CHAPTER 1 SETS 7.pdf · 1. QUADRATIC E Q UATIONS eLearn.Punjab 1. QUADRATIC E Q UATIONS...

CHAPTER

1 SETS

Animation 1.1: Sets-mathSource & Credit: elearn.punjab

Version 1.1

http://elearn.punjab.gov.pk/

1. Quadratic EQuations eLearn.Punjab 1. Quadratic EQuations eLearn.Punjab

2 3

1. SetseLearn.Punjab eLearn.Punjab1. Sets

Version 1.1 Version 1.1

Student Learning Outcomes

After studying this unit, students will be able to:• Expressasetin:

• descriptiveform,• setbuilderform,• tabularform.

• Defineunion,intersectionanddifferenceoftwosets• Find:

• unionoftwoormoresets,• intersectionoftwoormoresets,• differenceoftwosets

• Defineandidentifydisjointandoverlappingsets• Defineauniversalsetandcomplimentofaset• Verifydifferentpropertiesinvolvingunionofsets,intersectionof

sets,differenceofsetsandcomplimentofaset,e.gAkA’=f.

• RepresentsetsthroughVenndiagram.• Performoperationofunion,intersection,differenceand

complementontwosetsAandB,when:• AissubsetofB,• BissubsetofA,• AandBaredisjointsets,• AandBareoverlappingsets,throughVenndiagram.

1.1 Introduction

Inourdailylife,weusethewordsetonlyforsomeparticularcollectionssuchaswaterset,teaset,dinnerset,sofaset,asetofbooks,asetofcoloursandsoon.Butinmathematics,thewordsethasbroadermeaningsthanthoseinourdailylifebecauseitprovidesusawaytointegratethedifferentbranchesofmathematics.

It also helps to solve manymathematical problems of bothsimpleandcomplexnature.Inshort,itplaysapivotalroleintheadvancedstudyofthemathematicsinthemodernage.Lookatthefollowingexamplesofaset.

A=Thesetofcountingnumbers.B=ThesetofPakistaniProvinces.C=Thesetofgeometricalinstruments

“Asetisacollectionofwelldefinedobjects/numbers.Theobjects/numbersinanysetarecalleditsmembersorelements”

“Settheory”isabranchofmathematicsthatstudiessets.It isthecreationofGeorgeCantorwhowasborn in Russia on March 03, 1845. In 1873, hepublishedanarticlewhichmakesthebirthofsettheory.GeorgeCantordiedinGermanyonJanuary06,1918

1.1.1 Expressing a Set

Therearethreewaystoexpressaset.1.Descriptiveform 2.Tabularform 3.Setbuilderform

• Descriptive form Ifasetisdescribedwiththehelpofastatement,itiscalledasdescriptiveformofaset.

For Example:N=setofnaturalnumbersZ=setofintegersP=setofprimenumbersW=setofwholenumbersS=setofsolarmonthsstartwithletter“J”

Recall

A set cannot consist of elements like moral values, concepts, evils or virtues etc.

http://en.wikipedia.org/wiki/Georg_Cantor


4 5



• Tabular FormIfwelistallelementsofasetwithinthebraces{}andseparateeachelementbyusinga comma”,” it is calledthetabularorrosterform.

For Example:A={a, e, i, o, u}C={3,6,9,...,99}M={football, hockey , cricket}N={1,2,3,4,...}W={0,1,2,3,...} X={a, b, c, ..., z}

• Set Builder Form If a set is described by using a common property of all itselements,itiscalledassetbuilderform.Asetcanalsobeexpressedinsetbuilderform.Forexample,“Eisasetofevennumber”inthedescriptiveform,whereE={0,±2,±4,±6,.....}isthetabularformofthesameset.Thissetinsetbuilderformcanbewrittenas;

E = {x | x is an even number}andwecanreaditas,Eisasetofelementsx,suchthatxisanevennumber.A={x|xisasolarmonthofayear}B={x|xdN/1<x<5}C={x|xdW/x74}

| such that d belongs to/and 0 or> greater than or equal to< less than or equal to

EXERCISE 1.1

1. Write the following sets in descriptive form.

(i) A={a, e, i, o, u) (ii) B={3,6,9,12,...}(iii) C={s, p, r , i, n, g} (iv) D={a, b, c, ... ,z}(v) E={6,7,8,9,10} (vi) F={0,±1,±2}(vii) G={x|xdN/x<3} (viii) H={x|xdN/x>99}

2. Write the following sets in tabular form.

(i) A=Lettersoftheword“hockey”(ii) B=Twocoloursintherainbow(iii) C=Numberslessthan18divisibleby3(iv) D=Multiplesof5lessthan30(v) E={xIxdW/x>5}(vi) F={xIxdZ/ –7<x<–1}

3. Write the following sets in the set builder form.

(i) A={1,2,3,4,5}(ii) B={2,3,5,7}(iii) N=setofnaturalnumbers(iv) W=setofwholenumber(v) Z=setofallintegers(vi) L={5,10,15,20,...}(vii) E=setofevennumbersbetween1and10(viii) O=setofoddnumbersgreaterthan15(ix) C=setofplanetsinthesolarsystem(x) S=setofcoloursintherainbow

1.2 Operations on Sets

1.2.1 Union, Intersection and Difference of Two Sets• Union of Two Sets TheunionoftwosetsAandBisasetconsistingofallthe

Do you Know

Some Important Symbols

The sets of natural numbers, whole numbers, integers, even numbers and odd numbers are denoted by the English letters N, W, Z, E and O respectively.


6 7



elementswhichareinsetAorsetBorinboth.TheunionoftwosetsisdenotedbyAjBandreadas“AunionB”

Example 1: IfA={a, e, i, o}andB={a,b,c},thenfindAjBSolution:A={a, e, i, o},B={a,b,c}AjB={a, e, i, o}j{a,b,c}={a, e, i, o, a,b,c}

Example 2: IfM={1,2,3,4,5}andN={1,3,5,7},thenfindMjNSolution:M={1,2,3,4,5},N={1,3,5,7}MjN={1,2,3,4,5}j{1,3,5,7}={1,2,3,4,5,7}

• Intersection of Two Sets TheintersectionoftwosetsAandBisasetconsistingofallthecommonelementsofthesetsAandB.TheintersectionoftwosetsAandBisdenotedbyAkBandreadas“AintersectionB”

Example 3: IfA={a, e, i, o, u}andB={a, b, c, d, e},thenfindAkBSolution:A={a, e. i, o, u},B={a, b, c, d, e}AkB={a, e, i, o, u}k{a, b, c, d, e}={a, e}

Animation 1.2: Intersection of two setsSource & Credit: elearn.punjab

Example 4: IfX={1,2,3,4}andY={2,4,6,8},thenfindXkYSolution:X={1,2,3,4},Y={2,4,6,8}XkY={1,2,3,4}k{2,4,6,8}={2,4}

• DifferenceofTwoSets ConsiderAandBaretwoanysets,thenAdifferenceBisthesetofallthoseelementsofsetAwhicharenottheelementsofsetB.ItiswrittenasA-BorA\B.Similarly,BdifferenceAisthesetofallthoseelementsofsetBwhicharenottheelementsofsetA.ItiswrittenasB-AorB\A.

Example 5: IfA={1,3,6}andB={1,2,3,4,5},thenfind: (i)A-B (ii)B-ASolution:A={1,3,6},B={1,2,3,4,5}(i)A-B={1,3,6}-{1,2,3,4,5}={6}(ii)B-A={1,2,3,4,5}–{1,3,6}={2,4,5}

1.2.2 Union and Intersection of Two or More Sets

Wehave learnt themethod forfinding theunionandintersectionoftwosets.Nowwetrytofindtheunionandintersectionofthreesets.

• Union of three setsFollowingarethestepstofindtheunionofthreesets

Step 1:Findtheunionofanytwosets.Step 2:Findtheunionofremaining3rdsetandthesetthatwegetastheresultofthefirststepFor three sets A,BandC theirunioncanbe taken inanyof thefollowingways. (i) Aj(BjC) (ii) (AjB)jC



8 9



It will be easier for us to understand the above method withexamples.Lookatthegivenexamples.

Example 6: Find Aj (Bj C) where A = {1, 2, 3, 4},B={3,4,5,6,7,8}andC={6,7,8,9,10}.

Solution:Aj(BjC)={1,2,3,4}j{3,4,5,6,7,8}j{6,7,8,9,10}={1,2,3,4}j{3,4,5,6,7,8,9,10}={1,2,3,4,5,6,7,8,9,10}

Example 7: IfA={1,3,7},B={3,4,5}andC={1,2,3,6}

Solution:(AjB)jC={1,3,7}j{3,4,5})j{1,2,3,6}={1,3,4,5,7}j{1,2,3,6}={1,2,3,4,5,6,7}

• Intersection of Three Sets For finding the intersection of three sets, first we find theintersectionofanytwosetsofthemandthentheintersectionofthe3rdsetwiththeresultantsetalreadyfound. (i) Ak(BkC) (ii) (AkB)kC

Example 8: FindAk(BkC)whereA={a, b, c, d},B={c, d, e}andC={c, e, f, g}

Solution:Ak(BkC)={a, b, c, d}k({c, d, e}k{c, e, f, g})={a, b, c, d}k{c, e}={c}

Example 9: IfA={1,2,3,4},B={2,3,4,5}andC={1,2},thenfind(AkB)kC

Solution:(AkB)kC=({1,2,3,4}k{2,3,4,5})k{1,2}

={2,3,4}k{1,2}={2}

EXERCISE 1.2

1. Findtheunionofthefollowingsets.(i) A= {1,3,5}, B= {1,2,3,4}(ii) S= {a, b, c}, T= {c, d, e}(iii) X= {2,4,6,8,10}, Y= {1,5,10}(iv) C= {i, o, u}, D= {a, e, o},E={i, e, u}(v) L= {3,6,9,12}, M= {6,12,18,24,},N={ 4,8,12,16}2. Findtheintersectionofthefollowingsets.(i) P= {0,1,2,3}, Q= {–3,–2,–1,0}(ii) M= {1,2,...,10}, N= {1,3,5,7,9}(iii) A= {3,6,9,12,15}, B= {5,10,15,20}(iv) U= {-1,-2,-3}, V= {1,2,3},W={0,±1,±2}(v) X= {a, l, m}, Y= {i, s, l, a, m},Z={l, i, o, n}

3.IfN=setofNaturalnumbersandW=setofWholenumbers,thenfindNUWandNkW4.IfP=setofPrimenumbersandC=setofCompositenumbers,thenfindPUCandPkC5.IfA={a,c,d,f},B={b,c,f,g}andC={c,f,g,h},thenfind(i)AU(BUC) (ii)Ak(BkC)

6.IfX={1,2,3,.....,10},Y={2,4,6,8,12}andZ={2,3,5,7,11},thenfind:(i)XU(YUZ) (ii)Xk(YkZ)

7.IfR={0,1,2,3},S=[0,2,4)andT={1,2,3,4},thenfind:(i)R\S (ii)T\S (iii)R\T (iv)S\R

1.2.3 Disjoint and Overlapping Sets

• Disjoint Sets Two setsA andBare said tobedisjoint sets, if there is nocommonelementbetweenthem.Inotherwordsthereintersectionisanemptyset,i.e.AkB=f.Forexample,A={1,2,3},B={4,5,6}aredisjointsetsbecausethereisnocommonelementinsetAandB.


10 11



• Overlapping Sets TwosetsAandBarecalledoverlappingsets, if thereisatleast oneelementcommonbetween them but none of themisasubsetoftheother.Inotherwords,theirintersectionisnon-emptyset.Forexample, A={0,5,10}andB={1,3,5,7}areoverlappingsetsbecause5isacommonelementinsetsAandBandnonissubsetoftheother.

1.2.4 Universal Set and Complement of a Set

• Universal Set Asetwhichcontainsallthepossibleelementsofthesetsunderconsiderationiscalledtheuniversalset.Forexample,theuniversalsetofthecountingnumbersmeansasetthatcontainsallpossiblenumbersthatwecanuseforcounting.TorepresentsuchasetweusethesymbolUandreaditas“Universalset”i.e.

Theuniversalsetofcountingnumbers:U={1,2,3,4,...}

• Complement of a SetConsiderasetBwhoseuniversalsetisU,thenthedifferenceset U \ B or U - B is called the complement of a set B, which isdenotedbyB’orBcandreadas“Bcomplement”.So,wecandefinethecomplementofasetBas:“BcomplementisasetwhichcontainsallthoseelementsofuniversalsetwhicharenottheelementsofsetB,i.e.B’=U\B.

Example 1: IfU={1,2,3,...,10}andB={1,3,7,9},thenfindB’.

Solution:U={1,2,3,...,10},B={1,3,7,9}B’=U–B={1,2,3,...,10}–{1,3,7,9}={2,4,5,6,8,10}

EXERCISE 1.3

1. Lookateachpairofsetstoseparatethedisjointandoverlappingsets.(i) A= {a, b, c, d, e}, B= {d, e, f, g, h}(ii) L= {2,4,6,8,10}, M= {3,6,9,12}(iii) P= SetofPrimenumbers, C= SetofCompositenumbers(iv) E= SetofEvennumbers, O= SetofOddnumbers

2. IfU={1,2,3,....,10},A={1,2,3,4,5},B={1,3,5,7,9},C={2,4,6,8,10}andD=3,4,5,6,7},thenfind:(i)A’(ii)B’(iii)C’(iv)D’3. IfU={a, b, c,...., i},X={a, c, e, g, i},Y={a, e, i},andZ={a, g, h},thenfind:(i)X’(ii)Y’(iii)Z’(iv)U’4.IfU={1,2,3,...,20},A={1,3,5,...,19}andB={2,4,6,...,20},thenprovethat:(i)B’=A(ii)A’=B(iii)A\B=A(iv)B\A=B5.IfU=setofintegersandW=setofwholenumbers,thenfindthecomplementofsetW.6.IfU=setofnaturalnumbersandP=setofprimenumbers,thenfindthecomplementofsetP.

1.2.5 Properties involving Operations on Sets

Wehave learnt the four operations of sets, i.e. union,intersection, difference and complement. Now we discuss theirproperties.

• Properties involving Union of Sets

• Commutative property

If A,Bareany twosets, then “AjB = BjA” iscalled thecommutativepropertyofunionoftwosets.


12 13



Example 1:IfA={1,2,3}andB={2,4,6},thenverifythat:AjB=BjA.Solution:AUB={1,2,3}j{2,4,6}={1,2,3,4,6}BUA={2,4,6}j{1,2,3}={1,2,3,4,6}Fromtheabove,itisverifiedthat:AjB=BjA

• Associative Property IfA,BandCareanythreesets,then“Aj(BjC)=(AjB)jC”iscalledtheassociativepropertyofunionofthreesets.

Example 2: IfA={1,2,3,4,5},B={1,3,5,7}andC={2,4,6,8},thenverifythat:Aj(BjC)=(AjB)jCSolution:L.H.S=Aj(BjC)={1,2,3,4,5}j({1,3,5,7}j{2,4,6,8})={1,2,3,4,5}j{l,2,3,4,5,6,7,8}={1,2,3,4,5,6,7,8}

R.H.S=(AjB)jC=({1,2,3,4,5}j{1,3,5,7})j{2,4,6,8}={1,2,3,4,5,7}j{2,4,6,8}={1,2,3,4,5,6,7,8}WeseethatL.H.S=R.H.S

• Identity Property with respect to Union Insets,theemptysetf actsasidentityforunion,i.e.Ajf=A

Example 3:IfA={a, e, i, o, u},thenverifythatAjf=A.Solution:Ajf=AL.H.S=Ajf

={a, e, i, o, u}U{}={a, e, i, o, u}=A=R.H.SHenceproved:L.H.S=R.H.S

• Properties involving Intersection of Sets

• Commutative Property IfA,Bareanytwosets,thenAkB=BkAiscalledthecommutativepropertyofintersectionoftwosets.Example 4: Ifa={a,b,c,d}andB={a,c,e,g},thenverifythatAkB=BkA.Solution:AkB={a,b,c,d}k{a,c,e,g}={a,c}BkA={a,c,e,g}k{a,b,c,d}={a,c}FromtheaboveitisverifiedthatAkB=BkA.

Example 5: If A= {1, 2, 3} and B = {4, 5, 6}, then verify thatAkB=BkA.Solution:AkB={1,2,3}k{4,5,6}={}BkA={4,5,6}k{1,2,3}={}FromtheaboveitisverifiedthatAkB=BkA.

• Associative Property IfA,BandCareanythreesets,thenAk(BkC)=(AkB)kCiscalledtheassociativepropertyofintersectionofthreesets.

Example 6: IfA={1,2,5,8},B={2,4,6}andC={2,4,5,7},thenverifythat:Ak(BkC)=(AkB)kC


14 15



Solution:A={1,2,5,8},B={2,4,6},C={2,4,5,7}L.H.S=Ak(BkC)={1,2,5,8}k({2,4,6}k{2,4,5,7})={l,2,5,8}k{2,4}={2}R.H.S=(AkB)kC=({l,2,5,8}k{2,4,6})k{2,4,5,7}={2}k{2,4,5,7}={2}ItisverifiedthatL.H.S=R.H.S

• Identity Property with respect to Intersection Insets,theuniversalsetUactsasidentityforintersection,i.e.AkU=A.

Example 7: IfU={a, b, c, ..., z}andA={a, e, i, o, u},thenverifythatAkU=A.

Solution:U={a,b,c,...,z},A={a,e,i,o,u}L.H.S=AkU={a,e,i,o,u}k{a,b,c,...,z)={a,e,i,o,u}=A=R.H.SHenceverifiedthatL.H.S=R.H.S

• PropertiesinvolvingDifferenceofSets IfAandBaretwounequalsets,thenA-B≠B-A,ForexampleifA={0,1,2)andB={1,2,3},thenA–B={0,1,2}-{1,2,3}={0}B–A={1,2,3}-{0,1,2}={3}WecanseethatA-B≠B-A

• Properties involving Complement of a Set PropertiesinvolvingthesetsandtheircomplementsaregivenbelowA’jA=UAkA’=fU’=ff’=U

Example 8: IfU={1,2,3,…,10}andA={1,3,5,7,9},thenprovethat: (i) U’=f (ii) AjA’=U (iii) AkA’=f (iv) f’=USolution: U={1,2,3,...,10},A={1,3,5,7,9}

(i): U’ = fL.H.S=U’WeKnowthatU’=U-U={1,2,3,...,10}-{1,2,3,...,10}={}=R.H.SHenceverifiedthatL.H.S=R.H.S

(ii): A U A’ = UWeknowthatA’=U-A={1,2,3,...,10}-{1,3,5,7,9}={2,4,6,8,10}Nowwefind,AUA’={1,3,5,7,9}U{2,4,6,8,10}={1,2,3,...,10}=R.H.SHenceverifiedthatL.H.S=R.H.S

(iii) A ∩ A’ = fL.H.S=A∩A’={1,3,5,7,9}∩{2,4,6,8,10}={}=f =R.H.SHenceverifiedthatL.H.S=R.H.S

(iv) f’ = UWeknowthatf’=U-f={1,2,3,…,10}-{}={1,2,3,…,10}=U=R.H.SHenceverifiedthatL.H.S=R.H.S

EXERCISE 1.4

1.IfA={a,e,i,o,u},B={a,b,c}andC={a,c,e,g},thenverifythat:


16 17



(i)A∩B=B∩A(ii)AUB=BUA(iii)BUC=CUB(iv)B∩C=C∩B(v)A∩C=C∩A(vi)AUC=CUA2.IfX={1,3,7},Y={2,3,5}andZ={1,4,8},thenverifythat:(i)X∩(Y∩Z)=(X∩Y)∩Z(ii)XU(YUZ)=(XUY)UZ3. IfS={–2,–1,0,1},T={–4,–1,1,3}andU={0,±1,±2},thenverifythat:(i)S∩(T∩U)=(S∩T)∩U(ii)Su(TjU)=(SjT)jU4. IfO={1,3,5,7.....},E={2,4,6,8......}andN={1,2,3,4....},thenverifythat:(i)O∩(E∩N)=(O∩E)∩N(ii)Oj(EjN)=(OjE)jN5. IfU={a,b,c,....,z},S={a,e,i,o,u}andT={x,y,z},thenverifythat:(i)SUf=S(ii)T∩U=T(iii)S∩S’=f(iv)TUT’=U6. IfA={1,7,9,11},B={1,5,9,13},andC={2,6,9,11},thenverifythat:(i)A-B≠B-A(ii)A-C≠C-A7. IfU={0,1,2,....,15},L={5,7,9,....,15},andM={6,8,10,12,14},thenverifytheidentitypropertieswithrespecttounionandintersectionofsets.

1.3 Venn Diagram

AVenndiagramissimpleclosedfigurestoshowsetsandtherelationshipsbetweendifferentsets.

Venndiagramwere introducedbyaBritishlogicianandphilosopher“JohnVenn” (1834- 1923). John himself did not use the term“Venn diagram” Another logician “Lewis”useditfirsttimeinbook“Asurveyofsymboliclogic”

Animation 1.3: Venn diagramSource & Credit: elearn.punjab

1.3.1 Representing Sets through Venn diagrams

InVenndiagram,auniversalsetisrepresentedbyarectangleandtheothersetsarerepresentedbysimpleclosedfiguresinsidetherectangle.Theseclosedfiguresshowanoverlappingregiontodescribetherelationshipbetweenthem.FollowingfiguresaretheVenndiagramsforanysetAofuniversalsetU,disjointsetsAandBandoverlappingsetsAandBrespectively.

Set A Disjoint Sets Overlapping Sets

http://en.wikipedia.org/wiki/John_Venn

http://elearn.punjab.gov.pk/home/read_book/21


18 19



In the Venndiagram, the shaded region is used to represent theresultofoperation.

