquarerootsofperfectsquares.Ifthenumbersarenotperfectsquares,thenwewillnotbeabletofindtheexactsquareroots.Inallsuchcasesweatleastneedtoestimatethesquareroot.Letusestimatethevalueof tothenearestwholenumber.300liesbetweentwoperfectsquarenumbers100and400

∴100<300<400102<300<202

i.e.10< <20Butstillwearenotveryclosetothesquarenumber.weknowthat172=289,182=324Therefore289<300<324

17< <18As289ismorecloserto300than324.Theapproximatevalueof is17.

Exercise-6.3

1.Findthesquarerootsofthefollowingnumbersbydivisionmethod.(i)1089(ii)2304(iii)7744(iv)6084(v)9025

2.Findthesquarerootsofthefollowingdecimalnumbers.(i)2.56(ii)18.49(iii)68.89(iv)84.64

3.Findtheleastnumberthatistobesubtractedfrom4000tomakeitperfectsquare4.Findthelengthofthesideofasquarewhoseareais4489sq.cm.5.Agardenerwishestoplant8289plantsintheformofasquareandfoundthattherewere8plantsleft.Howmanyplantswere

plantedineachrow?6.Findtheleastperfectsquarewithfourdigits.7.Findtheleastnumberwhichmustbeaddedto6412tomakeitaperfectsquare?8.Estimatethevalueofthefollowingnumberstothenearestwholenumber

(i) (ii) (iii) CubesandCubeRoots6.10CubicNumbers

Weknowthatacubeisasolidfigurewithsixidenticalsquares.

NowletusmakecubicshapesusingtheseunitcubesS.No.FigureLengthofthesideNo.ofunitcubesused

1 11

2 28

3 327

Canyoumakenextcube?Guesshowmanyunitcubesarerequiredtomakeacubewhosesideis5units?So,werequire1,8,27,64……..unitcubestomakecubicshapes.Thesenumbers1,8,27,64……arecalledcubicnumbersorperfectcubes.

As1=1×1×1=13

8=2×2×2=23

27=3×3×3=33

64=……..×……×……=So,acubenumberisobtainedwhenanumberismultipliedbyitselfforthreetimes.Thatis,cubeofanumber‘x’isx×x×x=x3

Is49acubenumber?No,as49=7×7andthereisnonaturalnumberwhichwhenmultipliedbyitselfthreetimesgives49.Wecanalsoseethat3×3×3=27and4×4×4=64.Thisshowsthat49isnotaperfectcube.

TryThese

1.Is81aperfectcube?2.Is125aperfectcube?

Observeandcompletethefollowingtable.NumberCube113=1×1×1=1223=2×2×2=8333=3×3×3=27443=4×4×4=64553=5×5×5=125663=6×6×6=………773=………………=…………..883=………………=…………..993=………………=…………..10103=………………=…………..

Think,DiscussandWrite

(i)Howmanyperfectcubenumbersarepresentbetween1and100,1and500,1and1000?(ii)Howmanyperfectcubesaretherebetween500and1000?

Followingarethecubesofnumbersfrom11to20NumberCube

111331121728132197142744153375164096174913185832Doyoufindanythinginterestinginthesumofthedigitsinthecubesof17and18?196859208000

Fromthetable,wecanseethatcubeofanevennumberisalwaysanevennumber.Doyouthinkthesameistrueforoddnumbersalso?Wecanalsoobservethat,ifanumberhas1intheunitsplace,thenitscubeendswith1.Similarly,whatcanyousayabouttheunitsdigitofthecubeofanumberhaving0,4,5,6or9astheunitsdigit?

TryThese:

1.Findthedigitinunitsplaceofeachofthefollowingnumbers.(i)753(ii)1233(iii)1573(iv)1983(v)2063

6.11Someinterestingpatterns:1.AddingconsecutiveoddnumbersObservethefollowingpatterns.