1.3.2 Performing Operation on Sets through Venn Diagram • Union of SetsNowwerepresenttheunionofsetsthroughVenndiagramwhen:

• A is subset of B WhenalltheelementsofsetAarealsotheelementsofsetB, thenwecanrepresentAUBby (figure i).HereshadedportionrepresentsAUB.

Figure (i)

• B is subset of A WhenalltheelementsofsetBarealsotheelementsofsetA,thenwecanrepresentAUBby(figureii).HereshadedportionrepresentsAUB.

Figure (ii)

• A and B are overlapping SetsWhenonlyafewelementsoftwosetsAandBarecommon,thentheyarecalledoverlappingsets.AUBisrepresentedby(figureiii).HereshadedportionrepresentsAUB.

Figure (iii)

• A and B are disjoint SetsWhennoelementoftwosetsAandBiscommon,thenwecanrepresentAUBby(figureiv).HereshadedportionrepresentsAUB.

Figure (iv)

• Intersection of Sets NowwecleartheconceptofintersectionoftwosetsbyusingVenndiagram. In thegivenfigures theshadedportionrepresentstheintersectionoftwosetswhen: • A is subset of BWhenalltheelementsofsetAarealsotheelementsofsetB,thenwecanrepresentA∩Bby(figurev).HereshadedportionrepresentsA∩B.

Figure (v)


20 21



• B is subset of AWhenalltheelementsofsetBarealsotheelementsofsetA, thenwecanrepresentAkBby (figurevi).HereshadedportionrepresentsAkB.

Figure (vi)

• A and B are overlapping SetsWhensomeelementsarecommon,thenwecanrepresentAkBbythefig(vii).

Figure (vii)

• A and B are disjoint SetsWhennoelementiscommon,thenwecanrepresentAkBbyfig(viii).SoAkBisanemptyset.

Figure (viii)

• DifferenceofTwoSetsAandB Itisrepresentedbyshadedportionwhen:

• A is subset of BWhenalltheelementsofsetAarealsotheelementsofsetB,thenwecanrepresentA-Bbyfig(ix). Thereisnoshadedportion.So,A-B={}

Figure (ix)

• B is subset of AWhenalltheelementsofsetBarealsotheelementsofsetA, thenwecanrepresentA -Bby (figurex).HereshadedportionrepresentsA-B.

Figure (x)

• A and B are overlapping Sets When some elements are common, then we can representA- Bbythefig(xi).

Figure (xi)


22 23



• A and B are disjoint Sets WhennoelementoftwosetsAandBiscommon,thenwecanrepresentA-Bby(figurexii).HereshadedportionrepresentsA-B.

Figure (xii)

1.3.3 Complement of a Set

ForcomplementofasetAForcomplementofasetB

U - A = A’ U - B = B’

EXERCISE 1.5

1. Shadethediagramsaccordingtothegivenoperations.(i) AkB(AissubsetofB) (ii) AjB(BissubsetofA)

(iii) A-B(Fordisjointsets) (iv) B’

(v) AkB(Overlappingsets) (vi) AjB(Fordisjointsets)

2. If,U={1,2,3,10},A={1,4,8,9,10}andB={2,3,4,7,10},then showthat: (i)A-B≠B-A(ii)A∩B=B∩A (iii)AUB=BUA(iv)A’≠B’ throughVenndiagram.

Review Exercise 1

1. Answerthefollowingquestions. (i) Namethreeformsfordescribingaset. (ii) Definethedescriptiveformofset. (iii) Whatdoesthesymbol“|“mean? (iv) Writethenameofthesetconsistingofalltheelementsof givensetsunderconsideration. (v) Whatismeantbydisjointsets?


24 25



2. Fillintheblanks. (i) Thesymbol“^”means_______. (ii) Thesetconsistingofonlycommonelementsoftwosets iscalledthe_______oftwosets. (iii) Asetwhichcontainsallthepossibleelementsofthesets underconsiderationiscalledthe_______set. (iv) Twosetsarecalled_______ifthereisatleastoneelement commonbetweenthemandnonofthesetsissubsetof theother. (v) Insets,theuniversalsetactsas_______forintersection.

3. Tick(p)thecorrectanswer.

4. Writethefollowingsetsinthesetbuilderform. (i) A={5,6,7,8} (ii) B={0,1,2} (iii) C={a, e, i, o, u} (iv) D=setofnaturalnumbersgreaterthan100 (v) E=setofoddnumbersgreaterthan1andlessthan10

5. Writethefollowingsetsindescriptiveandtabularform. (i) A={x|xdW/x<7} (ii) B={x|xdE/3<x<12} (iii) C={x|xdZ/-2<x<+2} (iv) D={x|xdP/x<15}

6. IfA={3,4,5,6}andB={2,4,6},thenverifythat: (i) AjB=BjA (ii) AkB=BkA

7. IfX={2,3,4,5}andY={1,3,5,7},thenfind: (i) X-Y (ii) Y-X

8. IfA={a, c, e, g},B={a, b, c, d}andC={b, d, f, h},thenverifythat: (i) Aj(BjC)=(AjB)jC (ii) Ak(BkC)=(AkB)kC

9. IfU=setofwholenumbersandN=setofnaturalnumbers, thenverifythat: (i) N’jN=U (ii) N’kN=f

10. If U={a, b, c, d, e},A={a, b, c}andB={b, d, e}, thenshow throughVenndiagram (i) A’ (ii) B’ (iii) AUB (iv) A∩B

Summary

• Therearethreeformstowriteaset.(i)Descriptiveform(ii)Tabularform(iii)Setbuilderform• Twosetsaresaidtobedisjoint if there isnoelementcommon

betweenthem.• IfAandBaretwosetsthenunionofAandBisdenotedbyAjB

andintersectionofAandBisdenotedbyAkB.• IfAandBaretwosetsthenBissaidtobesubsetofAifevery

elementofsetBistheelementofsetA.• Twosetsarecalledoverlappingsetsifthereisatleastoneelement

commonbetweenthembutnoneofthemisasubsetoftheother.• Asetwhichcontainsallpossibleelementsofagivensituationor

discussioniscalledtheuniversalset.

CHAPTER

2 Rational numbeRs

Animation 2.1: Rational NumbersSource & Credit: elearn.punjab

Version 1.1

http://www.youtube.com/user/makemegenius%253Ffeature%253Dwatch


2 3

2. Rational Numbers 2. Rational NumberseLearn.Punjab eLearn.Punjab

2Version 1.1 Version 1.1


After studying this unit, students will be able to:• Definearationalnumberasanumberthatcanbeexpressedin

theform wherepandqareintegersandq≠0.• Representrationalnumbersonnumberline.• Addtwoormorerationalnumbers.• Subtractarationalnumberfromanother.• Findadditiveinverseofrationalnumbers.• Multiplytwoormorerationalnumbers.• Dividearationalnumberbyanon-zerorationalnumber.• Findmultiplicativeinverseofanon-zerorationalnumber.• Findreciprocalofanon-zerorationalnumber.• Verifycommutativepropertyofrationalnumberswithrespectto

additionandmultiplication.• Verify associativepropertyof rationalnumberswith respect to

additionandmultiplication.• Verifydistributivepropertyof rationalnumberswithrespect to

multiplicationoveraddition/subtraction.• Comparetworationalnumbers.• Arrangerationalnumbersinascendingordescendingorder.

2.1 Rational Numbers

In previous class, we have learnt that the difference of twocountingnumbersisnotalwaysanaturalnumber.Forexample,

2-4=-s2...............(i)1-5=-4...............(ii)

In(i)and(ii),wecanobservethat-2and-4arenotnaturalnumbers.Thisproblemgaveustheideaofintegers.Nowinintegers,whenwemultiplyanintegerbyanotherinteger,theresultisalsoaninteger.Forexample,

-1x2=-2............(iii)-2x(-3)=6............(iv)

From theabove (iii) and (iv),we cannotice that -2and6arealsointegers.But in caseofdivisionof integers,wedonotalwaysgetthesameresult, i.e. arenot integers.So, itmeansthatthe division of integers also demands another number systemconsisting of fractions, as well as, integers that is fulfilled by therationalnumbers.

2.1.1 Defining Rational Numbers

Anumberthatcanbeexpressedintheformof wherepandqareintegersandqm0,iscalledarationalnumber,e.g., areexamplesofrationalnumbers.

Thesetofrationalnumbersisthesetwhoseelementsarenaturalnumbers, negative numbers, zero and all positive and negativefractions.

2.1.2 Representation of Rational Numbers on Number line Wealreadyknowthemethodofconstructinganumberlinetorepresenttheintegers.Nowweusethesamenumberlinetorepresenttherationalnumbers.Forthispurpose,wedrawanum-berlineasgivenbelow.

Nowwe divide each segment of the above number line into twoequalparts,asgiveninthefollowingdiagram.

Inthefigure2.2,thenumberlinerepresentstherationalnumberswhicharegivenbelow.


4 5



Now we divide further each small segment of the above drawnnumberlineintotwomoreequalparts.

Inthefigure2.3,thenumberlinerepresentsthefollowingrationalnumbers.

Similarly,wecandivideeachsegmentofanumberlineintothree,fiveandevenmoreequalpartsandwecanalsorepresentanyrationalnumberonanumberlinebyusingtheabovegivenmethod.

Example 1: Draw a number line and represent the rational

number103

-

Solution:Step 1: Drawanumberlineasgivenbelow.

Step 2: Convert103

-tomixedfraction

133

-

Step 3: Divide the line segment of the number line between- 4and-3inthreeequalpartsandstartcountingfromthepoint-3

to-4onthefirstpartis133

- whichisourrequirednumber.

EXERCISE 2.1

1. Write“T”foratrueand“F”forafalsestatement. (i) Positivenumbersarerationalnumbers. (ii) “0”isnotarationalnumber. (iii) Anintegerisexpressedin form. (iv) Negativenumbersarenotrationalnumbers. (v) Inanyrationalnumber qcanbezero.

2. Representeachrationalnumberonthenumberline.

(i)5

2-

(ii)23 (iii)

415 (iv)

324

-

2.2 Operations on Rational Numbers

Inthissection,weperformoperationofaddition,subtraction,multiplicationanddivisiononrationalnumbers.

2.2.1 Addition of Rational Numbers

(a)If and areanytworationalnumberswiththesamedenominators,thenweshalladdthemasgivenbelow.

Example 1: Simplifythefollowingrationalnumbers.

Solution:

12 11 10 9 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12..., , , , , , , , , , , , ,0, , , , , , , , , , , , ,...4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4

- - - - - - - - - - - - + + + + + + + + + + + +


6 7



(b)If and areanytworationalnumbers,whereq,sm 0,whosedenominatorsaredifferent,thenwecanaddthembythefollowingformula.

Example 2:Findthesumofthefollowingrationalnumbers.

Solution:

2.2.2 Subtraction of Rational Numbers

(a)Considertowrationalnumberswiththesamedenominators.Thedifference isasunder:

Example 3:Simplifythefollowing.

Solution:

(b)Considertworationalnumbers withdifferentdenomina-tors.Thedifferent isasunder:

Example 4: Simplify.

Solution:


8 9



2.2.3 Additive Inverse

Considerthat areanytworationalnumbers,thenwecanaddthembythefollowingmethod.

Wecanexaminethatthesumofthesetworationalnumbersiszero.

Hence,tworationalnumbers arecalledadditiveinverse

of each other and 0 is known as additive identity. For example,

etc.allareadditiveinverseofeachother.

Example 5: Write the additive inverse of the following rationalnumbers.

(i) 3 (ii)12

- (iii)74

Solution: (i) Tofindtheadditiveinverseof3,changeitssign. Additiveinverseof3is-3 Check: 3+(-3)=3-3=0

(ii) Tofindtheadditiveinverseof12

- changeitssign.

(iii) Tofindtheadditiveinverseof changeitssign.

2.2.4 Multiplication of Rational Numbers

Wecanfindtheproductoftwoormorerationalnumbersbythegivenrule.Rule: Multiply the numerator of one rational number by thenumerator of the other rational number. Similarly, multiply thedenominatorsofbothrationalnumbers,i.e.

Example 6: Findtheproductofthefollowingrationalnumbers.

Solution:

2.2.5 Multiplicative Inverse Considertworationalnumbers wherepm0andqm0.Wefindtheirproductbythefollowingformulaasunder

Wecannoticethattheproductofthesetworationalnumbersis1.Hence, tworationalnumbers areknownasmultiplicativeinverse of each other and 1 is called the multiplicative identity.Forexample,2and1

2,-5and etc.allaremultiplicative

inverseofeachother.


10 11



Example 7: Findthemultiplicativeinverseofthefollowingrationalnumbers. (i) -4

Solution: (i) -4To find the multiplicative inverse of -4 , write the numerator asdenominatoranddenominatorasnumerator.Multiplicativeinverseof-4is

• Foranynon-zerorationalnumber therationalnumber iscalleditsreciprocal.

• Thenumber0hasnoreciprocal.• Themultiplicative inverseofanon-zerorationalnumber is its

reciprocal.

2.2.6 Division of Rational Numbers

Weknowthatdivisionisaninverseoperationofmultiplication.So,wecandotheprocessofdivisioninthefollowingsteps.Step 1:Findthemultiplicativeinverseofdivisor.

Step 2: Multiply it by the dividend, according to the rule ofmultiplication,i.e.

Example 8: Simplify:

Solution:

2.2.7 Finding Reciprocal of a Rational Number

Consideranon-zerorationalnumber whichismadeupoftwointegers3,asnumeratorand7asdenominator.Ifweinterchangetheintegersinnumeratoranddenominator,wegetanotherrationalnumber Ingeneralforanynon-zerorationalnumber wehaveanother non-zero rational number This number is called thereciprocalof Thenumber isthereciprocalof Likewise,

is the reciprocal of and is the reciprocal of

Weobservefromherethatif isthereciprocalof then isthereciprocalof Inotherwords, and arereciprocalsofeachother.


12 13



EXERCISE 2.2

1. Find the additive inverse and multiplicative inverse of thefollowingrationalnumbers.(i)-7(ii)23(iii)-11

(vi)6(vii)1

2. Simplifythefollowing.

3. Simplify:

• Properties of Rational NumbersTherationalnumbersalsoobeycommutative,associativeanddistributiveproperties likewholenumbers,fractions, integers,etc.Letusverifyitwithexamples.

2.2.8 Commutative Property

• Commutative Property of Rational Numbers w.r.t Addition Considerthat and areanytworationalnumbers,thenaccordingtothecommutativepropertyofaddition,wehave:

Example 1: Provethat

Solution:

• Commutative Property of Rational Numbers w.r.t Multiplication

Accordingtocommutativepropertyofmultiplication,foranytworationalnumbers and wehave:


Solution:

Result: Commutative property with respect to addition andmultiplicationholdstrueforrationalnumbers.


14 15



2.2.9 Associative Property

• Associative Property of Rational Numbers w.r.t Addition Considerthat and arethreerationalnumbers,thenaccordingtotheassociativepropertyofaddition,wehave:


Solution:

• Associative Property of Rational Numbers w.r.t Multiplication Accordingtoassociativepropertyofmultiplication, foranythreerationalnumbersandwehave:


Solution:

Result: Associative property with respect to addition andmultiplicationholdstrueforrationalnumbers.

2.2.10 Distributive Property of Multiplication over Addition and Subtraction Nowagainconsiderthethreerationalnumbers and thenaccordingtothedistributiveproperty:


Solution:


16 17



2.3.11 Comparison of Rational Numbers

Wehavestudiedthecomparisonofintegersandfractionsinourpreviousclass.Similarly,wecancomparetherationalnumbersbyusingthesamerulesforcomparison.Weshallmakeitclearwithexamples.

• Case I: Same Denominators

Example 6: Comparethefollowingpairsofrationalnumbers.

Solution:

• CaseII:DifferentDenominators

Example 7: Put thecorrect sign>or<between the followingpairsofrationalnumbers.

Solution:

Writeothertworationalnumbersfromthegivenrationalnumberssuchthattheirdenominatorsmustbeequal.

Nowcomparethenumeratorsofrationalnumberswiththesamedenominators.

Bymakingtheirdenominatorsequal

Nowcomparethenumeratorsofrationalnumberswiththesamedenominators.


18 19



2.3.12 Arranging Rational Numbers in Orders

Wecanalsoarrangethegivenrationalnumbersinascendingorder(lowesttohighest)andindescendingorder(highesttolowest)inthefollowingsteps.Step 1:FindtheL.C.Mofthedenominatorsofgivenrationalnumbers.Step 2:Rewritetherationalnumberswithacommondenominator.Step 3:Comparethenumeratorsandarrangetherationalnumbersinascendingordescendingorder.

Example 8: Arrange the rational numbers indescendingorder.Solution:Step 1: TheL.C.Mofdenominators2,3and8is24.Step 2: Rewritetherationalnumberswithacommondenominatoras,

Step 3: Comparethenumerators12,16and21andrearrangetherationalnumbersindescendingorder.

21>16>12

Thus,arrangingindescendingorder,weget

Example 9: Arrange the rational numbers inascendingorder.Solution:Step 1: TheL.C.Mofdenominators4,3and12is12.Step 2: Rewritetherationalnumberswithacommondenominatoras,

Step 3: Compare the numerators 3, 8 and 1 and rearrange therationalnumbersinascendingorder.

1<3<8

Thus,arranginginascendingorder,weget

EXERCISE 2.3

1. Putthecorrectsign>,<or=betweenthefollowingpairsofrationalnumbers.

2. Arrangethefollowingrationalnumbersindescendingorder.

3. Arrangethefollowingrationalnumbersinascendingorder.

4.Provethat:


20 21



REvIEw EXERCISE 2

1. Answerthefollowingquestions. (i) Definearationalnumber. (ii) Writetheadditiveinverseoftherationalnumbers“a”. (iii) Whatisthereciprocaloftherationalnumber qm0?

(iv) Writethesumoftworationalnumbers and q, rm 0? (v) What is the rule to find the product of two rational numbers? (vi) What are the inverse operations of addition and multiplication?

2. Fillintheblanks. (i) The________consistsoffractionsaswellasintegers. (ii) Therationalnumbers and arecalled____inverseof eachother. (iii) Anumberthatcanbeexpressedintheformof where p and q are integers and qm? 0 is called the________ number. (iv) 0 is calledadditive identitywhereas1 is called ________ identity. (v) Therationalnumber0hasno________. (vi) The________inverseofarationalnumberisitsreciprocal.


4. Draw thenumber lines and represent the following rational numbers.

5. Find the additive andmultiplicative inverse of the following rationalnumbers. (i)-14

6. Put the correct sign > or < between the following pairs of rationalnumbers.

7. Solvethefollowing.

8. Simplifythefollowing.

9. Provethat:

SuMMARy

• Every integer can be divided by another non-zero integer, thenumber obtained is called a rational number and is writtensymbolicallyas .

1. Quadratic EQuations eLearn.Punjab

22

2. Rational Numbers eLearn.Punjab

22Version 1.1

• Additionofrationalnumberswith:Samedenominators.Differentdenominators.

• Subtractionofrationalnumberswith:Samedenominators.Differentdenominators.

• To find the product of two rational numbers, multiply thenumeratorofonerationalnumberbythenumeratoroftheother.Similarly,multiplythedenominators.

• Divisionisaninverseoperationofmultiplication.So,foranytworationalnumbers.

• 0iscalledadditiveidentityand1iscalledmultiplicativeidentity.• iscalledthereciprocalof

• If aretworationalnumbers,thenaccordingtothecommutativeproperty:

• If and arethreerationalnumbers,thenaccordingtotheassociativeproperty.

• Now again consider the three rational numbers and thenaccordingtothedistributiveproperty:

CHAPTER

3 Decimals

Animation 1.1: Introduction to DecimalsSource & Credit: eLearn.Punjab

Version 1.1



3

3. DecimalseLearn.Punjab eLearn.Punjab

2

3. Decimals



After studying this unit, students will be able to:• Convertdecimalstorationalnumbers.• Defineterminatingdecimalsasdecimalshavingafinitenumber

ofdigitsafterthedecimalpoint.• Definerecurringdecimalsasnon-terminatingdecimalsinwhicha

singledigitorablockofdigitsrepeatsitselfaninfinitenumberoftimesafterdecimalspoint(e.g.=0.285714285714285714....)

• Usethefollowingruletofindwhetheragivenrationalnumberisterminatingornot.

• Rule:Ifthedenominatorofarationalnumberinstandardformhasnoprimefactorotherthan2,5or2and5,thenandonlythentherationalnumberisaterminatingdecimal.

• Expressagivenrationalnumberasadecimalandindicatewhetheritisterminatingorrecurring.

• Getanapproximatevalueofanumber,calledroundingoff,toadesirednumberofdecimalplaces.

Introduction

Inthepreviousclasses,wehavelearntthatadecimalconsistsoftwoparts,i.e.awholenumberpartandadecimalpart.Toseparatethese parts in a number, we place a dot between themwhich isknownasthedecimalpoint.

Decimal point

Whole number part Decimal part

So,wecandefineadecimal;anumberwithadecimalpointiscalledadecimal.

3.1 Conversion of Decimals to Rational Numbers

We take the following steps to convert decimals to rationalnumbers.Step 1:Write“1”belowthedecimalpoint.Step 2:Addasmanyzerosasthedigitsafterthedecimalpoint.Step 3:Reducetherationalnumbertothelowestform.

Example 1: Convert 0.12 to arationalnumber.

Solution:

Example 2: Convert2.55toarationalnumber.

Solution:

Example 3: Convert-1.375toarationalnumber.

Solution:

The word “decimal” has beendeduced from a latin word“decimus”whichmeansthetenth.

Do you Know


5


4

3. Decimals


EXERCISE 3.1

1. Convertthefollowingdecimalsintorationalnumbers. (i) 0.36 (ii) 0.75 (iii) –0.125 (iv) –6.08 (v) 6.46 (vi) 15.25 (vii)8.125 (viii) –0.00625 (ix) –0.268

3.2 Terminating and Non-Terminating Decimals

Decimalscanbeclassifiedintotwoclasses.(i)TerminatingDecimals(ii)Non-terminatingDecimals

3.2.1 Terminating Decimals Look at the conversion of rational numbers intodecimals.

Intheaboveexample,weobservethatafterafinitenumberofsteps,weobtainazeroasremainder.Suchrationalnumbers,forwhich longdivision terminatesafterafinitenumberof steps, canbeexpressedindecimalformwithfinitedecimalplacesandthesedecimalsarecalledterminatingdecimalswhichcanbedefinedas;“Adecimalinwhichthenumberofdigitsafterthedecimalpointisfinite,iscalledaterminatingdecimal.”

Example 1: Expresseachrationalnumberasadecimal.

Solution:

3.2.2 Non-Terminating Decimals In somecaseswhile convertinga rationalnumber intoa decimal, division never ends. Such decimals are called non-terminationdecimalsasshowninthefollowingexamples.


7


6

3. Decimals


So,wecandefineanon-terminatingdecimalas;“Adecimalinwhichthenumberofdigitsafterthedecimalpointareinfinite,iscalledanon-terminatingdecimal”.Fromtheaboveexamples,itcanalsobeobservedthatasingledigitorablockofdigitsrepeatsitselfaninfinitenumberoftimesafterthedecimalpointinsuchdecimals.i.e.• In0.3333...,thedigit3repeatsitselfaninfinitenumberfortimes.• In0.2727...,theblockofdigits27repeatsitselfaninfinitenumber

oftimes.• In0.1666...,thedigit6repeatsitselfaninfinitenumberoftimes.Thenon-terminationdecimals inwhichasingledigitorablockofdigitsrepeatsitselfinfinitenumberoftimesafterthedecimalpointarealsocalledrecurringdecimals.

Example 2: Changetherationalnumbersintonon-terminatingdecimals.

Solution:

3.2.3 Rule to find whether a given rational is terminating or not

Wehave learntthatthedivisionprocessterminatesforsomerationalnumbersanddoesnotterminateforcertainotherrationalnumbers.• Terminating Decimals

• Non-terminating Decimals

Fromtheaboveexamples,itcanbeobservedthatarationalnumbercanbeexpressedasa terminatingdecimal if itsdenominatorhasonlyprimefactors2and5,otherwiseitisanon-terminatingdecimal.So,wecanusethefollowingruletofindwhetherthegivenrationalnumberisterminatingornot.

Rule:Ifthedenominatorofarationalnumberinstandardformhasnoprimefactorotherthan2,5or2and5,thenandonlythentherationalnumberisaterminatingdecimal.

Example 3: Without actual division, separate terminating andnon-terminatingdecimals.


9


8

3. Decimals


Solution:

(i)97

97

isanon-terminatingdecimalbecauseitsdenominatoris7.

(ii)178

178

is a terminating decimal because its denominator has prime

factors2x2x2=8

(iii)206

Writeinthestandardformofthegivenrationalnumber.20 20 2 106 6 2 3

÷= =

÷

206

is a non-terminating decimal because the denominator of its

standardformis3.

(iv)4525

Thestandardformof45 45 5 925 25 5 5

÷= =

÷.

4525

isaterminatingdecimalbecausethedenominatorofitsstandard

formis5.

3.2.4 Expressing a Rational Number as a Decimal to indicate whether it is Terminating or Recurring

Example 4: Express the rational numbers as decimals. Alsoseparateterminatingandrecurringdecimals.

Solution:

3.3 Approximate Values

Wheneverwecomeacrossthenon-terminatingdecimals,itisverydifficulttosolvetheproblemswithoutthehelpofacalculator.Evencalculatorsalsohave limitations.Therefore, inordertosolvesuchkindsofproblems,weroundoffthedecimals.• RoundoffHerethetermroundoffisusedtoleavethedigitsafterthedecimalpoint.Thefollowingarethestepstoroundoffadecimal.Step 1:Decidehowmanydigitsweneedafterthedecimalpoint.Step 2:Droptheremainingdigitsoff,ifthefirstmostdigitwewanttoleaveislessthan5.Andifitis5ormore,thenadd1tothe

Thesymbol“c“means“approximatelyequalto”.


11


10

3. Decimals


requiredlastdigitbeforedroppingtheremainingdigits.Itwillbeeasierforustounderstandthismethodwithsomeexampleswhicharegivenbelow.

Example 4: Round off the following decimals up to: (a) 3-decimalplaces (b) 2-decimalplaces (i) 2.3427 (ii) 4.7451 (iii) 1.5349

Solution: (i) 2.3427(a) Thedigitnextto3-decimalplacesis7(greaterthan5).So,we increasethedigit2byone.i.e.2.3427c2.343(b) Thedigitnextto2-decimalplacesis2(lessthan5).So,weignore theremainingdigitswithoutanychange.i.e.2.3427c2.34 (ii)4.7451(a) The digit next to 3-decimal places is 1 (less than 5). So, we ignore the remaining digits without any change. i.e. 4.7451c4.745(b) The digit next to 2-decimal places is 5 (equal to 5). So, we increasethedigit4byone.i.e.4.7451c4.75 (iii)1.5349(a) Thedigitnextto3-decimalplacesis9(greaterthan5).So,we increasethedigit4byone.i.e.1.5349c1.535(b) The digit next to 2-decimal places is 4 (less than 5). So, we ignore the remaining digits without any change. i.e. 1.5349c1.53

EXERCISE 3.2

1. Without actual division, separate the terminating and non-terminatingdecimals.

2. Expressthefollowingrationalnumbersinterminatingdecimals.

3. Express the following rational numbers in non-terminating decimalsuptothreedecimalplaces.

4. Roundoffthefollowingdecimalsuptothreedecimalplaces.

REvIEW EXERCISE 3

1. Answer the following questions. (i) Definetheterminatingdecimals. (ii) Writethenamesoftwoclassesofdecimals. (iii) Which of the non-terminating decimals are called recurringdecimal? (iv) How many digits after a decimal point show a non- terminatingdecimal? (v) Writetheruletofindwhetheragivenrationalnumberis terminatingornot. (vi) Whatismeantbythetermroundoffindecimals?

2. Fill in the blanks. (i) A________decimalmayberecurringornon-recurring. (ii) Twopartsofdecimalnumberseparatedbyadotiscalled the________. (iii) In terminating decimals, division ________ after a finite numberofsteps.

1. Quadratic EQuations eLearn.PunjabeLearn.Punjab

12

3. Decimals

Version 1.1

(iv) Indecimals,thetermroundoffisusedtoleavethedigits afterthe________. (v) Afractionwillbeterminatingifthe________has2or5or bothasfactors.

3. Tick (p) the correct answer.

4. Convert the following decimals into rational numbers. (i) 0.375 (ii) 0.25 (iii) 0.5 (iv) 4.75 (v) 0.79 (vi) 1.29 (vii) 2.34

5. Convert the following into decimal fractions and identify terminating and non-terminating fractions.

(i)45 (ii)

1112

(iii)89 (iv)

17

(v)227 (vi)

216 (vii)

310

6. Roundoffthefollowingupto2-decimalplaces. (i) 4.5723 (ii) 107.328 (iii) 5.7395 (iv) 6.7982 (v) 25.4893

SummARy

• Everydecimalwithfinitedigitsafterthedecimalpointiscalledaterminatingdecimal.

• Aterminatingdecimalrepresentsarationalnumber.• A decimal with infinite digits after a decimal point is called a

non-terminatingdecimal.• Anonterminatingdecimalmayberecurringornon-recurring.• Decimals can be reduced by rounding off the digits after the

decimalpoint.• Afractionwillbeterminatingifthedenominatorinstandardform

has2or5orbothasfactors.

CHAPTER

4 EXPONENTS

Animation 4.1: ExponentsSource & Credit: elearn.punjab

Version 1.1



2 3

4. Exponents 4. ExponentseLearn.Punjab eLearn.Punjab



After studying this unit, students will be able to:• Identifybase,exponentandvalue.• Userationalnumberstodeducelawsofexponents.• Product Law: whenbasesaresamebutexponentsaredifferent:

amxan=am+n

whenbasesaredifferentbutexponentsaresame:anxbn=(ab)n

• Quotient Law: whenbasesaresamebutexponentsaredifferent:

whenbasesaredifferentbutexponentsaresame:

• Power Law: (am)n=amn

Forzeroexponent:a0=1Forexponentasnegativeinteger:

• Demonstratetheconceptofpowerofintegerthatis(–a)nwhennisevenoroddinteger.

• Applylawsofexponentstoevaluateexpressions.

4.1 Exponents/Indices

4.1.1 Identification of Base, Exponent and Value

We have studied in our previous class that the repeatedmultiplication of a number can be written in short form, usingexponent.Forexample,• 7×7×7canbewrittenas73.

Wereaditas7tothepowerof3where7isthebaseand3istheexponentorindex.

Similarly,• 11×11canbewrittenas112.Wereaditas11tothepowerof2

where11isthebaseand2istheexponent.Fromtheaboveexampleswecanconcludethatifanumber“a” ismultipliedwithitselfn–1times,thentheproductwillbean,i.e.an=axaxax...................xa(n-1timesmultiplicationsof“a”withitself)Wereaditas“atothepowerofn”or“nthpowerofa”where“a”isthebaseand“n”istheexponent.Example 1: Expresseachofthefollowinginexponentialform.(i) (-3)x(-3)x(-3) (ii) 2x2x2x2x2x2x2

(iii)1 1 1 14 4 4 4

× × ×

(iv)7 7

12 12- - ×

Solution:(i) (-3)x(-3)x(-3)=(-3)3 (ii) 2x2x2x2x2x2x2=(2)7

(iii)41 1 1 1 1

4 4 4 4 4 × × × =

(iv)27 7 7

12 12 12- - - × =

Example 2: Identifythebaseandexponentofeachnumber.

(i) 1325(ii) 97

11-

(iii) am(iv) (- 426)11(v) na

b

(vi)t

xy

-

Solution:(i) 1325 (ii)

9711

-

(iii) am

base=13 base=a

exponent=25 base=7

11-

exponent=m

exponent=9

(iv) (-426)11 (v) na

b

(vi)t

xy

- base=- 426

exponent=11 base=ab base=

xy

-

exponent=n exponent=t

Theexponentofanumberindicatesus,howmanytimesanumber(base)ismultipliedwithitself.


4 5



Example 3: Writethefollowinginthesimplestform.

(i) (-5)3 (ii)22

3

(iii)41

4-

Solution:(i) (-5)3=(-5)x(-5)x(-5)(ii) =(+25)x(-5)=-125Thus,(-5)3=-125

(iii)

EXERCISE 4.1

1. Identifytheexponentandbaseineachofthefollowing. (i) (-1)9 (ii) 2100 (iii) (-19)22 (iv) 3-5

(v) (ab)n (vi) 86

11-

(vii) a-mn (viii) 72

9

(ix)4

pq

(x)61

x -

(xi)m

xy

(xii)1113

b

2. Expresseachofthefollowinginexponentialform.

(i) 5x5x5x5 (ii)3 3 3 3

7 7 7 7- - - -

× × ×

(iii) pxpxpxpxp (iv) 1 1 1

10 10 10× ×

(v) xyxxyxxy (vi) 31x31x31x31x31 (vii) (-a)x(-a)x(-a)x(-a)x(-a)x(-a)x(-a)

3. Provethat: (i) (5)3=125 (ii) (-1)11=-1 (iii) (-3)5=-243

(iv)23 9

7 49 =

(v)31 1

8 512 - = -

(vi)62 64

3 729- =

(vii)41 1

10 10000 =

(viii)34 64

3 27- - =

(xi)

42 165 625

=

4. Expresseachrationalnumberusinganexponent. (i) 121 (ii) 81 (iii) -625

(iv)1

1000 (v)

8343

(vi)1

32-

4.2 Laws of Exponents/Indices

Exponentsareusedinsolvingmanyproblems,soitisimportantthat we understand the laws for working with exponents. Let usdiscusstheselawsonebyone,andseesomeexamples.

4.2.1 Using Rational Numbers to Deduce Laws of Exponents

• Product Law• WhenbasesaresamebutexponentsaredifferentConsider the following examples

23x22=(2x2x2)x(2x2)=2x2x2x2x2=25

Fromtheabove,wecannoticethatthesameresultcanbeobtainedbyaddingtheexponentsoftwonumbers.

23x22=23+2=25

Similarly,

2

2

2 2 23 3 3

2 2 4 3 3 92 4Thus, = 3 9

= ×

×= =

×

4

4

1 1 1 1 14 4 4 4 4

1 1 1 1 1 4 4 4 4 2561 1Thus, =

4 256

- = - × - × - × -

- × - × - × -= =

× × ×

-

2 23 3 3 3 3 3 3 3 34 4 4 4 4 4 4 4 4

- - - - - - - - - × = × × × × × ×


6 7



734

- =

Againusetheshortmethodtofindtheresult.

2 5 2 5 73 3 3 34 4 4 4

+- - - - × = =

Fromtheaboveexamples,wecandeducethefollowinglaw:“Whilemultiplying two rational numbers with the same base, weaddtheirexponentsbutthebaseremainsunchanged, i.e. foranynumber“a”withexponentsmandn,thislawiswrittenas,

amxan=am+n

• WhenbasesaredifferentbutexponentsaresameWeknowthat

23x53=(2x2x2)x(5x5x5)=(2x5)x(2x5)x(2x5)=(2x5)3

Similarly,

3 31 3 1 1 1 3 3 34 4 4 4 4 4 4 4- - - - × = × × × × ×

31 3 1 3 1 3 1 34 4 4 4 4 4 4 4

= - × × - × × - × = - ×

Fromtheaboveexamples,wecandeducethefollowinglaw:“Whilemultiplyingtworationalnumbershavingthesameexponent,theproductof the twobases iswrittenwith thegivenexponent.”Suppose two rational numbers are “a” and “b” with exponent “n”then,

anxbn=(ab)n

Example: Simplifythefollowingexpressions. (i) 53x54 (ii) (-3)3x(-2)3

(iii)2 21 2

4 3- ×

(iv)

3 43 32 2

- - ×

Solution: (i) 53x54

=53+4=57 [a amxan=am+n]

(ii)(-3)3x(-2)3

=[(-3)x(-2)]3=[6]3 [a anxbn=(ab)n]

(iii)2 21 2

4 3- ×

21 2

4 3[ ( ) ]n n na b ab - = ×

× =

2 21 2 1

4 3 6- × - = = ×

(iv)3 43 3

2 2[ ]m n m na a a +- - ×

× =

3 4 73 32 2

+- - = =

EXERCISE 4.21. Simplify theusingthe lawsofexponent intotheexponential form.

(i) (-4)5x(-4)6 (ii) m3xm4 (iii)3 22 2

7 7 ×

(iv)4 51 1

10 10 ×

(v) p10xq10 (vi)3 32 5

5 7 ×

(vii)6 51 1

2 2- - ×

(viii) (-3)7x(-5)7 (ix)

10 72 23 3

×

(x)7 610 10

11 11- - ×

(xi)

8 811 217 22

×

(xii)11

x xy y

- -×


8 9



2. Verifythefollowingbyusingthelawsofexponent. (i) (3x5)4=34x54 (ii) (7x9)8=78x98

(iii) (2)6x(2)3=29 (iv) (xxy)m=xmym

(iv) (8)5x(8)7=(8)12 (v) (p)rx(p)s=pr+s

• Quotient Law• WhenbasesaresamebutexponentsaredifferentConsiderthefollowing.

7

3

2 2 2 2 2 2 2 22 2 2 2

× × × × × ×=

× ×

42 2 2 2 2= × × × =Letusfindthesamequotientbyanotherway.

77 3 4

3

2 2 22

-= =

Similarly,

Accordingtotheshortmethodthatweusedforfindingthequotient:

Thus, from the above examples we can suggest another law;“Thedivisionof tworationalnumberswith thesamebasecanbeperformedbysubtractingtheirexponents”.Suppose‘a’isthebaseof any two rationalnumberswithexponents ‘m’ and ‘n’ such thata≠ 0andm>n,then,

am'an=am-n

• WhenbasesaredifferentbutexponentsaresameWeknowthat:

44 4

4

2 2 2 2 2 2 33 3 3 3 3

× × ×= = = ÷

× × × Similarly,

5x x x x x xy y y y y y

= × × × ×

55 5

5

x x x x x x x yy y y y y y

× × × ×= = = ÷

× × × ×Thus,thislawcanbewrittenas:Foranytworationalnumbers“a ”and“b”,whereb≠0and“n”istheirexponent,then,

Example: Simplify.

(i) 98' 38(ii)7 43 3

11 11 - ÷ -

(iii)9 23 3

7 7 ÷

(iv) (14)11'(63)11

Solution:

(i) 98'38 (ii)

(iii) (iv) (14)11'(63)11

5 2

3

2 2 2 2 22 2 3 3 3 3 3

2 23 33 3

2 2 2 2 3 3 3 3

- - - - - × × × × - - ÷ = - - ×

- - - - = × × =

5 2 5 2 32 2 2 23 3 3 3

- - ÷ - = - = -

42 2 2 2 23 3 3 3 3

= × × ×

889 3

3

nn n aa b

b = = ÷ =

7 43 311 11

- ÷ -

7 4 33 3 11 11

m n m na a a-

-- = = - ÷ =

9 2 23 37 7

- ÷ 9 2 73 3

7 7m n m na a a

-- = = ÷ =

11 1114 2 63 9

nn n aa b

b = = ÷ =

nn n aa b

b ÷ =


10 11



EXERCISE 4.31. Simplify

(i) 27'22 (ii) (-9)11'(-9)8 (iii) (3)4'(5)4

(iv) (m)3'(n)3 (v) (a)7'(a)2 (vi) (b)p'(b)q

(vii)7 23 3

4 4 ÷

(viii)15 111 1

6 6 ÷

(ix) (2)5'(3)5

(x)17 83 3

10 10- - ÷

(xi) (x)a'(y)a (xii)

23p pq q

÷

2. Provethat

(i)4

4 4 22 77

÷ =

(ii)3

3 3 4( 4) (5)5

- - ÷ =

(iii) 38'3=37

(iv)6

6 6 aa bb

÷ =

(v)7 3 421 21 21

22 22 22- - - ÷ =

(vi)5 49 9 9

13 13 13- - - ÷ =

• Power LawWehavestudiedthatam×an=am+n.Letususethislawtosimplifyanexpression(34)2.(34)2=34x34

=34+4=38isthesameas34x2

Wesolveanotherexpressionusingthesamelaw.

Thus, from the above examples, we can deduce that the baseremainsthesamewithanewexponentequaltotheproductofthetwoexponents,thatis: (am)n = amxn = amn

• Zero ExponentBythequotientlaw,weknowthatanythingdividedbyitselfis1asshownbelow.

Thiscanalsobewrittenas32-2=30=1Similarly,

Thiscanalsobewrittenas(-2)4-4=(-2)0=1.Thus,wecandefinethislawas:Any non-zero rational number with zero exponent is equal to 1.Suppose “a” be any non-zero rational numberwith exponent “0”,then a0=1• Negative ExponentsLookatthepatterngivenbelow.102=10x10101=10100=1

...................................................

...................................................

...................................................

Wecanalsodeducethis lawfromamxan=am+n.Supposen=-m,thenwewillget,

Dividedbyamonbothsides.

Thus,wehaveanotherlaw:

27 7 71 1 12 2 2

- - - = ×

7 7 14 7 21 1 1 is also the same as 2 2 2

+ ×- - - = =

2

2

3 3 3 13 3 3

×= =

×

4

4

( 2) ( 2) ( 2) ( 2) ( 2) 1( 2) ( 2) ( 2) ( 2) ( 2)

- - × - × - × -= =

- - × - × - × -

1 11010

- =

22

1 1 1 11010 10 10 10 10

- = × = =×

1 11010 10 10( times) 10

mmm

- = =× ×⋅⋅ ⋅×

1In general, it can be written as; mma

a- =

1 1m mm

m m m

a a aa a a

--×

= ⇒ =

0 0 1 1m m m m m m m ma a a a a a a a a- - - -× = ⇒ × = ⇒ × = =


12 13



Anynon-zeronumberraisedtoanynegativepower isequalto itsreciprocalraisedtotheoppositepositivepower.i.e.

Ifpq is anon-zero rational number, thenaccording to the above

givenlaw,wehave:

Thus,

1 1m mm

m m m

m

p p ppq q qpqq

- = = = =

m mp pq q

- =

Example 1: Expressthefollowingasasingleexponent.

(i) (34)5 (ii)

2323

-

(iii)

6517

Solution:

(i) (34)5a(am)n=amn (ii) (iii)

=34x5

=320

Example 2: Change the following negative exponents intopositiveexponents.

(i)33

4

-

(ii)42

5

--

(iii)6a

b

-

-

Solution: (i)33

4

-

(ii) (iii)

4.2.2 Demonstration of the concept of Power of an Integer

Weknowthatwhenwemultiplyanegativenumberbyitself,itgivesapositiveresultbecauseminustimeminusisplus.Forexample, (-3)x(-3)=(-3)2=+9 (-5)x(-5)=(-5)2=+25Butdoyouknowithappenstoallevenexponentsthatcanbeseeninthepatterngivenbelow.