1=1=13

3+5=8=23

7+9+11=27=33

13+15+17+19=……=…….Canyouguesshowmanynextconsecutiveoddnumberswillbeneededtoobtainthesumas53?2.Considerthefollowingpattern

23−13=1+2×1×3=733−23=1+3×2×3=1943−33=1+4×3×3=3753−43=……………..=..............

Usingtheabovepatternfindthevaluesofthefollowing(i)103−93(ii)153−143(iii)263−253

3.Observethefollowingpatternandcompleteit13=12

13+23=(1+2)2=(3)2

13+23+33=(1+2+3)2=()2

13+23+33+43=(_____)2

………………….=(1+2+3+....+10)2

Hencewecangeneralizethat,Thesumofthecubesoffirst‘n’naturalnumbersisequaltothesquareoftheirsum.i.e.13+23+33+………+n3=(1+2+3+....+n)2.

6.12CubesandtheirPrimeFactors:

Considerthenumbers64and216Resolving64and216intoprimefactors64=2×2×2×2×2×2216=2×2×2×3×3×3Inboththesecaseseachfactorappearsthreetimes.Thatistheprimefactorscanbegroupedintriples.Thus,ifanumbercanbeexpressedasaproductofthreeequalfactorsthenitissaidtobeaperfectcubeorcubicnumber.Is540aperfectcube?

Resolving540intoprimefactors,weget540=2×2×3×3×3×5Here,2and5donotappearingroupsofthree.Hence540isnotaperfectcube.

DoThese

1.Whichofthefollowingareperfectcubes?

(i)243(ii)400(iii)500(iv)512(v)729Example13:Whatisasmallestnumberbywhich2560istobemultipliedsothattheproductisaperfectcube?

Solution:Resolving2560intoprimefactors,weget2560=2×2×2×2×2×2×2×2×2×5ThePrimefactor5doesnotappearinagroupofthree.So,2560isnotaperfectcube.Hence,thesmallestnumberbywhichitistobemultipliedtomakeitaperfectcubeis5×5=25

Example14:Whatisthesmallestnumberbywhich1600istobedivided.sothatthequotientisaperfectcube?Solution:

Resolving1600intoprimefactors,weget1600=2×2×2×2×2×2×5×5Theprimefactor5doesnotappearinagroupofthreefactors.So,1600isnotaperfectcube.Hence,thesmallestnumberwhichistobedividedtomakeitaperfectcubeis5×5=25

Exercise-6.4

1.Findthecubesofthefollowingnumbers(i)8(ii)16(iii)21(iv)30

2.Testwhetherthegivennumbersareperfectcubesornot.(i)243(ii)516(iii)729(iv)8000(v)2700

3.Findthesmallestnumberbywhich8788mustbemultipliedtoobtainaperfectcube?4.Whatsmallestnumbershould7803bemultipliedwithsothattheproductbecomesaperfectcube?5.Findthesmallestnumberbywhich8640mustbedividedsothatthequotientisaperfectcube?6.Ravimade a cuboid of plasticine of dimensions 12cm, 8cm and 3cm.Howmanyminimumnumber of such cuboidswill be

neededtoformacube?7.Findthesmallestprimenumberdividingthesum311+513.6.13CuberootsWeknowthat,werequire8unitcubestoformacubeofside2units(23=8)similarly,weneed27unitcubestoformacubeofside3units(33=27)Suppose,acubeisformedwith64unitcubes.Thenwhatcouldbethesideofthecube?Letusassume,thelengthofthesidetobe‘x’

∴so064=x3Tofindthesideofacube,itisnecessarytofindanumberwhosecubeis64.Therefore,findingthenumberwhosecubeisknowniscalledfindingthecuberoot.Itistheinverseoperationofcubing.As,43=64then4iscalledcuberootof64Wewrite .Thesymbol denotescuberoot.Hence,anumber‘x’isthecuberootofanothernumbery,ify=x3thenx= .