(-2)2=(-2)x(-2)=+4....................................................................(even)(-2)3=(-2)x(-2)x(-2)=-8..............................................................(odd)(-2)4=(-2)x(-2)x(-2)x(-2)=+16................................................(even)(-2)5=(-2)x(-2)x(-2)x(-2)x(-2)=-32........................................(odd)(-2)6=(-2)x(-2)x(-2)x(-2)x(-2)x(-2)=+64.............................(even)

Fromtheaboveitcanalsobenoticedthatanegativenumberwithanoddexponentgivesanegativeresult.So,wecanexplainitas:

Let “a” be any positive rational number and “n” be any non-zerointeger,thanaccordingtothislaw:

1mma

a- =

232 ( ) = 3

m n mna a -

651 ( ) = 7

m n mna a

3 2 62 23 3

×- - = =

5 6 301 17 7

× = =

31 1 34

mma

a-= =

3 3 33

3 33

1 4 4 3 4= = Thus, 3 3 3 4 3

4

- = =

425

--

6ab

-

-

41 1 2

5

mma

a-= =

-

4 44

4 4

4

1 5 5 5= = or ( 2) 2 2( 2)

5

- = - - -

4 42 5Thus, =5 2

-- -

61 1 m

maaa

b

-= =

-

66

6 6

6

1 ( )=

( )

b ba a ab

- - = =

-6 6

Thus, =a bb a

- -

-


14 15



• If“n”isaneveninteger,then(–a)nispositive.• If“n”isanoddinteger,then(–a)nisnegative.

4.2.3 Applying Laws of Exponent to Evaluate Expressions

Example 3: Simplifyandexpresstheresultinthesimpleform.

(i) (47' 45)x22 (ii)3 3 5 52 2 3 3

5 5 5 5

- - × × ×

(iii)

15 2 22 2 27 7 7

-- - - - × ×

Solution:

(i) (47' 45)x22 (iii)

=47-5x22

=42x22aam ' an = am-n

=(4x2)2aan x bn = (ab)n

=82=64

(ii)

EXERCISE 4.4

1. Expressthefollowingassingleexponents. (i) (23)5 (ii) (102)2 (iii) (-34)5

(iv) (p2)3 (v) (-m7)4 (vi) (x a)b

(vii)

3313

-

(viii)

6329

(ix)

nmpq

2. Change the following negative exponents into positive exponents. (i) (12)-3 (ii) (-a)-2 (iii) (100)-5

(iv)42

3

-

(v)91

10

--

(vi)bx

y

-

3. Evaluatethefollowingexpressions. (i) (12)3x(23)2 (ii) [(-3)7]0x[(-3)2]2

(iii)

3 20 23 34 4

- - ×

(iv)3

6 3

22 2

÷

(v)

3 6

5

1 12 2

12

- -

-

×

(vi)

5 5

4 4

2 29 93 32 2

-

-

- - ×

×

(vii)

3 5

4 6

1 13 31 13 3

- -

- -

-

(viii)

5 4

4 4

2 23 32 23 3

-

- -

× ×

(ix)3 0 32 2 2

3 3 3

- × ×

(x)2 3 41 1 1

2 3 4

- - -- + +

15 2 22 2 27 7 7

-- - - - × ×

3 3 5 52 2 3 35 5 5 5

- - × × ×

0 02 35 5

+

3+3 5+( 5)2 35 5

m n m na a a- -

+ + × =

01 1 2 1a+ = =

5 (2) 2 ( 1)2 27 7 ( )

m n m n

m n mn

a a aa a

+ × - +× =- - = × =

3 22 27 7

-- - = ×

3 ( 2)2 7

m n m na a a+ -

+- = × =

3 22 27 7

-- - = =


16 17



REvIEw EXERCISE 4

1. Answerthefollowingquestions. (i) Whatismeantbytheexponentofanumber? (ii) Whatistheproductlawwiththesamebase? (iii) Definethepowerlawofexponent.

(iv) Whatisthereciprocalofpq?

2. Fillintheblanks. (i) 5×5×5×5canbewritteninexponentialformas_________. (ii) anxbn=__________. (iii) an'bn=___________. (iv) Anynon-zerorationalnumberwith__________exponent equalsto1. (v) (-a)nispositive,if‘n’isan_______integer. (vi) _________isreadas‘nthpowerofa’.


4. Findthevalueof:

(i) (4)-3 (ii) (-5)4 (iii) (2)-9

(iv)51

3

--

(v)33

10

(vi)211

13 -

5. Usethelawsofexponentstofindthevalueof x.

(i) [(-7)3]6 =7x (ii)

523 34 4

x

x

=

(iii)413 134

8 8

x

x

=

(iv)5 11 85 5 5

3 3 3

x × =

(v)2 9 2 12 2 2

9 9 9

x- ÷ =

6. Simplifyandwritetheanswerinsimpleform.

(i)22 3 23 3 3

4 4 4 - - - × ÷

(ii)3 410 2 45 5 5

19 19 19 × ÷

(iii)5 23 2 218 18 18

11 11 11 ÷ ÷

(iv)8 52 34 4 4

9 9 9 - - - ÷ ×

(v)2 33 6 251 1 1

10 10 10 × ÷

SummARy

• Theexponentofanumberindicatesushowmanytimesanumber(base)ismultipliedwithitself.

• While multiplying two rational numbers with the same base,we add their exponents but the base remains unchanged. i.e.am×an=am+n

• Whilemultiplyingtworationalnumbershavingsameexponent,theproductoftwobasesiswrittenwiththegivenexponent.i.e.an×bn=(ab)n


18

4. Exponents eLearn.Punjab

18Version 1.1

• Thedivisionoftworationalnumberswiththesamebasecanbeperformedbysubtractingtheirexponents.i.e.am'an=am–n

• Toraiseapowertoanotherpower,wejustwritetheproductoftwoexponentswiththesamebase.i.e.(am)n=amn

• Anynon-zero rationalnumberwith zeroexponentequals to1,i.e.a0=1

• Anynon-zerorationalnumberwithanegativeexponentequalsto

itsreciprocalwiththesamebutpositiveexponent.i.e.1mma

a- =

• (-a)nispositive,ifnisanevenintegerand(-a)nisnegative,ifnisanoddinteger.

CHAPTER

5 SQUARE ROOT OF POSITIVE NUMBER

version: 1.1


2 3

5. Square Root of Positive Number 5. Square Root of Positive NumbereLearn.Punjab eLearn.Punjab

version: 1.1 version: 1.1


After studying this unit, students will be able to:

• Defineaperfectsquare.• Testwhetheranumberisaperfectsquareornot.• Identifyandapplythefollowingpropertiesofperfectsquareofa

number.• Thesquareofanevennumberiseven.• Thesquareofanoddnumberisodd.• Thesquareofaproperfractionislessthanitself.• Thesquareofadecimallessthan1issmallerthanthedecimal.

• Define the square root of a natural number and recognize itsnotation.

• Findsquare root,bydivisionmethodand factorizationmethodofa• Naturalnumber,• Fraction,• Decimal,

Whichareperfectsquares.• Solvereallifeproblemsinvolvingsquareroots.

5.1 Introduction

Inpreviousclasses,wehavelearntthattheareaofasquarecanbecalculatedbymultiplyingitslengthbyitselfasshownbelow.Areaofthesquare= length×length=x×x=x2

Itmeansx2isanareaofasquarewhosesidelengthisxorsimplywecansaythat“x2isthesquareofx”.i.e.Thesquareofx=x2

Thus,thesquareofanumbercanbedefinedas:“Theproductofanumberwithitselfiscalleditssquare.”

5.1.1 Perfect Squares

Anaturalnumberiscalledaperfectsquare,ifitisthesquareofanynaturalnumber.Tomakeitclear,letusfindthesquaresofsomenaturalnumbers.12=1x1=162=6x6=36

22=2x2=472=7x7=4932=3x3=982=8x8=6442=4x4=1692=9x9=8152=5x5=25102=10x10=100andsoon

Here,“1isthesquareof1”,“4isthesquareof2”,“9isthesquareof3”andsoon.Itcanbenoticedthatallthesearenaturalnumbers.So,theseareperfectsquareswhichcanberepresentedbydrawingdotsinsquares.

Whenwehaveanumberofrowsequaltonumberofdotsinarow,thenitshowsaperfectsquare.

5.1.2 To Test whether a number is a Perfect Square or not

Tocheckwhetheragivennumberisaperfectsquareornot,writethenumberasaproductofitsprimefactors,ifallthefactorscanbegroupedinpairs,thenthegivennumberisaperfectsquare.


4 5



Example 1: Check whether the following numbers are perfectsquaresornot.(i) 3969 (ii) 6084 (iii) 3872

Solution:(i) 3969

Theprimefactorsof3969=

We can see that each factor forms a pair. Hence,3969isaperfectsquare.

(ii) 6084


Here, each factor of 6084 forms a pair. So, it is aperfectsquare.

(iii) 3872


Wecanseethat2isafactorwhichcannotbepairedwithanyequalfactor.So,3872isnotaperfectsquare.

5.1.3 Properties of Perfect Squares of Numbers

Therearesomeinterestingpropertiesaboutperfectsquares.Letusdiscusssomeofthem.• ThesquareofanevennumberisevenWe know that natural numbers can be divided into two groups:evennumbersandoddnumbers.Lookatthesquaresoftheevennumbersgivenbelow.22=2x2=462=6x6=36102=10x10=100

42=4x4=1682=8x8=64122=12x12=144

Noticethatthesquaresofallevennumbersareevennumbers.

• The square of an odd number is oddNowwefindthesquareofsomeoddnumbers.

12=1x1=152=5x5=2592=9x9=81

32=3x3=972=7x7=49

112=11x11=121

Hence,thesquaresofalloddnumbersarealsooddnumbers.

Example 2: Withoutsolving,separatetheperfectsquaresofevennumbersandoddnumbers(i) 3481 (ii) 2704 (iii) 49284 (iv) 12321

Solution:(i) 3481Thesquareofanoddnumberisalsoodd.Q 3481isthesquareofanoddnumber.(ii) 2704Thesquareofanevennumberisalsoeven.Q 2704isthesquareofanevennumber.(iii) 49284Thesquareofanevennumberisalsoeven.Q 49284isthesquareofanevennumber.(iv) 12321Thesquareofanoddnumberisalsoodd.Q 12321isthesquareofanoddnumber.

• The square of a proper fraction is less than itselfTosquareafraction,wemultiplythenumeratorbyitselfanddothesameforthedenominator.

Nowletuscomparethefractionwithitssquarebyusingthemethodofcrossmultiplication.


6 7



From the above it can be observed that the square of a proper

fractionislessthanitself,i.e. .Similarly,

• The square of a decimal less than 1 is smaller than the decimal Tofindthesquareofadecimal,wecanusethefollowingmethod.

Is0.09smaller than0.3orgreater?Certainly,0.09 is smaller than0.3.i.e.0.09<0.3,

Again0.0004issmallerthan0.02i.e.0.0004<0.02.Itmeansthesquareofadecimallessthan‘1’isalwayssmallerthanthegivendecimal.

EXERCISE 5.1

1. Findthesquaresofthefollowingnumbers. (i) 6 (ii) 5 (iii) 10 (iv) 7 (v) 13 (vi) 8 (vi) 41 (vii) 19 (ix) 100 (x) 9 (xi) 11 (xii) 252. Test whether the following numbers are perfect squares ornot. (i) 59 (ii) 625 (iii) 225 (iv) 196 (v) 425 (vi) 81 (vi) 121 (vii) 25003. Withoutsolving,separatetheperfectsquaresofevenandodd numbers. (i) 441 (ii) 144 (iii) 2401 (iv) 6561 (v) 2025 (vi) 11236 (vi) 7569 (vii) 12544

4. Findthesquaresofproperfractions.Alsocomparethemwith itself.

(i)34 (ii)

56 (iii)

411

(iv)17

5. Findthesquaresofdecimalsandcomparethemwithitself. (i) 0.4 (ii) 0.6 (iii) 0.12 (iv) 0.05

5.2 Square Roots

5.2.1 Defining square root of a natural number and recognizing its notation

Theprocessoffindingthesquarerootisanoppositeoperationof“squaringanumber”.Tounderstandit,againwefindsomeperfectsquares.

22=4(2squaredis4)52=25(5squaredis25)72=49(7squaredis49)

Theseequationscanalsobereadas,“2isthesquarerootof4”,“5isthesquarerootof25”and“7isthesquarerootof49”. Similarly,wecanfindthesquarerootofanysquarenumber.

Forthispurpose,weusethesymbol“ “torepresentasquare

root, i.e. 2x x= where “ “ is called redical sign. Here, x2 iscalledredicand. Ifxisanynumberthatcanbewrittenintheformofx=y2,thenxiscalledthesquareofyandyitselfiscalledthesquarerootofx.

5.2.2 Finding square roots by prime factorization

Wehavelearntthat:


8 9



Thesquarerootof4is, 24 2 2= =



andsoon.But in caseof largeperfect squares, itbecomesmoredifficultforustoguesstheirsquareroots.Tosolvethisproblem,weuseamethodwhich is called theprime factorizationmethod.Thestepsforfindingthemethodaregivenbelow,

Step 1: Findtheprimefactorsofthegivennumber. Supposethegivennumberis36,then. 36=2x2x3x3

Step 2: Takethesquarerootonbothsides.

36 2 2 3 3= × × ×

Step 3: Writethemasapairofprimefactorsofaperfectsquare.

2 2

36 2 2 3 3

2 3

= × × ×

= ×

Step 4: Writethesquarerootofeachperfectsquare,i.e. 2 =x x

andfindtheirproduct.

36 2 3 6= × =Hence,6isthesquarerootofthegivennumber36.

Theprimefactorsofaperfectsquarearealwaysinthepairs.

Example 1: Writethesquarerootof900.

Solution:• Findtheprimefactorsof900.

Factorizationof900=2×2×3×3×5×5

• Takesquarerootonbothsides.

900 2 2 3 3 5 5= × × × × ×Writethemasapairofprimefactorsofaperfectsquare.

2 2 2

900 2 2 3 3 5 5

2 3 5

= × × × × ×

= × ×

Writethesquarerootofeachperfectsquare,i.e. 2 =x x andfindtheirproduct.

900 2 3 5 30= × × =Hence,30isthesquarerootof900.

• Finding Square Roots of FractionsWeknowthattherearethreetypesofcommonfractions.

• Properfraction • Improperfraction • Compoundfraction

Example 2: Findthesquarerootofacommonfraction144256

Solution:

• Wehavetofindthesquarerootof144256

.So,

wecanwriteitas:144 144256 256

=

• Find separately the prime factorsof144and256asgiven.

Ifa,bbeanytwonumbers,then(i) × = ×a b a b

(ii) =a ab b

2 362 183 9

3

2 9002 4503 2253 755 25

5

2 1442 722 362 183 9

3

2 2562 1282 642 322 162 82 4

2


10 11



Therefore,1216

istherequiredanswer.

Example 3: Findthesquarerootofthecompoundfraction63181

Solution: (i) Changethemixedfractionintoanimproperfractionas:

63 144181 81

=

Nowfindthesquareroot.Thus,

Thus,319 isthesquarerootof

63181

• Finding Square Roots of DecimalsInthecaseofdecimalsfirstwechangethemintocommonfractionsandthenwefindthesquareroot.Afterfindingthesquareroot,weagainwrite theanswer indecimal form.Wemake itclearwithanexample.

Example 4: Findthesquarerootofthedecimal0.64

Solution:• Change the decimal into a

fractionas,640 64

100. =

• Now find the square root as aproperfraction.

2 2 2

2 2

64 2 2 2 2 2 2 2 2 2 2 2 2100 2 2 5 5 2 2 5 5

2 2 2 2 2 2 8 0 82 5 102 5

.

× × × × × × × × × ×= =

× × × × × ×

× × × ×= = = =

××Thus,0.8istherequiredsquarerootof0.64

EXERCISE 5.2

1. Findthesquarerootsofthefollowingnumbers. (i) 4 (ii) (9)2 (iii) 36 (iv) (25)2

(v) 16 (vi) c2 (vii) 49 (viii) a2 (ix) 25 (x) 81 (xi) y2 (xii) 1002. Find the square roots of the following numbers by prime factorization. (i) 144 (ii) 256 (iii) 576 (iv) 324 (v) 441 (vi) 729 (vii) 196 (viii) 1225 (ix) 10000 (x) 1764 (xi) 43563. Findthesquarerootsofthefollowingfractions.

(i)4981

(ii) 2.25 (iii)144196

(iv) 0.0196

(v)784441

(vi)13136

(vii) 3.24 (viii) 12.25

2 2 2

2 2 2 2

144 2 2 2 2 3 3 2 2 2 2 3 3256 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 3 2 2 3 12 2 2 2 2 162 2 2 2

× × × × × × × × × ×= =

× × × × × × × × × × × × × ×

× × × ×= = =

× × ×× × ×

3 813 273 9

3

2 1442 722 362 183 9

3

2 642 322 262 82 4

2

2 1002 505 25

5

2 2 2

2 2

144 2 2 2 2 3 3 2 2 2 2 3 381 3 3 3 3 3 3 3 3

2 2 3 2 2 3 12 3 = 13 3 9 93 3

× × × × × × × × × ×= =

× × × × × ×

× × × ×= = =

××


12 13



(ix)3253900

(x) 59.29 (xi)2521324

(xii)1.5626

4. Proveeachofthefollowingbyprimefactorization.

(i) 9 36 9 36× = × (ii) 144 4 144 4× = ×

(iii) 64 25 64 25× = × (iv) 81 100 81 100× = ×

(v)144 144

9 9= (vi)

256 2564 4

=

(vii)484 484121 121

= (viii)576 576144 144

=

• Finding Square Root by Division MethodWehave already learnt the process of finding the square root ofnatural numbers by prime factorization method. Now we learnanothermethod for finding the square roots of natural numberswhichisknownas‘divisionmethod’.

Example 1: Findthesquarerootof324bydivisionmethod.Solution: 324

Step 1:Fromrighttotheleftmakepairsofthedigitsandshowthembyputtingabarovereachofthem.

Step 2:

Try to guess the greatest number whosesquaremustbeequaltoorlessthanthefirstpairordigit(fromlefttoright).Herewecanseetherequiredgreatestnumberis1.

Step 3:Subtractthesquareofthenumberfromthepairordigit. i.e.12 =1and3–1=2.Nowbringdownthe2ndpairasshown.

Step 4:Double the quotient and use it as2nddivisor.

Step 5: Again try to guessthegreatestnumberwhose product withdivisormustbeequalto or less than the2nddividendasgivenintheopposite.

Thequotientistherequiredsquareroot.Itcanbecheckedbyfindingitssquare.Thustherequiredsquarerootis18.

Example 2: Findthesquarerootof585225bydivisionmethod.

Solution: 585225

Thus,therequiredsquarerootis765.• Finding Square Roots of FractionsWehavelearntthemethodoffindingthesquarerootoffractionsbyprimefactorization.Nowwefindthesquarerootofafractionbydivisionmethod.

Example 1:Findthesquarerootsof4096

15129bydivisionmethod.

Solution: 4096

15129

Weknowthat:4096 4096

15129 15129=


14 15



• Finding Square Roots of DecimalsTo learn the process of finding the square roots of decimals, weexaminethefollowingexampleanditssteps.

Example 2: Findthesquarerootof333.0625bydivisionmethod.Solution: 333.0625Step 1: Makethepairsofthewholenumberpartofthedecimalas

number.(from right to left) 333.0625Step 2: Make the pairs of the decimal part. (from left to right)

3 33 . 06 25Step 3: Usethesamedivisionmethodasnumbers.

Step 4: Putthedecimalpointinthequotientbeforebringingdown

thepairafterdecimalpoint.

Thus, 333 0625 18 25. .= Example 3: Find the square root of the following by divisionmethod.(i) 0.119025 (ii) 199.9396

Solution:(i) 0.119025Makepairsofthewholenumberpartanddecimalpartrespectively:0.1190 25

(ii) 199.9396Makepairsofthewholenumberpartanddecimalpartrespectively:

199 .9396

EXERCISE 5.3

1. Findthesquarerootsofthefollowingbydivisionmethod.(i) 729 (ii) 2304 (iii) 4489 (iv) 7056(v) 9801 (vi) 14400 (vii) 15625 (viii) 18496(ix) 207936 (x) 321489 (xi) 5499025 (xii) 4986289


16 17



2.Findthesquarerootsofthefollowingcommonfractionsbydivisionmethod.

(i) (ii) (iii) (iv)

(v) (vi) (vii) (viii)

(ix) (x) (xi) (xii)

3.Findthesquarerootsofthefollowingdecimalsbydivisionmethod.(i) 0.0529 (ii) 1.5625 (iii) 9.7344 (iv) 0.4761(v) 0.001369 (vi) 32.1489 (vii) 0.002025 (viii) 131.1025(ix) 508.5025 (x) 799.7584 (xi) 1082.41 (xii) 4596.84

5.2.3 Solving Real Life Problems involving Square Root

Wesolvereallifeproblemsinvolvingsquarerootsbyusingthemethodoffindingthesquareroot.

Example 1: Theareaofarectangularparkisequaltoanothersquareshapedpark.Findthelengthofasquareshapedparkifthelengthandbreadthoftherectangularparkare81mand25mrespectively.Solution:Areaoftherectangularpark = lengthxbreadth

= 81mx25m=2025m2

Asweknowthat,Areaofsquareshapedpark = AreaofrectangularparkLengthofside =

=

=

=

=

Thus,therequiredlengthis45m.

Example 2: Find the length of a boundary of a square fieldwhoseareais784m2.Solution:Areaofthesquarepark=784m2

Lengthofside = 784=

=

= (2x2x7)m=28mThelengthoftheboundaryorperimeterofthesquarefield:

= 4(length)= 4(28m)=112m

Example 3: Find the perimeter of a rectangular park whoselength is three times of its width and the area is 720.75m2. Alsocalculate the cost of fencing the park at the rate of Rs.195/m.(use division method for finding square root)Solution: WehaveLengthofthepark = 3(widthofthepark)Areaoftherectangularpark = 720.75m2

(i) perimeter=? (ii) Costoffencing=?Weknowthat:Areaoftherectangularpark=Lengthxwidth

720.75m2 = 3(width)xwidth720.75m2 = 3(width)2

= (width)2

240.25m2 = (width)2

= width


18 19



perimeter = 2(length+width)= 2(46.5m+15.5m)=2x62m=124m

Costoffencingof1m = Rs.195Costoffencingof124m = Rs.(195x124)=24,185

EXERCISE 5.4

1. Theareaofasquareis73.96m2.Calculatethelengthofitsside.2. 324 soldiers queued up such that the number of queues is equal to the number of soldiers in each queue. Find the numberofqueues.3. Bywhichsmallestnumbercan275bemultipliedtogetaperfect square?4. Bywhichsmallestnumbercan648bedividedtogetaperfect square?5. The lengthandbreadthofa rectangular swimmingpoolare 243mand27mrespectively.Findthelengthofasquareshaped swimming pool which has the same area as rectangular swimmingpool.6. Thebaseandheightofatriangleare8cmand4.5cmrespectively. Findthelengthofthesideofasquarewhoseareaisthedouble ofthegiventriangle.7. Theareaofasquarefieldis617796m2.Findthelengthofits side.8. Anurseryownertriestoarrange89500plantsintotheshape of solidsquare.Buthefinds thathehas99plants leftover.

Find how many plants did the owner arrange in a row. (Hint=89500–99=?)9. Whichsmallestnumbercanbesubtractedfrom15198togeta perfectsquare?10. Findthenumberthatgives992.8801aftermultiplyingitself.11. Findthemeasurementofthesidesofarectanglewhoselength isfourtimesofitswidthandareais51.84cm2.12. Theareaofacircularswimmingpool is154m2. Findout the radiusoftheswimmingpool.

Review Exercise 5

1. Answerthefollowingquestions. (i) Whatismeantbythesquareofanumber? (ii) Defineaperfectsquare. (iii) Whichsmallestnumbercanbesubtractedfrom50toget aperfectsquare? (iv) Namethetwomethods forfindingthesquarerootsof largenaturalnumbers.2. Fillintheblanks. (i) 4,9,16,25,...arecalled___________. (ii) Ifx=y2,thenyiscalledthe___________ofx. (iii) While finding the square root by divisionmethod, the digitsarepairedfrom___________. (iv) Thenumberwhosesquarerootisnon-terminitingand non-recurringdecimaliscalledthe___________number.

(v)121 121144

=

(vi)169100

=



20 21



4. Findthesquarerootofthefollowing.

(i) 1024 (ii) 484 (iii)19649

(iv) 6 . 2 5

(v) 0.0225 (vi)12253025

(vii)14225

(viii)40181

(ix) 10.89 (x)231

121 (xi)

225324

(xii) 3.0625

(xiii) 29.16 (xiv)5391

1225

5. Proveeachofthefollowingbyprimefactorization.

(i) 16 81 16 81× = ×

(ii) 0 25 0 04 0 25 0 04. . . .× = ×

(iii)5625 5625625 625

=

(iv)5 76 5 761 44 1 44. .. .

=

6. 10201 soldiers have queued up for an attack such that the numberofqueues isequal to thenumberof thesoldiers in eachqueue.Findthenumberofsoldiersineachqueue.

7. A businessman bought a square shaped park whose area is 50625m2.Hewants to fix light poles after thedistanceof each metre on its surroundings. For this he calculated the perimeter of the park. Do you know what perimeter he calculated?

8. The length and breadth of a rectangular swimming pool in abungaloware125mand45mrespectively.Findthelengthof another square shaped swimmingpoolwhichhas the same areaasrectangularswimmingpool.

9. Ateacherdrewatriangleof8cmheightand18cmbase.Now hewantstodrawasquarewhoseareamustbethetwicethat of the triangle. Calculate the length of the each side of the squarethathehastodraw.

10. Solve: (i) Bywhichsmallestnumbercan605bemultipliedtogeta perfectsquare? (ii) Bywhichsmallestnumbercan3675bedividedtogeta perfectsquare? (iii) Theareaofasquareis94.09m2.Whatisthelengthofits side? (iv) The lengthofasideofasquare is55.5m.What is the areaofthesquare?

Summary

• Theproductofanumberwithitselfiscalleditssquare.• Anaturalnumberiscalledaperfectsquare,ifitisasquareofany

naturalnumber.• Thesquareofanevennumberisevenandofanoddnumberis

odd.• Thesquareofaproperfractionislessthanitself.• Thesquareofadecimallessthan1issmallerthanitself.• Theprocessoffindingthesquarerootisthereverseoperationof

‘squaringanumber’.• Ifxisanumbersuchthatx=y2,thenxisknownasthesquareof

yandyisknownassquarerootosx.

• Torepresentthesquareroot,weusethesymbol“ ”whichiscalledradical.

• Tofindthesquarerootofamixedfraction,weconvertitintoanimproperfraction.

• Wefindthesquarerootofadecimalbychangingitintoafraction.• Wefindthesquarerootofadecimalbychangingitintoafraction.

CHAPTER

6 DIRECT AND INVERSE VARIATION

version: 1.1

Animation 6.1: Continued RatioSource & Credit: eLearn.Punjab



2 3

6. Direct & Inverse Variation 6. Direct & Inverse VariationeLearn.Punjab eLearn.Punjab

2version: 1.1 version: 1.1


After studying this unit, students will be able to:• Definecontinuedratioandrecalldirectandinverseproportion.• Solvereallifeproblems(involvingdirectandinverseproportion)

usingunitarymethodandproportionmethod.• Solvereallifeproblemsrelatedtotimeandworkusingproportion.• Findrelationbetweentimeanddistance(i.e.speed).• Convertunitsofspeed(kilometerperhourintometerpersecond

andviceversa).• Solvevariationrelatedproblemsinvolvingtimeanddistance.

Introduction

Supposesomeoneuses3cupsofwaterand1cupofmilktopreparetea.Wecancomparethesetwoquantitiesbyusingthetermofratio.Waterandmilkareintheratioof3:1.Thus,theratioisacomparisonbetweenthetwoormorequantitiesofthesamekindwhichcanbewrittenbyputtingacolon(:)amongthem.Aratioisthenumericalrelationbetweentwoormorequantitieshavingthesameunit.

6.1 Continued Ratio

Ifthetworatiosa : bandb : caregivenforthreequantitiesa,bandc,thentheratioa : b : ciscalledcontinuedratiowhichcanbewrittenas,

Here, ratioa : b : c isa continuedratiowhich is formed fromtheothertworatiosa:bandb:ctoexpresstherelationbetweenthreequantitiesa,bandc.

Fromtheaboveexplanation,wecanobserve thatb isa commonelementof two ratioswhich is thecause to combine them.Suchtypeofanelementiscalledthecommonmemberofthegivenratios.Always write the common member in the middle of the otherelementsaccordingtothemethodgivenabove.

Example 1: Thetworatiosofthreequantitiesa,bandcareas,a:b=1:2andb:c=2:3.Findtheircontinuedratio.Solution:Theratiosare:

Thecommonmemberoftworatiosisb,so

Hence,a:b:c =1:2:3Therefore,1:2:3istherequiredcontinuedratio.Ifthecorrespondingelementsforthetworatiosarenotequal,thenthesearemadeequalbymultiplyingboththeratiosbythenumberswhichmakethemequalasshownbelow.

Example 2: TheratioofSaleem’sincometoHaider’sis2:3andImran’s incometoSaleem’sis1:5.Findthecontinuedratioamongtheirincomes.Solution: Theratiosare: Saleem’sincometoHaider’s=2:3 Imran’sincometoSaleem’s=1:5


4 5



Saleemisthecommonmemberbutthevalueofhisincomeisnotthesameinbothratios.Thus,firstfindthesamevaluesofcommonmemberasgivenbelow:

Thus,15:10:2istherequiredcontinuedratio.

Example 3: Ifa:b=1:3andb:c=2:5,thenfinda:c.

Solution: Theratiosare:a:b=1:3,b:c=2:5Wecanseethatbisthecommonmemberso,

Thus,a:b:c =2:6:15Fromtheabove,wecanobservethatthevalueofa =2andc =15.So,a : c=2:15.

EXERCISE 6.1

1. Ifa:b=3:5andb:c=5:6,thenfinda:b:c.2. Ifr:s=1:4ands:t=2:3,thenfindr:s:t.3. Ifp:q=1:2andq:r=1:2,thenfindp:q:r.4. Ifx:z=3:2andy:z=1:2,thenfindx:y:z.5. Ifl:m=1:7andl:n=5:6,thenfindl:m:n.6. Inabakery,theratioofthesaleofbreadtoeggsis2:3andthe saleofeggstomilk is3:1.Findthecontinuedratioofbread, eggsandmilk.7. AhmadandIrfangotaprofitinabusinessintheratioof5:4 andIrfanandWaseemgotaprofitintheratioof8:9.Findthe ratioofprofitamongAhmad,IrfanandWaseem.

8. According to a survey, the people’s liking for chicken and muttonareintheratioof2:1andthepeople’slikingforchicken andbeef is in the ratioof5:2.Find the ratioamongpeople’s likingforchicken,muttonandbeef.9. InamathstestZara,MoonaandKomalgotmarksintheratio asgivenbelow:

ZaratoMoona=4:5MoonatoKomal=4:3

Findcontinued ratioofmarksobtainedbyZara,Moonaand Komal.• Proportion Wehavelearntinourpreviousclassesthatfourquantitiesaresaidtobeinproportion,iftheratioofthe1sttothe2ndisequaltothe ratioof the3rd to the4th. Inotherwords,the fourquantitiesa,b,canddare inproportionifa:b=c:d.Letusrecaponwhatwestudiedinourpreviousclassaboutproportion.• In aproportion, the secondand the thirdelements are called

“meansofaproportion”andthefirstandthefourthelementsarecalled“extremesofaproportion”i.e.

• Ifsecondandthirdelementsofaproportionhavethesamevaluesuchas:a:b::b:cHere‘b’iscalledmeanproportional.

• Oneratioisproportionaltotheotherratio,ifandonlyif,productofmeans=productofextremes• The4thelementofaproportionisknownasthefourthproportional

e.g.,inproportion,a : b :: c :d,discalledthefourthproportionalofa,bandc.

• A relation inwhich one quantity increases or decreases in thesameproportionbyincreasingordecreasingtheotherquantity,iscalledthedirectproportion.

• Arelationinwhichonequantityincreasesinthesameproportionbydecreasingtheotherquantityandviceversa,iscalledinverseproportion.


6 7



• Amethodwhich is used to calculate the valueof a numberofthingsbyfindingthevalueofone(unit)thingiscalledtheunitarymethod.

Example 1: GhaziearnsRs.7500in2weeks.Whatwillheearnin2daysifheworks6daysaweek?Solution:

Unitary Method

Ghaziearnsin12days =Rs.7500Q 2weeks=12days

Ghaziearnsin1day

Ghaziearnsin2days.

GhaziearnsRs.1250in2days.Proportion Method

Daysaredirectlyproportionaltotherupees.

GhaziearnsRs.1250in2days.

Example 2: 10boyscompleteawork in4days. Inhowmanydayswill20boyscompletethesamework?Solution:

Unitary Method• Moreboyswillcompletetheworkinlessnumberofdays.10boyscompletethework=4days.1boycompletethework=(4x10)days

20boycompletethework=

Proportion Method• Boysareinverselyproportionaltothedays.

Example 3: 125men can construct a road in 120 days. Howmanymencandothesameworkin100days?Solution:

Unitary Method• Todotheworkinlessdays,weneedmoremen.

In120days,mencanconstructtheroad=125menIn1day,mencanconstructtheroad=(125x120)men

In100days,mencanconstructtheroad.

Proportion Method• Menareinverselyproportionaltothedays.

150mencandothesameworkin100days.

EXERCISE 6.2

1. Findthevalueofminthefollowingproportion. (i) 13:3=m:6 (ii) m:5=3:10 (iii) 35:21=5:m (iv) 9:m=54:42 (v) 0.21:6.3=0.06:m (vi) 1.1:m=0.55:0.272. Whatisthefourthproportionalof2,5and6?3. Findmeanproportionalof4and16.


8 9



4. AworkerispaidRs.2130for6days.Ifhistotalwageduringa month isRs.9230,find thenumberofdaysheworked in the month.5. Uzair takes75 steps to coveradistanceof50m.Howmuch distancewillbecoveredin375steps?6. If2boxesoccupyaspaceof500cm3,thenhowmuchspacewill berequiredforsuch175boxes?7. Anarmycampof200menhasenoughfoodfor60days.How longwillthefoodlast,if: a. Thenumberofmenisreducedto160? b. Thenumberofmenisincreasedto240?

6.2 Time, Work and Distance

6.2.1 Time and Work

• Whilesolving theproblemsrelated to timeandwork, it canbeobservedthat:timeisdirectlyproportionaltowork,becausemoreworktakesmoretimeandlesstimegiveslesswork.

• Numberofworkersisinverselyproportionaltothetime,becausemoreworkinghandstakelesstimetocompleteaworkwhereasmoretimegivenforaworkneedslessworkinghands.

Example 1: Ifagirlcanskiparope720times in1hour.Howmanytimescansheskipin35minutes?Solution: Skippingaropeisdirectlyproportionaltothetime.So,wecanwritethissituationas;

Example 2: If theheartofahumanbeingbeats72times in1minute.Find,inwhattimewilltheheartbeat204times.Solution: (Heartbeatingisdirectlyproportionaltothetime)Thesituationcanbewrittenas:

170secondsor2min50sec

Example 3: If36mencanbuildawallin21days,findhowmanymencanbuildthesamewallin14days.Solution:Menareinverselyproportionaltothetime.So,thissituationcanbewrittenas:

EXERCISE 6.3

1. Ifamancanweave450mclothin6hours.Howmanymetresof clothcanheweavein14hours?2. Ifa162kmlongroadcanbeconstructedin9months.Howmany numberofmonthsare required to construct a306km long road?


10 11



3. 540mencanconstructabuildingin7months.Howmanymen should be removed from work to finish the building in 9 months?4. Asmacaniron5shirtsin14minutes.Howlongwillshetaketo iron35shirts?5. 12waterpumpscanmakeawatertankemptyin20minutes. But 2 pumps are out of order.How longwill the remaining pumpstaketomakethetankempty?6. 14horsesgrazeafield in25days. Inhowmanydayswill35 horsesgrazeit?7. Amasoncanrepaira744mlongtrackin24days.Ifherepairs 589mtrack,thenfindhowmanydayswillhetaketorepairthe remainingtrack.8. Afarmercanploughanareaof40acresin16hours.Howmany acreswillheploughin36hours?9. Adishwasherdeems1350dishesin1hour.Howmanydishes willitwashin16moreminutes?

6.2.2 Relation between Time and Distance

Inourdailylife,weobservemanymovingthingslikevehicle,birds,humanbeings,ships,animals,etc.inoursurroundings.Whilemovingthesethingscoveracertaindistance inacertaintimeatacertainspeed.Tounderstandtherelationbetweenthesethreequantitieswecanuseaformulawhichisgivenbelow:Distance=SpeedxTimeFromthegivenformula,itcanbeexaminedthat:• Distanceisdirectlyproportionaltothetimeandspeed.• Timeisinverselyproportionaltothespeed.Anintervalbetweentwohappeningsiscalledtime.Itsbasicunitis

second.• Units of Speed“Thedistancecoveredperunittimeiscalledspeed.”Speed ismeasured indifferentunits that is, kilometresperhour,metrespersecond,etc.Wewritetheseunitsbydividingtheunitsofdistance(km,m)bytheunitsoftime.

(hr,min,sec).

Theunitsof speedaremutually convertable. Letusmake it clearwiththehelpofsomeexamples.

Example 1: Convert thespeedof54kilometersperhour intometrespersecond.Solution: Speed =54km/hour Q km=1000m Distance =54km Q 1hour=60min Time =1hour 1min=60sec

54 1000Speed60 60××

54000 15 metre / second3600

= =

Example 2: Convert the speed of 10meters per second intokilometresperhour.Solution: Speed =10m/sec Q 1000m=1km

Distance =10meter 11m km

1000=

Time =1second Q 3600sec=1hour

11 sec hour

3600=

10 1000Speed1 3600

//

=

10 3600 36 km/hour1 1000×

= =×

Example 3: A truck covers a distance of 360 kilometres in 5hours.Finditsspeedin: (i) kilometresperhour (ii) metrespersecond


12 13



Solution:Distance=360km,Time=5hours,Speed=?(i) kilometresperhourByusingtheformula,

Distance 360Speed 72 km/hour

Time 5= = =

(ii)metrespersecondDistanceinmetres=360x1000=360000mTimeinseconds=5x60x60=18000secNowchangetheunitofspeedintometrespersecond.Speed=72km/hour

72 1000 72000 20 metre/second1 60 60 3600

×= = =

× ×

EXERCISE 6.4

1. Converttheunitofspeedintometrespersecond. (i) 72km/hour (ii) 144km/hour (iii)216km/hour (iv) 360km/hour (v) 180km/hour (vi)1152km/hour2. Converttheunitofspeedintokilometresperhour. (i) 10m/sec (ii) 25m/sec (iii) 5m/sec (iv) 15m/sec (v) 30m/sec (vi) 20m/sec3. Iramwalksuptoherschoolataspeedof4km/hour.Ittakes45 minutestoreachtheschool.Howfar isherschool fromher home?4. Anon-stoptrainleavesLahoreat4:00p.mandreachesKarachi at10:00a.mnextday.Thespeedofthetrainwas70km/hour. FindthedistancebetweenLahoreandKarachi.5. Acyclistcrossesa30metrelongbridgein minutes.Find thespeedofthecyclist.6. Acarcovers201kilometresin3hours.Howmuchdistancewill itcoverin7hours?7. Atruckmovesatthespeedof36kilometresperhour.Howfar willittravelin15seconds?

8. AbusleavesIslamabadat11:00a.mandreachesLahoreat 3:00p.m.IfthedistancebetweenLahoreandIslamabadis380 km,findthespeedofthebus.

REvIEw EXERCISE 6

1. Answerthefollowingquestions. (i) Definedirectproportion. (ii) Whatiscontinuedratio? (iii) Write the formula to show the relation between time, speedanddistance. (iv) Definespeed.

2. Fillintheblanks. (i) Distanceisdirectlyproportionaltothe_______andspeed. (ii) Numberofworkersis_______proportionaltothetime. (iii) Thecombinationoftworatiosofthreequantitiesiscalled a_______ratio. (iv) Distance=_______xtime.

(v) SpeedTime

=

(vi) Intworatiosa : bandb : c,biscalledthe________.

3 Tick(p)thecorrectanswer.

4. Findthemissingtermsinthetable, if isdirectlyproportional tothey.

x 2 4 6 8 10y 4 12 16

5. Findthemissingtermsinthetable,ifxisinverselyproportional toy.


14

6. Direct & Inverse Variation eLearn.Punjab

14version: 1.1

x 1 2 4 6 8y 24 12 3

6. Inaclass,8icecreamsareservedforeverygroupof5students. Howmanyicecreamswillbeservedifthereare40studentsin theclass?7. Inahostelof50girls,therearefoodprovisionsfor40days.If 30moregirlsjointhehostel,howlongwilltheprovisionslast?

8. Howmanydayswill1648personstaketoconstructabridge,if 721personscanbuildthesamein48days?

9. Arickshawtravelsatthespeedof36kmperhour.Howmuch distancewillittravelin20seconds.

10.Abuscoversadistancein3hoursataspeedof60kmperhour. Howmuch timewill it take to cover the samedistance at a speedof80kmperhour?

SUMMARy

• Tworatiosofthreequantitiescanbecombinedintoacontinuedratiotoexpresstherelationofthesequantities.

• A relation inwhich one quantity increases or decreases in thesameproportionbyincreasingordecreasingtheotherquantity,iscalledthedirectproportion.

• Arelationinwhichonequantityincreasesinthesameproportionby decreasing the other quantity and vice versa, is called theinverseproportion.

• Time is directly proportional to the work, and the number ofworkersisinverselyproportionaltotime.

• Tounderstandtherelationbetweendistance,speedandtime,weusetheformula:

Distance=SpeedxTime• Anintervalbetweentwohappeningsiscalledtime.• Thedistancecoveredperunittimeiscalledspeed.

CHAPTER

7 Financial arithmetic

version 1.1


3

7. Financial Arithmetic 7. Financial ArithmeticeLearn.Punjab eLearn.Punjab

2 version 1.1version 1.1


After studying this unit, students will be able to:• Explainpropertytaxandgeneralsalestax.• Solvetax-relatedproblems.• ExplainProfitandmarkup.• Findtherateofprofit/markupperannum.• Solvereallifeproblemsinvolvingprofit/markup.• Definezakatandushr.• Solveproblemsrelatedtozakatandushr.

7.1 Taxes

Governmentneedsmoneytorunastate.Forthispurpose,governmentcollectsanamountfromthepublicandprovidesthemfacilities like security, hospitals, education, defense, roads, parks,etc.Thisamountiscalledatax.WepaydifferenttypesoftaxesinPakistanbuthereweshall,discussonlypropertyandgeneralsalestax.