(1unitcube)(2unitcube)(3unitcube)(4unitcube)

Completethefollowingtable:CubesCuberoots13=1 =123=8 =233=27 =343=64 =453=125 =563=…... =673=……. =783=…… =8……=...................=...........……=.......................=.........6.14FindingcuberootthroughPrimeFactorizationmethod:Letusfindthecuberootof1728byprimefactorizationmethod.

Step1:Resolvethegivennumber1728intoprimefactors.1728=2×2×2×2×2×2×3×3×3

Step2:Makegroupsofthreeequalfactors:1728=(2×2×2)×(2×2×2)×(3×3×3)

Step3:Chooseonefactorfromeachgroupandmultiplybydoingso,weget=2×2×3=12

∴ =2×2×3=12LetusseesomemoreexamplesExample15:Findthecuberootof4096?Solution:

Resolving4096intoPrimeFactors,weget4096=(2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)

=2×2×2×2=16∴ =16

6.15Estimatingthecuberootofanumber

Ifweknowthat,thegivennumberisacubenumberthentofinditscuberootthefollowingmethodcanbeused.Letusfindthecuberootof9261throughestimation.Step1:Startmakinggroupsofthreedigitsstartingfromtheunitplace.

i.e.9261secondfirstgroupgroup

Step2:Firstgroupi.e.261willgiveustheunitsdigitofthecuberoot.As261endswith1,itscuberootalsoendswith1.So,theunitsplaceofthecuberootwillbe1.

Step3:Now,takesecondgroupi.e.9.Weknowthat23<9<33.Asthesmallestnumberis2,itbecomesthetensplaceoftherequiredcuberoot

Exercise-6.5

1.Findthecuberootofthefollowingnumbersbyprimefactorizationmethod.(i)343(ii)729(iii)1331(iv)2744

2.Findthecuberootofthefollowingnumbersthroughestimation?(i)512(ii)2197(iii)3375(iv)5832

3.Statetrueorfalse?(i)Cubeofanevennumberisanoddnumber(ii.)Aperfectcubemayendwithtwozeros(iii)Ifanumberendswith5,thenitscubeendswith5(iv)Cubeofanumberendingwithzerohasthreezerosatitsright(v)Thecubeofasingledigitnumbermaybeasingledigitnumber.

(vi)Thereisnoperfectcubewhichendswith8(vii)Thecubeofatwodigitnumbermaybeathreedigitnumber.

4.Findthetwodigitnumberwhichisasquarenumberandalsoacubicnumber.

Whatwehavediscussed

•Estimatingnumberofdigitsinsquareofanumber.•Squarenumberswrittenindifferentpatterns.•a,b,carepositiveintegersandifa2+b2=c2then(a,b,c)aresaidtobePythagoreantriplets.•Findingthesquarerootsbyprimefactorisationanddivisionmethod.•Squarerootistheinverseoperationofsquaring.•Estimatingsquarerootsofnonperfectsquarenumbers.•Ifanumberismultipliedthreetimesbyitselfiscalledcubenumber.•Findingcuberootbyprimefactorisationmethod.•Estimatingcuberootsofanumber.•Thesquareofintegerisaintegerandasquarenumber,whereassquareofrationalnumberisaperfectsquare.

EternaltriangleTheformulaethatwillgiveintegralsidesofaright-angledtrianglehavebeenknownsincethetimeofDiophantusandtheearlyGreeks.TheyareonelegX=m2−n2

secondlegY=2mnHypotenuseZ=m2+n2

Thenumbersmandnareintegerswhichmaybearbitarilyselectus.ExamplemnX=m2−n2Y=2mnZ=m2+n2

2134532512135221202943724254115827

T

=2×11=2224842242

1112111111212962648232421623813273933152025540538132739331

272023602180290345315551240022002100250525551260002300021500275033755125525551

Observe

Observe

Y=2mnX=m2−n2

Z=m2+n2