7.1.1 Property Tax and General Sales Tax

• Property Tax Thetaxwhichisreceivedonapropertyiscalledthepropertytax.Propertytaxisaprovincialtaxpaidonthevalueofaproperty.Itisgenerallypaidataflatrateof2%butthetaxratesvary,dependingontheprovince.

Example 1: FindthepropertytaxonapropertyofRs.6,200,000attherateof0.8%.Solution:Worthoftheproperty=Rs.6,200,000Taxrate=0.8%Propertytax=?Propertytax=0.8%ofRs.6,200,000

8 6,200,000

1000= ×

=Rs.49,600

Example 2: RaheempaidRs.8,676asapropertytaxattherateof2%.FindtheworthofRaheem’sproperty.Solution: Propertytax=Rs.8,676Taxrate=2%Worthoftheproperty=?Byusingtheunitarymethod2%oftheworthofproperty=Rs.8,676

1%oftheworthofproperty8 676Rs.

2, =

100%oftheworthofproperty8 676Rs. 100 Rs. 433,800

2, = × =

TheworthofRaheem’spropertyisRs.433,800.

• General Sales Tax “The tax abuyerpays to the seller at the timeof buyingthings iscalledgeneralsalestax”.Generalsalestax is imposedbythegovernmentonthepercentageofthesellingpricesofthings.InPakistan,itsratevariesfrom0%to25%dependingonexemptionsandtypesofindustry.In Pakistan some basic items including wheat, rice, pulses,vegetables,meat,poultry,books,drugs, etc. areexempted fromthegeneralsalestax.

Example 3: SaleemboughtacarforRs.875,000andpaid16%asatax.Howmuchtaxdidhepay?Solution:PriceofthecarTaxGeneralsalestax

===

Rs.875,00016%?

RememberThestandardrateofsalestaxinPakistanis16%.


5



GST=16%ofRs.875,000

16 875,000100

= ×

=(16x8750)=Rs.140,000

Example 4: ThepriceofamobileisRs.8,800inclusiveofa10%GST.Whatistheoriginalpriceofthemobile?Solution: Priceofthemobile=Rs.8,800GSTrate=10%Originalprice=?Price%ofthemobile=100%+10%=110%Byusingtheunitarymethod110%priceofthemobile=Rs.8,800

1%priceofthemobile8 800Rs. 110, =

100%priceofthemobile8 800Rs. 100 Rs. 8,000110, = × =

Thus,theoriginalpriceofthemobileisRs.8,000.

EXERCISE 7.1

1. Calculatethepricethatacustomerhastopayforeacharticle witha16%generalsalestaximposedonit. (i) Football=Rs.800 (ii) Rackets=Rs.1,250 (iii) Hockey=Rs.1,650 (iv) Bat=Rs.2,1002. FindthepropertytaxonapropertyofworthRs.948,000atthe rateof1.5%.3. HarispaidthepropertytaxofRs.2,068attherateof0.8%.Find theworthofproperty.4. PropertytaxRs.18,720waspaidwhentheworthofpropertyis Rs.1,560,000.Findthepercentageofpropertytax.5. Thepriceofatoyincluding5%generalsalestaxisRs.945.Find theoriginalpriceofthetoy.

6. FindthepropertytaxonapropertyofRs.650,000attherate of1.8%.7. FarahpaidthepropertytaxofRs.9,240attherateof2%.Find theworthofherproperty.8. ThepriceofabicycleisRs.6,480inclusive8%GST.Whatisthe originalpriceofthebicycle?

7.2 ProfitandMarkup

Weknowthatinabusiness,generallygoodsareboughtatacertainpriceandsoldatahigherprice.Insuchacase,thereisagain,i.e. Saleprice-Costprice=GainWhilediscussingthisgain,weoftenusetwodifferentterms,profitandmarkup.Tounderstandthedifferencebetweenthesetwoterms,letuslearnthemonebyone.• Profit Aprofitmeanswhatwehaveearnedaftersellingathing.Itiscalculatedasapercentageofthecostpriceasshownbelow.

GainProfit% = × l00%Cost price

• Markup Inourdailylife,weoftenborrowmoneyfromourfriendsandrelativestobuyathingthatwerepaythemafteracertainperiod.Somebanksandretailorganizationsalsoprovidethesameservicesandchargeanadditionalamountcalledmarkup.“Amarkupisanamountaddedtoacostpricetocalculatethesaleprice.”Usually, we calculate the markup as a percentage of the actualamount paid for a thing. This is called themarkup rate andpaidamountitselfiscalledtheprincipal.Suppose“P”istheprincipal,“T”isthetimeperiodand“R” isthemarkuprate,thentheamountofmarkupwillbe:


7



RPTMarkup =100

Example 1: AdaboughtajewelrysetforRs.84,000andsoldforRs.855,00.Findthepercentageofprofit.

Solution: Costprice(C.P)=Rs.84,000Saleprice(S.P)=Rs.85,500Gain=Saleprice-Costprice=Rs.85,500-Rs.84,000=Rs.1,500

Gain%Profit = ×l00%Cost price

1 500 100 1 7984 000

, % . %,

= × =

Example 2: Aleem bought a television for Rs. 15,000 oninstallmentsatthemarkuprateof12%perannum.Findthesellingpriceofthetelevisioniftimeperiodis3years.Solution: Costprice(P)=Rs.15,000;Markuprate=12%perannum Timeperiod(T)=3years; PriceoftheTelevision=?Usingtheformula,

RPT 12 15,000 3Amount of the markup Rs. 5 400100 100

,× ×= = =

Priceofthetelevision=costprice+markup=Rs.15,000+Rs.5,400=Rs.20,400

Example 3: ImransoldabicycleforRs.3,978andgot17%profit.Findthecostpriceofthebicycle.Solution: Saleprice(S.P)=Rs.3,978%Profit=17%Byusingformula,

Sale priceCost price (C.P)(100% + Profit%)

=

3978Cost price (C.P) 100 Rs. 3,400117

= × =

Example 4: HatimboughtabikeforRs.135,000andsoldat62%profit.Findthesalepriceofthebike.Solution:

Method I %profit=62%Costprice=Rs.135,000Saleprice=?Saleprice=(100%+62%)x135,000rupees=162%x135,000rupees

162 135 000 Rs. 218,700100

, = × =

Method II

Profit%Profit 100Cost price

= ×

Profit62 100135,000

= ×

62 135,000Profit rupees100

× =

Rs. 83,700=Weknowthat;Saleprice=costprice+profit=Rs.135,000+Rs.83,700=Rs.218,700


9



Example 5: FindthemarkuponathingwhosepriceisRs.45,000for73daysattherateof10%perannum.Solution: Principal (P)=Rs.45,000,Markuprate (R)=10%perannum,Markup=?

73 1Time period = 73 days = year = year365 5

RPTBy using the formula: Markup = 100

110 45 0005Rs.

100

,× ×=

10 45 000 1Rs. Rs. 900100 5

,× ×= =

×

Example 6: Themarkupon a principal amount is Rs. 820 for6monthsat the rateof12.5%perannum.Calculate theprincipalamount.Solution: Markup=Rs.820Markuprate(R)=12.5%

6 1Time period (T) = 6 months year year12 2

= =

Principalamount(P)=?

RPTBy using the formula: Markup = 100

112 5 P5820

100

. × ×=

820 100 2Principal amount = Rs. 13,12012.5 1× ×

=×

EXERCISE 7.2

1. Findthemissingquantitiesbyusingtheformula

Markup Principal Time Period Markup rate

(i) Rs.500 2years 12%

(ii) Rs.205 1year 8%

(iii) Rs.528 Rs.1,650 10years

(iv) Rs.350 Rs.3,500 2.5%

(v) Rs.100,000 3years 1.25%

(vi) Rs.1,050 5years 4.5%

2. Adnanbought96eggsattherateofRs.40perdozenandsold attherateofRs.4peregg.Findthepercentageofprofit, if3 eggswererotten.3. If16%profitonamobilesetisRs.832.Findthecostpriceofthe mobileset.4. ZiaboughtanoutoforderclockforRs.750andgotitrepaired forRs.425.WhatshouldbethesellingpriceoftheclockifZia wantstoearn25%profit?5. Find themarkuponaprincipal amountofRs. 75,500at the rateof9%perannumfor4years.6. UjalaboughtacarforRs.280,000andspentRs.12,000moreon it.Whatshould bethesellingprice ifshewantstoget7.5% profit?7. ThepriceofabicycleincludingmarkupisRs.5610.Ifthemarkup rateis5%perannum,findtheamountofmarkupfor146days.8. KhushiboughtacomputerforRs.100,000andpaidamarkup ofRs.25,000for2years.Whatmarkupratedidshepay?

7.3 ZakatandUsher

ZakatandUshrareleviedasorderedintheHolyQur’anandSunnah.Letusdiscussthemonebyone.


11



• Zakat ZakatisoneofthefivepillarsofIslamwhichisordrsbyAlmightyAllahwhichispaidonthewealthwhichremainswithapersonforacompleteyear.Islamhasfixeditsrate,thatis2.5%.

Nisab (minimum limitofwealth thatattracts liabilityofZakat) incaseofgoldis7.5tolasandincaseofsilveris52.5tolas.

Example 1: CalculatetheamountpayableasZakatbyHaleemwhosavesRs.949,000foroneyear.Solution: TotalSaving=Rs.949,000 RateofZakat=2.5% AmountofZakat=? AmountofZakat=2.5%ofRs.949,000

2.5 949,000 rupees100 = ×

25 949,000 rupees = Rs. 23,7251,000×

=

Thus,HaleemwillpayRs.23,725asZakat.

Example 2: Find thewealth of Ibrahim if he paid Rs.7,500 asZakat.Solution: 2.5%ofIbrahim’swealth=Rs.7,500

7 5001 % of Ibrahim Rs.

2 5,.

=

7 500100 % of Ibrahim Rs. 100 Rs.300,0002 5,.

= × =

• Ushr Ushrmeansone-tenth.It ispaidonagriculturalproducts.Itis paid at the rate of 10%of theproduce in case a piece of landirrigatedbynaturalsourceslikerain,springs,streams,etc.However,therateofUshrisone-half,i.e.5%oftheentireproduceincase

apieceof land wateredby artificialmeansof irrigation suchaswells,buckets,tubewell,etc.Nisab(minimumamountofagriculturalproduce)whichisliabletoUshris948kginweight.Iftheproduceislessthanthat,noUshrischargeable.

Example 3: AfarmersoldhiscropofwheatforRs.995,400.FindtheamountofUshrattherateof10%.Solution: TotalAmount=Rs.995,400RateofUshr=10%AmountofUshr=?AmountofUshr=10%ofRs.995,400

10 995,400 = Rs. 99,540100 = ×

Thus,amountofUshrisRs.99,540

Example 4: Adnan sold mangoes and paid Rs. 3,675 as theamountofUshrattherateof5%.Findthesalepriceofthemangoes.Solution: AmountofUshr=Rs.3,675 RateofUshr=5% Amountofmangoes=? 5%ofmangoesamount=Rs.3,675

3,6751% of mangoes amount = Rs.

5

3,675Price of mangoes = Rs. 100

5 ×

Price=Rs.73,500Thus,theamountofmangoesisRs.73,500


13



EXERCISE 7.3

1. AnamountofRs.62,480remainedwithNosheenforacomplete year.HowmuchZakatwillshepay?2. SabapaidRs.2,250asZakat.Whatistheworthofherwealth?3. NadeempaidRs.6,075asZakat.Howmuchwealthdidhehave?4. SaleemearnedRs.114,700fromaricecropandpaidUshrat therateof5%.WhatamountdidhepayasUshr?5. NabeelsoldapplesforRs.398,160andpaid10%asUshr.Find theamountofUshr.6. Shama’sannualsavingisRs.222,000.Whatistheamountof Zakattobepaidbyher?7. NahalpaidRs.7,895asUshrattherateof10%.Whatamount didsheearn?8. CalculatetheamountpayableasUshrbyafarmerwhoearned Rs.88,460.Findtheactualamount,ifrateofUshris5%.

Review Exercise 7

1. Answerthefollowingquestions. (i) Whatismeantbythetax? (ii) Definethegeneralsalestax. (iii) Whatisthedifferencebetweenprofitandmarkup? (iv) WhatrateofZakathasIslamfixed? (v) WhatisUshr?2. Fillintheblanks. (i) Thetaxwhichisreceivedonapropertyiscalledthe_______. (ii) Thetaxabuyerpaystotheselleratthetimeofbuying thingsiscalled_______. (iii) An amount added to a cost price to calculate the sale priceiscalleda_______. (iv) Zakat and Ushr are levied as ordered in the____ and Sunnah. (v) Amarkupisanamountaddedto_______tocalculatethe saleprice.


4. Calculatetheamountofpropertytaxofahouseattherateof 2%.ThevalueofthehouseisRs.1,450,000.5. AdnanhaspaidRs.16,000asapropertytaxattherateof1.6%. Findthevalueofhisproperty.6. ThepriceofatoyisRs.500.FindthesalepriceofthetoyifGST is16%.7. NabeelboughtabagforRs.4,000andpaidRs.560moreasGST. FindthepercentageofGST.8. NadeemsoldabicycleforRs.4,500withmarkupof25%.Find thecostpriceofbicycle.9. A shopkeeper sold a calculator for Rs.900 and earned 22% profit.Findtheactualpriceofacalculator.10. Komal saves Rs.96,000 in a year.Howmuchwill shepay as Zakat?11. Saleemhas2,400kgwheat.ThepriceofwheatisRs.30kg.Find theUshrthathewillpay.

Summary

• Ataxisafeechargedonthepublicattheratefixedbyagovernmenttorunitsaffairs.

• Thetaxwhichisreceivedonpropertyiscalledthepropertytax.• Thetaxabuyerpaystotheselleratthetimeofbuyingthingsis

calledgeneralsalestax.• An amount added to a cost price to calculate the sale price is

calledamarkup.• Aprofitmeanswhatwehaveearnedaftersellingathing.• ZakatisoneofthefivepillarsofIslamwhichisordrsbyAlmighty

Allahwhichispaidonthewealthwhichremainswithapersonforacompleteyear.Islamhasfixeditsrate,thatis2.5%.

• Ushrmeansone-tenth.Itispaidonagriculturalproducts.

CHAPTER

8 ALGEBRAIC EXPRESSIONS

Animation 8.1: Algebraic expressionSource & Credit: eLearn.Punjab

version 1.1



2 3

8. Algebraic Expressions 8. Algebraic ExpressionseLearn.Punjab eLearn.Punjab

2version 1.1 version 1.1


After studying this unit, students will be able to:• Defineaconstantasasymbolhavingafixednumericalvalue.• Recallavariableasaquantitywhichcantakevariousnumerical

values.• Recallaliteralasanunknownnumberrepresentedbyaletterof

analphabet.• Recallanalgebraicexpressionasacombinationofconstantsand

variablesconnectedbythesignoffundamentaloperations.• Define a polynomial as an algebraic expression in which the

powersofvariablesareallwholenumbers.• Identifyamonomial,abinomialandatrinomialasapolynomial

havingoneterm,twotermsandthreetermsrespectively.• Addtwoormorepolynomials.• Subtractapolynomialfromanotherpolynomial.• Findtheproductof:

1. monomialwithmonomial.2. monomialwithbinomial/trinomial.3. binomialswithbinomial/trinomial.

• Simplifyalgebraicexpressionsinvolvingaddition,subtractionandmultiplication.

• Recognizeandverifythealgebraicidentities:1. (x+a)(x+b)=x2+(a+b)x+ab,2. (a+b)2=(a+b)(a+b)=a2+2ab+b2,3. (a-bf=(a-b){a- b)=a-2ab+b2,4. a2-b2=(a-b)(a+b).

• Factorizeanalgebraicexpression(usingalgebraicidentities).• Factorizeanalgebraicexpression(makinggroups).

8.1 Algebraic Expressions

Algebraisoneoftheusefultoolsofmathematics. Itusesmathematical statements to describe the relationships betweenthingsthatvaryovertime.Inourpreviousclass,wehavelearntthe

introduction to the basic ideas of algebra including the effects ofsomebasicoperations,conceptofvariablesandsimplificationofanalgebraicexpressionwithitsevaluation.

Do you KnowAlgebra is an Arabic word which means ‘‘bringing together

broken parts”.

8.1.1 Literals

Thelettersoralphabetsthatweusetorepresentunknownsarecalledliteralnumbers.Forexample,areaofarectanglecanbecalculatedbymultiplyingitslengthandbreadth,i.e.Area=l x bWhere, l = length andb = breadth. Clearly, l andb represent theunknowns.So,thesearecalledliteralnumbers.

8.1.2 Constant

Asymbolhavingafixednumericalvalueiscalledaconstant.Forexample,2,7,11,etc.areallconstants.

8.1.3 Variable

A symbol representedby a literal and can take variousnumericalvaluesiscalledavariable,i.e.inx+1,xisavariableand1isaconstant.

8.1.4 Algebraic Expressions

Acombinationofconstantsandvariablesconnectedbythesignsof fundamentaloperations (+,',-, x) is calledanalgebraicexpression,i.e.8,4x+y,x2+y2,a2-2ab+b2,etc.


4 5



• Algebraic Terms Thepartsofanalgebraicexpressionseparatedby theoperationalsigns“+”and“-”arecalleditsterms,i.e.inx + y,xandyareitstwoterms.

8.1.5 Polynomial

Normally,thewordpolyisusedformorethanonethingsbutinalgebrapolynomialrepresentsanalgebraicexpressioncontainingasingletermaswellastwoormorethantwoterms.Forapolynomial,theexponentsofthevariablesmustbethewholenumbers.Forinstance,9,3x,x2+2,x3+2x +1,etc.allexpressionsarepolynomialsbutx -2+1,x1/2+3x+2,etc.arenotpolynomialsbecausetheirexponents(-2;1/2)arenotwholenumbers.“Analgebraicexpressioninwhichtheexponentsofvariablesareallwholenumbersiscalledapolynomial”.

8.1.6 Identification of a monomial, bionomial and trinomial

Monomial:Apolynomialhavingonetermiscalledamonomial,i.e.5,3x,2ab,etc.aremonomials.Binomial:Apolynomialhavingtwoterms iscalledabinomial, i.e.6x+a,a -3b,etc.arebinomials.Trinomial:Apolynomialhavingthreetermsiscalledatrinomial,i.e.x2+3x+5,2a+3b+c,etc.aretrinomials.

Inroutine,wewriteapolynomialindescendingorderandarrangea polynomial with respect to one variable, e.g. we arrange thepolynomialx3y2+y4+x4-x2y3withrespecttoxas,x4+x3y2-x2y3+y4.

EXERCISE 8.11. Addthetermstowriteanalgebraicexpression.

(i) 2ab, 3bc, ca (ii) 7l2, 3m2, -8 (iii) p2, -q2, -r2

(iv) 5xyz, 2yz, -8xy (v) -2ab, a,-bc, c (vi) 9 , 8 , 10 , 2lm mm ml- -

2. Writeconstantsandvariablesusedineachexpression. (i) x + 3 (ii) 3a + b - 2 (iii) l2 + m2 + n2

(iv) 5a (v) 2x2 - 1 (vi) 3l2 - 4n2

3. Identifymonomials,binomialsandtrinomials. (i) x + y - z (ii) -6l (iii) 2x2 - 3 (iv) abc (v) x2+2xy+y2 (vi) (-a)3 (vii) l - m (viii) 7a2 - b2 (ix) lm + mn + nl (x) 2a - 3b - 4c (xi) 11x2y2 (xii) a3+a2b+ab2

8.2 Operations with Polynomials

Recall that inourpreviousclass,wehave learnt theapplicationofsomebasicoperationsinalgebra.Nowwelearnmoreaboutthem.

8.2.1 Addition and Subtraction of Polynomials

Inpolynomials,weusethesamemethodforadditionandsubtractionthatweuseforliketerms,whichisgivenbelow.• We can arrange the polynomials in any order but usually we

arrange them in descending order and write the like termsverticallyinasinglecolumnforadding.

• For subtraction, we just change the signs of the terms of thepolynomialwhicharetobesubtractedandsimplyaddthem.

Example 1: Addthefollowingpolynomials. (i) 2x4y2 + x3y + x2y - 5, 2x2y - x4y2 + x3y + 1, 2 - x4y2 + x3y - 7x2y (ii) x2 + y2 + 2xy, y2 + z2 + 2yz, 2x2 +3y2 + z2 , z2 - 2xy - 2yz

Solution: (i) 2x4y2 + x3y + x2y - 5, 2x2y - x4y2 + x3y + 1, 2 - x4y2 + x3y - 7x2yArrangethepolynomialsindescendingorderandwriteallliketermsinasinglecolumn.


6 7



2x4y2 + x3y + x2y - 5-x4y2 + x3y + 2x2y + 1-x4y2 + x3y - 7x2y + 2

• 2x4y2 - x4y2 - x4y2 = (2-1-1)x4y2 = 0x4y2

• x3y + x3y + x3y = (1+1+1) = 3x3y• x2y + 2x2y - 7x2y = (1+ 2- 7) = -4x2y• -5 + 1 + 2 = -20x4y2 + 3x3y - 4x2y - 2

Thusx4y2 + 3x3y - 4x2y - 2 istherequiredpolynomial.(ii) x2 + y2 + 2xy, y2 + z2 + 2yz, 2x2 + 3y2 + z2 , z2 - 2xy - 2yzArrangethepolynomialsindescendingorderandwriteallliketermsinasinglecolumn.x2+y2+2xy y2 + z2 + 2yz2x2+3y2+z2

z2 - 2xy - 2yz

• x2+2x2=(1+2)x2=3x2

• y2 + y2 + 3y2=(1+1+3)y2=5y2

• z2 + z2 + z2 = (1+1+1)z2 = 3z2

• 2xy - 2xy = (2- 2)xy = 0• 2yz - 2yz = (2-2)yz = 03x2+5y2+3z2+0xy +0yz

Thus,3x2+5y2+3z2istherequiredpolynomial.

Example 2: Whatshouldbeaddedto3+2x-x3y2+4x2ytoget2x3y2+x2y-3x-1?Solution:Arrangethepolynomialsindescendingorder.1stpolynomial=2x3y2 + x2y - 3x - 12ndpolynomial=-x3y2 + 4x2y + 2x + 3Ifwesubtractthe2ndpolynomialfrom1stpolynomial,wecangettherequiredpolynomial.

• 2x3y2 + x3y2 = (2+1) = 3 x3y2

2x3y2 + x2y - 3x - 1 • x2y - 4x2y = (1-4) = -3x2y

*x3y2 + 4x2y + 2x + 3 • -3x -2x = (3-2)x = -5x

3x3y-3x2y-5x-4 • -1-3=-4Thus,3x3y-3x2y-5x-4istherequiredpolynomial.

Example 3:Whatshouldbesubtractedfrom3x4y2+11+4x6y4-6x2ytoget1+x4y2-x2y+x6y4?Solution:Arrangethepolynomialsindescendingorder.1stpolynomial=4x6y4+3x4y2-6x2y+112ndpolynomial=x6y4+x4y2-x2y+1

Ifwesubtractthe2ndpolynomialfrom1stpolynomial,wecangettherequiredpolynomial.4x6y4 + 3x4y2 - 6x2y + 11+x6y4 + x4y2 " x2y + 1

• 4x6y4 - x6y4 = (4 -1) = 3x6y4

• 3x4y2 - x4y2 = (3 -1) = 2x4y2

• -6x2y+x2y= (-6+1) = -5x2y• 11 -1=10

3x6y4 + 2x4y2 - 5x2y + 10

Thus,3x6y4 + 2x4y2 - 5x2y + 10 istherequiredpolynomial.

EXERCISE 8.2

1. Addthefollowingpolynomials. (i) x2 + 2xy + y2, x2 - 2xy + y2

(ii) x3 + 3x2y - 2xy2 + y3, 2x3 - 5x2y -3xy2 - 2y3

(iii) a5 + a3b - 2ab3 + b3, 4a5 + 3a3b + 2ab3 + 5b3

(iv) 2x4y - 4x3y2 + 3x2y3 - 7xy4, x4y - 4x3y2 - 3x2y3 + 8xy4

(v) ab5 + 12a2b4 - 6a3b3 + 10a4b2 - a5b, 4ab5 - 8a2b4 + 6a3b3 - 6a4b2 + 4a5b2. IfA = x - 2y + z, B = -2x + y + zandC = x + y - 2zthenfind.

(i) A-B (ii) B-C(iii)C- A (iv) A -B-C (v) A+B-C (vi)A-B +C

3. Whatshouldbeaddedtox7 - x6 + x5 - x4 + x3 - x2 + x + 1toget x7 + x5 + x3 - 1?4. What should be added to 2x4y3 - x3y2 - 3x2y - 4 to get 5x4y3 + 2x3y2 + x2y - 9?5. Whatshouldbesubtractedfrom5x5y5 - 3x3y3 + 10xy - 9toget 3x5y5 + 7x3y3 - 11xy + 19?

8.2.2 Multiplication of Polynomials

Whilemultiplyingtwopolynomialsinadditiontothecommutative,associativeanddistributivelaws,wealsousethelawsofexponentsthatcanbeseeninthegivenexamples• Multiplying monomial with monomialExample 1: Findtheproductof:(i)4a2 and 5a3 (ii)5x2 and 3y2 (iii)3l4m2n and 7l5m8n6


8 9



Solution:(i) 4a2 and 5a3

4a2 x 5a3 = (4x5)(a2x a3) = (20)(a)(2+3) a productlaw = 20a5 am x an = am+n

(ii) 5x2 and3y2

5x2 x 3y2 = (5x3)(x2 x y2) = (15)(x2y2) = 15x2y2

(iii) 3l4m2n and7l5m8n6

3l4m2n x 7l5m8n6 = (3x7)(l4 x l5)(m2 x m8)(n x n6)= 21 x l4x5 x m2+8 x n1+6 = 21l9m10n7

• Multiplying monomial with Binomial / Trinomial

Example 2: Simplify:(i) 3x2(x2 - y2) (ii) -6a2(2a + 3b) (iii) 2l2m2n2 (3lm - 2mn + 5nl)Solution:(i) 3x2(x2 - y2)

= (3x2 x x2) - (3x2 x y2) = 3(x2+2) - 3(x2 x y2) = 3x4 - 3x2y2

(ii) -6a2 (2a +3b) = (-6a2 x 2a) + (-6a2 x 3b) =(-6 x 2 )(a2 x a) + (-6x 3)(a2 x b) = (-12)(a2+1) +(-18) (a2b) = -12a3 - 18a2b

(iii) 2l2m2n2(3lm -2mn + 5nl) = (2l2m2n2 x 3lm) - (2l2m2n2 x 2mn) + (2l2m2n2 x 5nl) = (2x3) (l2m2n2 x lm)- (2x2)(l2m2n2 x mn) +(2x5)(l2m2n2 x nl) = (6)(l2+1m2+1n2) - (4)(l2m2+1n2+1)+ (10)(l2+1m2n2+1) = (6)(l3m3n2) -(4)(l2m3n3) +(10)(l3m2n3) = 6l3m3n2 - 4l2m3n3 + 10l3m2n3

EXERCISE 8.3

1. Multiply (i) 7m and -8 (ii) 2ab and 3a2b2 (iii) 4xy and 2x2y (iv) - 4ab and -2bc (v) 3lm3 and 3mn (vi) -6x2y and 3xyz2 (vii) 2a2b and 5a2b3

(viii) l2mn and lm3n6 (ix) -4x2yz7 and 8xy4z3

2. Simplify (i) lm (l + m) (ii) 2p(p + q) (iii) 3a (a - b) (iv) 2x (3x + 4y) (v) 2a (2b - 2c) (vi) 2lm (l2m2 - n) (vii) a(a + b - c ) (viii) 3x( x - 2 y - 2z) (ix) 3p2q (p3 + q2 - r4)

• Multiplying binomial with Binomial / TrinomialExample 3: Multiply: (i) (x + 3)(x - 1) (ii) (2a + 3b)(2a - 3b) (iii) (m + 2)(m2 - 2m + 3) (iv) (2x - 1)(x2 - 5x + 6)Solution:(i) (x + 3)(x - 1)

x + 3x x - 1

(ii) (2a + 3b) (2a - 3b)(2a + 3b)

x (2a-3b)a2ax3b=6ab

(iii) (m + 2)(m2-2m + 3) (m + 2)

x (m2 - 2m + 3)

x2 + 3x - x - 3

4a2 + 6ab - 6ab - 9b2

m3 - 2m2 + 3m + 2m2 - 4m + 6

x2 + 2x - 3 4a2-9b2 m3 - m + 6Thus,(x + 3) (x - 1) = x2 + 2x - 3

Thus,(2a + 3b)(2a - 3b)= 4a2 - 9b2

Thus, (m + 2)(m2 - 2m + 3) = m3 - m +6

Example 4: Simplify: (i) 2x2 (x3 - x) - 3x (x4 - 2x)+ 2 (x4 - 3x2) (ii) (5a2 - 6a + 9)(2a - 3) -(2a2 - 5a + 4)(5a + 1)Solution: (i) 2x2 (x3 - x) - 3x (x4 - 2x) + 2 (x4 - 3x2)

= (2x2 % x3 - 2x2 % x) - (3x % x4 - 3x % 2x)+ (2x4 - 6x2) = (2x2+3 -2x2+1) - (3x1+4 - 6x1+1) + (2x4 - 6x2) = (2x5 - 2x3) - (3x5 - 6x2) + (2x4 - 6x2) = 2x5 - 2x3 - 3x5 + 6x2 + 2x4 - 6x2


10 11



= (2x5 - 3x5) + 2x4 - 2x3 + (6x2 - 6x2)= -x5 + 2x4 - 2x3

(ii)(5a2 - 6a + 9)(2a - 3)- (2a2 -5a + 4)(5a + 1)5a2 - 6a + 9% 2a - 3

2a2 - 5a + 4% 5a + 1

10a3 - 12a2 + 18a - 15a2 + 18a - 27

10a3 - 25a2 + 20a + 2a2 - 5a + 4

10a3 -27a2 + 36a - 27 10a3 -23a2 + 15a + 4 (5a2 - 6a + 9)(2a - 3) - (2a2 -5a + 4)(5a + 1)= (10a3 - 27a2 + 36a - 27) - (10a3 - 23a2 + 15a + 4)= 10a3 - 27a2 + 36a - 27 - 10a3 + 23a2 -15a - 4= (10a3 - 10a3) + (-27a2 + 23a2)+(36a - 15a) +(-27-4)= - 4a2 + 21a - 31

EXERCISE 8.4

1. Multiply (i) (3a + 4)(2a - 1) (ii) (m + 2)(m - 2) (iii) (x - 1)(x2 + x + 1) (iv) (p - q )( p2 + pq + q2) (v) (x + y)(x2 - xy + y2) (vi) (a + b)(a - b) (vii)(l - m)(l2 - 2lm + m3) (viii) (3p - 4q)(3p + 4q) (ix)(1 - 2c)(1 + 2c) (x) (2x - 1)(4x2 + 2x + 1) (xi)(a + b)(a2 - ab + b2) (xii) (3 - b)(2b - b2 + 3)

2. Simplify (i) (x2 + y2)(3x + 2y ) + xy (x - 3y) (ii) (4x + 3y)(2x - y)-(3x- 2y)(x + y) (iii) (2m2 - 5m + 4)(m + 2) - (m2 + 7m - 8)(2m - 3) (iv) (3x2 + 2xy - 2y2)(x + y) -(x2 - xy + y2)(x - y)

8.3 Algebraic Identities

An algebraic identity is a simplified form consisting of thealgebraic termswhich provide uswith a rule for solving the longcalculationsinashortandeasyway.Forexample,tocalculatetheareaoffourrectangularwallsweusethefollowingidentityasashort

method.Areaoffourwalls=2(l + b )x hNowwelearnsomeimportantalgebraicidentities.Identity 1:(x + a)(x + b) = x +(a+ b)x + abProof:L.H.S.=(x + a)(x + b)

= x(x + b)+ a(x + b)= x2 + bx + ax + ab= x2 + (b + a)x + ab= x2 + (a + b)x + ab =R.H.S.

ThusL.H.S.=R.H.S.Identity 2:(a + b)2 = a2 + 2ab + b2

Proof:L.H.S.=(a + b)2 = (a + b)(a + b)

= a(a + b)+ b(a + b)= a2 + ab + ba + b2

= a2 + ab + ab + b2

= a2 + 2ab + b2

ThusL.H.S.=R.H.S.Identity 3:(a - b)2 = a2 - 2ab + b2

Proof:L.H.S.=(a - b)2 =(a - b)(a - b)

= a(a - b) - b(a - b) = a2 - ab - ba + b2

= a2 - ab - ab + b2

= a2 - 2ab + b2

ThusL.H.S.=R.H.S.Identity 4:(a -b)(a + b)= a2 - b2

Proof:L.H.S.=(a - b)(a + b)

=a(a+b)-b(a+b)=a2+ab-ba-b2

=a2+ab-ab-b2

=a2 - b2

ThusL.H.S.=R.H.S.


12 13



Example 1: Simplifythebinomialsbyusingtheidentity.(i) (x + 6)(x + 5) (ii) (x - 4)(x - 8) (iii) (2x + 9)(2x - 3)Solution:(i) (x + 6)(x + 5)Byusingtheidentity,

(x + a)(x + b)= x + (a+ b)x + ab(x + 6)(x + 5) = x + (6 + 5)x +(6x 5) = x2 + 11x + 30

(ii) (x - 4)(x - 8)By using the identity,

(x + a)(x + b) = x2 + (a + b)x + ab(x - 4)(x - 8) = x2 + (-4-8)x + (-4) x(-8) = x2 - 12x + 32

(iii) (2x+9)(2x-3)Byusingtheidentity,

(x + a)(x + b) = x2 + (a + b)x + ab(2x+9)(2x-3)=(2x)2+(9-3)2x+9%(-3) =4x2+(6)2x+(-27) =4x2+12x-27

Example 2: Findthesquareofthefollowingbyusingidentity.(i) (4a + 3b ) (ii) (2x - 3y)Solution:(i) (4a + 3b)Byusingtheidentity,

(a + b)2 = a2 + 2ab + b2

(4a + 3b)2 = (4a)2 + 2 % (4a) % (3b)+ (3b)2

= 16a2 + 24ab + 9b2

(ii) (2x - 3y)Byusingtheidentity,

(a - b)2 = a2 - 2ab + b2

(2x - 3y)2 = (2x)2 + 2 %(2x) % (3y)+ (3y)2

= 4x2 - 4xy + 9y2

Example 3: Write the product of the following binomials byusingidentity,(i) (3x - 4y),(3x + 4y) (ii) (7a - 9b), (7a + 9b)(iii) (6x2y2 + 8a2b2),(6x2y2 - 8a2b2)

Solution:(i) (3x - 4y),(3x + 4y)Byusingtheidentity, (a + b)(a - b) = a2 - b2

(3x + 4y)(3x - 4y) = (3x)2 - (4y)2

= 9x2 - 16y2

(ii) (7a - 9b),(7a + 9b)Byusingtheidentity, (a + b)(a - b) = a2 - b2

(7a+9b)(7a-9b)=(7a)2-(9b)2

=49a2-81b2

(iii) (6x2y2 + 8a2b2),(6x2y2 - 8a2b2)Byusingtheidentity, (a + b)(a - b) = a2 - b2

(6x2y2+8a2b2)(6x2y2-8a2b2)=(6x2y2)2-(8a2b2)2

=36x4y4-64a4b4

EXERCISE 8.5

1. Simplifythefollowingbinomialsbyusingtheidentity.(i)(x + 1)(x + 2) (ii)(x - 2)(x - 4) (iii)(a + 5)(a + 3)(iv)(b + 6)(b - 9) (v)(2x + 3)(2x - 7) (vi)(2y + 1)(2y + 5)(vii)(3b - 1)(3b - 7)(viii)(4x + 5)(4x + 3) (ix)(5y - 2)(5y + 6)(x)(8a + 7)(8a - 3)

2. Byusingidentity,findthesquareofthefollowingbinomials.(i)x+y(ii)3a+4 (iii)x-y (iv)a+2b (v)2x+3y(vi)2a-b(vii)3x-2y (viii)4x+5y (ix)7a-8b

3. Findtheproductofthefollowingbinomialsbyusingidentity.(i)(x+y)(x-y) (ii)(3a-8)(3a+8) (iii)(2a+7b)(2a-7b)(iv)(x+3y)(x-3y) (v)(6a-5b)(6a+5b) (vi)(9x-11y)(9x+11y)

8.4 Factorization of Algebraic Expressions

Inarithmetic,wehave learnt that theprimenumberswhicharemultipliedwitheachothertogetaproductarecalledfactors.Forexample,


14 15



18=1x2x3x3....................(i)Similarlyinalgebra,ifanalgebraicexpressionisaproductoftwoormorethantwootheralgebraicexpressions,thenthetwoormorethan twootheralgebraicexpressionsarecalled the factorsof theproduct.Forexample,3xy - 3xz = 3x (y - z )..............(ii)Herein(ii),3,xand(y - z)arethefactorsof3xy - 3xzand3&xareknownascommonfactorsofthewholeexpression.So,wecandefinethefactorizationofanalgebraicexpressionas,“Theprocessofwritinganalgebraicexpressionastheproductoftwoormoreexpressionswhichdivideitexactlyiscalledthefactorization”.Inalgebra,theoppositeofthefactorizationiscalledtheexpansion.This is the process of multiplying the factors to get the samealgebraicexpression.

Example 1: Resolvethefollowingexpressionsintofactors.(i) 3a + 6b + 9c (ii) a(x - y) -b(x - y)Solution:(i) 3a + 6b +9c(taking3ascommon)

=3(a+2b+3c)(ii) a(x - y) -b(x - y)(takingx-yascommon)

=(x-y)(a-b)Example 2: Factorize.(i)(ax - y) - (ay - x) (ii)(x2 + yz ) - (y + z)xSolution:(i) (ax - y) - (ay - x)= ax - y - ay + x= ax + x - ay - y= x(a + 1) - y(a + 1)= (x - y)(a + 1)

(ii)(x2 + yz) - (y + z)x=x2 + yz - yx - zx = x2 - zx - yx + yz = x(x - z) - y(x - z) = (x - y)(x - z)

• Factorization of a2 - b2 type expression Ifwehavethedifferenceoftwosquaredterms,thenwecanfactorizethemasonefactoristhesumoftwotermsandtheotherfactoristhedifferenceoftwoterms.Forexample,thedifferenceof

twosquaredtermsisa2-b2.So,a2-b2=a2+ab-ab-b2

=a(a+b)-b(a+b)=(a-b)(a+b)

Example 1: Factorize.(i) 49x2 - 81y2 (ii) 18a2x2 - 32b2y2 (iii)(6a - 8b)2 - 49c2

Solution:(i) 49x2 - 81y2

=(7x)2-(9y)2

=(7x-9y)(7x+9y)[Usingtheidentity,a2-b2=(a-b)(a+b)](ii) 18a2x2 - 32b2y2

=2[9a2x2-16b2y2]=2[(3ax)2-(4by)2]=2(3ax-4by)(3ax+4by)

[Usingtheidentity,a2-b2=(a-b)(a+b)](iii) (6a - 8b)2 - 49c2

=(6a-8b)2-(7c)2

=(6a -8b-7c)(6a-8b+7c)[Usingtheidentity,a2-b2=(a-b)(a+b)]

EXERCISE 8.6

1. Resolveintofactors.(i)5x2y - 10xy2 (ii)2a - 4b + 6c (iii)9x4 + 6y2 + 3 (iv) a3b + a2b2 + ab3 (v) x2yz + xy2z + xyz2 (vi) bx3 + bx2 - x - 1(vii) x2 + qx + px+ pq (viii) ab - a - b + 1 (ix)(pm + n) + (pn + m) (x)(a2 + bc) - (b + c)a (xi) x2 - (m + n) x + mn (xii) x3- y2 + x - x2y2

2. Factorizebyusingidentity.(i)4a2 - 25 (ii)4x2 - 9y2 (iii)9a2 - b2 (iv)9m2 - 16n2 (v)16b2 - a2 (vi)-1 + (x + 1)2 (vii)8x2 - 18y2 (viii)(a + b)2 - c2 (ix) x2-(y + z)2 (x)7x2 - 7y2

(xi)5a2 - 20b2 (xii) x4 - y4


16 17



• Factorization of a2 ± 2ab + b2 type expressionsWe know that the square of a binomial can be expanded as thesquareof1sttermplusthesquareof2ndtermplusthetwiceoftheproductofthetwoterms,i.e.

• (a+b)2=a2+2ab+b2

• (a-b)2=a2-2ab+b2

Example 3: Resolveintofactors.(i)8x2 -56x +98 (ii) 16a4 + 14a2b2 + 9b2

Solution:(i)8x2 - 56x + 98 = 2[4x2 - 28x + 49]Itcanbewrittenas:

=2[(2x)2 - 2(2x)(7) + (7)2] a28x = 2(2x)(7)= 2(2x - 7)2

[Usingtheidentity,a2 -2ab + b2 = (a - b)2]Thus,therequiredfactorsare2and(2x - 7)2.(ii)16a4 + 14a2b2 + 9b2

=(4a2)2+2(4a2)(3b2)+(3b2)2 a2(4a2)(3b2)=24a2b2

Byusingtheidentity,a2 + 2ab + b2 = (a + b)2, wehave(4a2)2+2(4a2)(3b2)+(3b2)2=(4a2+3b2)2

Thus,therequiredfactorsare(4a2+3b2)2

Example 2:Factorize

Solution:

Itcanbewrittenas,

Byusingtheidentity,a2 - 2ab + b2 = (a - b)2

Thus,therequiredfactorsare

EXERCISE 8.7

1. Resolveintofactorsbyusingidentity.(i)x2 + 8x + 16 (ii) x2 - 2x + 1 (iii) a4 - 14a2 + 49 (iv)1 + 10m + 25m2 (v)4x2 - 12xy + 9y2 (vi)9a2 + 30ab + 25b2

(vii)16a2 + 56ab + 49b2 (viii)36x2 + 108xy + 81y2 (ix)49m2 + 154m + 121 (x)64a2 - 208ab + 169b2 (xi)3x4 + 24x2 + 48 (xii)11x2 + 22x + 11(xiii)44a4 - 44a3b + 11a2b2 (xiv)a4 + 16a2b + 64b2 (xv)1-4xyz +4x2y2z2 (xvi) 16x3y -40x2y2 + 25xy3

2. Factorizebyusingtheidentity.

(i)a2x2 + 2abcx + b2c2 (ii)

(iii) (iv)

(v) (vi)

(vii)a2b2c2x2 - 2a2b2cdxy + a2b2d2y2 (viii)

• Factorization by making groupsLookatthefollowingalgebraicexpressions.

• x2 + ax + 4x + 4a• al + bm + bl + am• pq - 2p - q + 2

Wecanobservefromtheabovegivenexpressions.Thattherearenocommonfactorsinthemandtheseexpressionsarealsonotanyof the three typesdiscussed inother sections. For factoring suchtypesofexpressions,werearrangethemandmaketheirgroupsasgivenintheexamples.Example 1: Factorize5a + xa + 5x + x2

Solution:5a +xa+5x+x2

Step 1:Rearrangetheexpression. x2 + 5x + xa + 5aStep 2:Maketheirgroups. (x2 + 5x) + (xa + 5a)


18 19



Step 3:Factoroutthecommonfactors. x(x + 5) + a(x + 5)Step 4:Factoroutthecommonexpression. (x + 5)(x + a)Thus,therequiredfactorizationis(x + 5)(x + a)

Example 2: Factorize:2a2b + 4ab2 - 2ab - 4b2

Solution:2a2b + 4ab2 - 2ab - 4b2

Step 1:Rearrangetheexpressionand 2a2b - 2ab + 4ab2 - 4b2

factoroutthecommonfactor. = 2b (a2- a + 2ab - 2b)Step 2: Maketheirgroups. = 2b [(a2 - a) +(2ab - 2b)]Step 3: Againfactoroutthecommonfactors.

= 2b [a (a -1)+ 2b(a - 1)]

Step 4:Factoroutthecommonexpression. = 2b [(a - 1)(a + 2b)]Thus,therequiredfactorizationis2b(a - 1)(a + 2b).

EXERCISE 8.8

1. Factorizethefollowingexpressions.(i)lx -my + mx - ly (ii)2xy - 6yz + x - 3z (iii)p2 + 2p - 3p -6 (iv) x2 + 5x - 2x - 10 (v)m2 - 7m + 2m - 14 (vi) a2 + 3a - 4a + 12(vii) x2 - 9x + 3x - 27 (viii) z2 - 8z - 4z + 32 (ix) t2 - st + t - s(x) n2 + 5n - n - 5 (xi) a2b2 + 7ab - ab - 7 (xii) l2m2-13Im-2lm+26

REvIEw EXERCISE 8

1. Answerthefollowingquestions. (i) Whatismeantbyliterals? (ii) Defineaconstant. (iii) Whatisabinomial? (iv) Whatisanalgebraicidentity? (v) Definethefactorizationofanalgebraicexpression.

2. Fillintheblanks. (i) (a + b)2=______. (ii) (a - b)2 =______. (iii) (x + a)(x + b) =_____. (iv) a2 - b2=______. (v) Asymbol representedbya literalandcan takevarious numericalvaluesiscalleda______. (vi) Apolynomialhavingonlyonetermiscalled______.


4. Resolveintofactors. (i) 10a2 - 200a4b (ii) 36x3y3z3 -27x2y4z + 63xyz4 (iii) 15x4y + 21x3y2 - 27x2y2 -33xy4

(iv) x(a2 + 11)-16(a2 + 11) (v) x2(ab + c) + xy(ab + c) + z2(ab + c)

5. IfA = 2(x2 + y2 + z2), B = -x2 + 3y2 - 2z2 and C = x2 - y2 - 3z2,then find:

(i) A + B + C (ii) B + C - A (iii) A - B + C (iv) A + B - C (v) A - B - C (vi) B - C - A

6. Simplifythefollowingpolynomials (i) (x - 2y)(x + 2y) (ii) (4x2)(3x +1) (iii) 2x(x + y) - 2y(x - y) (iv) (a2b3)(2a - 3b) (v) (a2 - b2)(a2 + b2) (vi) (a2 + 1)(a2 - a - 1) (vii) x(y + 1) - y(x + 1) - (x - y) (viii) a2(b2 - c2) + b2(c2 - a2) + c2 (a2 - b2)

7. Simplifythefollowingbyusingidentity. (i) (3x - 4)(3x + 5) (ii) (2a - 5b)2

8. Factorize. (i) a2 - 26a + 169 (ii) 1 - 6x2y2z + 9x4y4z2 (iii) 7ab2 - 343a (iv) 75 - 3(x - y)2

(v) 49(x + y)2- 16(x - y)2 (vi) 2 29 416 9

a ab b+ +

(vii)2 2

2 22 2

2a ac cl lm mb bd d

- +

(viii) 2 29 36( )5 25

a m- -


20

8. Algebraic Expressions eLearn.Punjab

20version 1.1

SuMMARy

• The letters or alphabets thatwe use to represent unknowns /numbersarecalledliterals.

• Asymbolrepresentedbyaliteralthatcantakevariousnumericalvaluesiscalledavariable.

• Asymbolhavingafixedvalueiscalledaconstant.•• Acombinationofconstantsandvariablesconnectedbythesigns

offundamentaloperationsiscalledanalgebraicexpression.• Thepartsofanalgebraicexpressionseparatedbytheoperational

signs‘+’and‘-’arecalleditsterms.• Analgebraicexpressioninwhichtheexponentsofvariablesare

allwholenumbersiscalledapolynomial.• Apolynomialcanbearrangedinanyorderbutusuallywearrange

itindescendingorder.• Analgebraicequationwhichistrueforallvaluesofthevariable

occurringintherelationiscalledanalgebraicidentity.• The process of writing an algebraic expression as the product

of twoormoreexpressionswhichdivide itexactly iscalledthefactorization.

CHAPTER

9 Linear equations

version 1.1

Animation 9.1: Linear EquationSource & Credit: eLearn.Punjab



3

9. Linear Equations 9. Linear EquationseLearn.Punjab eLearn.Punjab



After studying this unit, students will be able to:• Definealinearequationinonevariable.• Demonstratedifferenttechniquestosolvelinearequations.• Solvelinearequationsofthetype: + = ax b c,

+=

+

ax b m .cx d n

• Solvereallifeproblemsinvolvinglinearequations.

9.1 Linear Equation

Theequationwhichcontainsasinglevariablewiththeexponentof1iscalledthelinearequationinonevariable.Forexample,

• 2x+4=6x• 3y-7=14-2y• z+5=0

(Linearequationinvariablex)(Linearequationinvariabley)(Linearequationinvariablez)

9.2 Solution of a Linear Equation

A linear equation in one variable is an open sentence. Theprocessoffindingthatvalueofthevariablewhichmakesitatruesentenceiscalleditssolution.Thatvalueofthevariablewhichmakestheequationatruesentenceiscalledasolutionoftheequation.Asolutionisalsocalledarootoftheequation.(i)x+2=5Heresolutionisx=3ortherootisx=3becausewhenweputx=3,weget5=5whichisatruestatement.(ii)2x=4Weputx=2andget4=4,atruestatement,thusthesolutionoftheequationisx=2.

Equation Left-hand side Right-hand sidex+3=6

2x-5=56=12+x

x+32x- 56

65

12+x

• Addition Wecanaddthesamenumbertobothsidesofanequation.Forexample,ifwearegivenanequation.x+2=4...(i)Wecanadd3tobothsidesof(i)toobtain:x+2+3=4+3orx+5=7...(ii)(i)and(ii)areequivalentequationswhichhavethesamesolutionorroot• Subtraction Wecansubtractthesamenumberfromthebothsidesofanequation.Forexample;x+5=3...(i)x+5-2=3-2orx+3=1...(ii)(i)and(ii)areequivalentequations.• Multiplication Wecanmultiplyboth sidesofanequationbyanon-zeronumber.Forexample:

• Division Wecandividebothsidesofanequationbyannon-zeronumber.Forexample:


5



6x=12 ...(i)Multiplybothsidesby4

4 8 4× = ×

x=32 ...(ii)

Example 1:Solvetheequation,x- 6=2.Solution:

x-6=2...........(i)Add6tobothsides,x- 6+6=2+6x=8Example 3:Solvetheequation,x+ 1=5.Solution:

x+ 1=5...........(i)Subtract1frombothsidesof(i),x+ 1-1=5-1 x=4

Example 2:Solvetheequation,

1 26

=x .Solution:

1 26

=x

Multiplybothsidesof(i)by4

16 6 26

× = ×x

orx=12

Example 4: Find the solution of the following equations andverifythesolution.

(i) 6 42 3+ +

=x x (ii) 8 4 1

16 4+

=-

xx

Solution:

(i) 6 42 3+ +

=x x

6 46 6

2 3+ +

× = ×x x (MultiplybothsidesbytheL.C.M6

of2and3) 3(x+ 6)=3(x+ 6) 3x+ 18=2x+ 8 3x- 2x=8- 18 x=- 10 (Separatevariablesandnumbers)

(ii) 8 4 116 4

+=

-x

x

or 8 4(16 4 ) 1 (16 4 )16 4

+- × = × -

-xx x

x

or 8x+4=16- 4x (MultiplybothsidesbytheL.C.M16- 4x)or 8x+4x=16- 4(Separatevariablesandnumbers)or 12x=12

or 12 112

= =x

EXERCISE 9.1

1. Solvethefollowingequations.

(i) 1 48

=x

(ii) x - 7= -15 (iii) x + 1= 5

(iv) 2x - 6= 0 (v) 11x - 2= 20 (vi) 17x = 255

(vii) 5x - 3= 12 (viii) 11 - x= 6 (ix) 2 85

=x

(x) 7 23

- =x (xi) 5 10

2=

x (xii) 9x + 11= 83

(xiii) 5 74-

=x (xiv) 2 5

4- =

x (xv) 7 3 192+

=x

2. Findthesolutionsofthefollowingequations. (i)5x-3= 3x-5

(ii)3x+8= 5x +2(iii) 12x-3= 5(2x+1)

(iv) 10(2-x)= 4(x-9) (v) 3 3

1 5-

=+

xx

(vi) 1 42 3

-=

-xx

(vii) 2 13 4 7

-=

+xx

(viii) 3 8 15 2

-=

-xx

(ix) 2 22 5 5

+=

-xx

(x) 3 62 3+ +

=x x (xi) 7 6 1

18-

=-x

x (xii) 4 3 7

3 2+ +

=x x


7



9.2.1 Solving Real Life Problems involving Linear Equations

Letussolvesomereallifeproblemsinvolvinglinearequations.

Example 1: A96cmlongwireisgiventheshapeofarectanglesuchthatitslengthis12cmmorethanthebreadth.Findthelengthandbreadthoftherectangle.Solution:Supposethatbreadthoftherectangle=xthenlengthoftherectangle=x+12lengthofthewire(perimeter)=96cmByusingtheformula2(length+breadth)=perimeteror2[(x+12)+x)]=96or2(2x+12)=96or4x+24=96or4x=96- 24or4x=72orx=18Thus,breadthoftherectangleis18cmLengthoftherectangle=x+12=18+12=30cm

Example 2: After32yearsfromnow,aboywillbe5timesasoldashewas8yearsback.Howoldistheboynow?Solution:Supposetheageoftheboy=xAfter32yearsagewillbe=x+328yearsbacktheagewas=x-8Accordingtothesituation, x+32=5(x-8) or x+32=5x-40 or 5x-x=40+32 or 4x=72 or x=72/4=18years

Thus,theboyis18yearsold.

EXERCISE 9.2

1. Hussain bought 10 ice creams. He gave Rs. 1,000 to the shopkeeper.Theshopkeeper returnedhimRs.250.Forhow muchdidhebuyoneicecream?2. Thelengthofarectangleis2cmmorethantwiceitsbreadth.If the perimeter of the rectangle is 28cm, find its length and breadth.3. ThepriceofapenisRs.42andofanotebookisRs.18.Calculate howmanypensandnotebooksyoucanbuyforRs.480ifyou wanttobuyanequalquantityofboth.4. Afather’sageistwicehisdaughter’sagebut16yearsagothe father’s age was 4 times his daughter’s age. Calculate their ages.5. DistributeanamountofRs.200betweenRaheemandUsman such that Raheem gets Rs.50 more than twice as much as Usmangets.6. The length of a marriage hall is 4 times its breadth. If the perimeter of the hall is 240m, find the length and the breadthofthemarriagehall.7. Aslam’sageishalfofhisfather’sagebut15yearsagohisage

wasjust 13offather’sage.Findhispresentagenow.

8. DistributeanamountofRs.500among2brothersand1sister suchthat,

a. sistergetstwiceasmuchasbrothersgets.b. eachbrothergetstwiceasmuchasthesisterdoes.

REvIEw EXERCISE 9

1. Answerthefollowingquestions. (i) Whatisalinearequation? (ii) Whatismeantbythesolutionofanequation? (iii) Definethelinearequationinonevariable.


9



2. Fillintheblanks. (i) The equationwhich contains a single variablewith the exponent1iscalledthelinearequationinone_______. (ii) Asolutionisalsocalleda______oftheequation. (iii) Theprocessoffindingthevalueofavariabletomakea sentencetrueiscalledits_______. (iv) Additionofthe_______tobothsidesofanequationdoes notaffectitsequality.3. Tick(p)thecorrectanswer.

4. Solveeachofthefollowingequations.

(i) 2x + 3= 5x + 7

(ii) 5 25 33 3

- = -x x

(iii) 3 5 5 72 3 2 3

- = +x x (iv) 33(3 1) 8( ) 02

- - + =x x

(v) 5 3 3 5( 2 ) + (2 ) 0

2 2 2 2- - =x x (vi)

2 2 3 13 3 2 3

- = -x x

(vii) 3 52 (1 )2 2

- = -x x (viii) 2 (3 1) 2 1

5- = -x x

(ix) 1 2 4 3( 3) + 3 3 6

-- =

xx

(x) 1 2 1 7( 3)+ (4 3)+3 3 3 2

- = -x x

5. Findthenumber. (i) -3addedtoanumberisequalto10. (ii) Threetimesanumberis15. (iii) 13subtractedfromthreetimesanumberis8. (iv) Anumberdividedby5gives9lessthantwicethenumber. (v) Thesumofthreeconsecutivenumbersis45.

SuMMARy

• Anequationwhichcontainsasinglevariablewiththeexponent“1”iscalledthelinearequationinonevariable.

• Thevalueofthevariablethatmakestheequationatruesentenceiscalledthesolutionoftheequation.

• Anumbernon-zeroincaseofdivisioncanbeadded,subtracted,multipliedanddividedon theboth sidesofanequationand itdoesnotaffecttheequalityoftheequation.

CHAPTER

10 Fundamentals oFgeometry

Version: 1.1

Animation 10.1: Circle radiansSource & Credit: .wikipedia

https://en.wikipedia.org/wiki/File:Circle_radians.gif

https://en.wikipedia.org/wiki/File:Circle_radians.gif


2 3

10. Fundamentals of Geometry 10. Fundamentals of GeometryeLearn.Punjab eLearn.Punjab



After studying this unit, students will be able to:• Defineadjacent,complementaryandsupplementaryangles.• Defineverticallyoppositeangles.• Calculate unknown angles involving adjacent angles,

complementary angles, supplementary angles and verticallyoppositeangles.

• Calculateunknownangleofatriangle.• Identifycongruentandsimilarfigures.• Recognizethesymbolofcongruency.• Applythepropertiesfortwofigurestobecongruentorsimilar.• Applyfollowingpropertiesforcongruencybetweentwotriangles.

1. SSSbSSS2. SASbSAS3. ASAbASA4. RHSbRHS

• Describeacircleanditscenter,radius,diameter,chord,arc,majorandminorarcs,semicircleandsegmentofthecircle.

• Drawasemicircleanddemonstratetheproperty;theangleinasemicircleisarightangle.

• Drawa segmentof a circleanddemonstrate theproperty; theanglesinthesamesegmentofacircleareequal.

Introduction

Geometry has a long and glorious history. It helped us tocreateart,buildcivilizations,constructbuildingsanddiscoverotherworlds.Therefore,theknowledgeofgeometryremainedthefocusofancientmathematicians.

Themost importantworkdone intheareaofGeometrywasofEuclid.Hisbook,“Euclid’sElements”hadbeentaughtthroughouttheworld.

10.1 Properties of Angles

Twodifferentrayswithacommonstartingpointformananglewhichisdenotedbythesymbol∠.Theunitofmeasuringanangleisdegree(°).Anglesareclassifiedbytheirdegreemeasures,e.g.rightangle,acuteangle,obtuseangle,etc

10.1.1 Adjacent, Complementary and Supplementary Angles

• Adjacent Angles Thewordadjacentmeans“nextorneighbouring”andbytheadjacentangleswemeananglesnexttoeachother.Twoanglesaresaidtobeadjacentif:(i) theyhavethesamevertex.(ii) theyhaveonecommonarm.(iii) otherarmsoftwoanglesextendon oppositesidesofthecommonarm.Forexample,inthefigure(10.1),itcanbeseenthatthetwoangles∠AOC and ∠BOC areadjacentbecausetheyhavethesamevertex“O”

andonecommonarm

OC .Otherarms

OA and

OB ontheopposite

sidesofthecommonarm

OC .

• Complementary Angles Twoanglesarecalledcomplementaryangleswhentheirsumofdegreemeasureisequalto90°.Forexample,inthefigure(10.2),

m∠BAD = 600

m∠SAD = 300

m∠BAD + m∠SAD = 600+300

=900

Thus∠BAD and∠SAD arecomplementaryangles.


4 5



• Supplementary AnglesTwoanglesarecalledsupplementaryangleswhentheirsumofdegreemeasureisequalto180°.Forexample,inthefigure(10.3), m∠BAD = 600

m∠SAD = 1200

m∠BAD + m∠SAD = 600+1200

=1800

Thesumofthetwoanglesis180°.Hence,thesearesupplementaryangles.

10.1.2 Vertically Opposite Angles

Apairofanglesissaidtobeverticallyopposite,iftheanglesare formed from two intersecting lines and the angles are non-adjacent.Suchanglesarealwaysequalinmeasurementasshowninthefigure(10.4).

HereitcanbeseenthattwolinesAB

andCD

intersecteachotheratpoint“O”andformfouranglesi.e.∠AOC,∠BOC,∠BODand∠AOD.Heretheangles∠AODand∠BOCarecalledverticallyoppositeangles.Similarlytheangles∠AOCand∠BODarealsoverticallyoppositetoeachother.Wecanprovethatverticallyoppositeanglesareequal inmeasureasgivenbelow. Inthefigure(10.4),wecanalsoseethat: m∠AOD+m∠AOC=1800 (supplementaryangles) m∠AOC+m∠BOC=1800 (supplementaryangles) m∠AOD+m∠AOC=m∠AOC+m∠BOC m∠AOD=m∠BOCSimilarly,wecanalsoprovethat: m∠AOC =m∠BOD

10.1.3 Finding Unknown Angles involving Adjacent, Complementary, Supplementary and Vertically Opposite Angles

Example 1: Look at the following diagram and name all thepairsof:

(a) Adjacentangles (b) VerticallyOppositeAngles

Solution:(a) Adjacent angles(i) ∠ u and∠ v (ii) ∠ v and∠ w (iii) ∠ w and∠ x(iv) ∠ x and∠ y (v) ∠ y and∠ z (vi) ∠ z and∠ uAllthesearepairsofadjacentangles(b) Vertically Opposite Angles(i) ∠ u and∠ x (ii) ∠ w and∠ z (iii) ∠ v and∠ yAllthesearepairsofverticallyoppositeangles.

Example 2: Writethemeasurementofmissingangles.

(i)(ii)

(iii)(iv)


6 7



Solution:(i) (ii)

Wehavem∠AOC=35°.Since, ∠AOC and ∠BOC arecomplementaryangles.So, m∠AOC+m∠BOC=900

350+m∠BOC=900

m∠BOC=900-350

=550

Thus,m∠BOC=550

We have to ∠AOB = 1350 andm∠BOC = 450. Since,∠AOB and∠COD are vertically oppositeangles.So,m∠COD=m∠AOBThus,m∠COD=1350

Similarly, ∠BOC and ∠AOD areverticallyoppositeangles.So,m∠AOD=m∠BOCThus,m∠AOD=450

(iii) (iv)

We have m∠BOC = 1280.Since, ∠AOC and ∠COB aresupplementaryangles.So,m∠AOC+m∠BOC=1800

m∠AOC+1280=1800

m∠AOC=1800-1280

=520

Thus,m∠AOC=520

We have m∠AOB = 750 and∠AOC = 450. Since, ∠AOC and∠COBareadjacentangles.So, m∠AOC+m∠BOC=∠AOB450+ m∠BOC=750

m∠BOC=750-450

=300

Thus,m∠BOC=300

10.1.4 Finding Unknown Angle of a Triangle

If themeasurementsof twoanglesof a triangle are known,thenthethirdanglecanbecalculated.

Example 3: Findthemissingangleineachtriangle.

Solution:Weknowthatthesumofthemeasuresofthreeanglesofatriangleisalwaysequalto180°.Letususethesameanglesumpropertyofatriangletofindthefollowingunknownangles.(i) (ii)

Wehave,m∠B=520,m∠C=480,m∠A=?

Weknowthat m∠A + m∠B + m∠C = 1800

m∠A + 520+480=1800

m∠A + 1000=1800

m∠A=1800-1000=800

Thus,m∠A=800

Wehave,m∠O=900,m∠A=400,m∠B=?

Weknowthat m∠O + m∠A + m∠B = 1800

900+400+ m∠B = 1800

1300 + m∠B = 1800

m∠B=1800-1300=500

Thus,m∠B=500

(iii)


8 9



Wehave,m∠D=1100

m∠E=300 (verticallyoppositeanglesareequal)

m∠F=?Weknowthat m∠D + m∠E + m∠F = 1800 (verticallyoppositeanglesareequal)

1100+300+ m∠F = 1800

1400 + m∠F = 1800

m∠F=1800-1400=400

Thus,m∠F=400

EXERCISE 10.1

1. Namealltheanglesinthefigurewhichareadjacent.

2. Inthefollowingfigure∠AOB and∠BOCareadjacentangles.i.e. m∠AOB = 200 andm∠AOC = 420.Findm∠BOC.

3. Identify the pairs of complementary and supplementary angles. (i) 500,400 (ii) 1200,600 (iii) 700,700

(iv) 1300,500 (v) 700,200 (vi) 500,1000

4. Inthegivenfigure,findalltheremainingangles.

5. Findtheunknownanglesofthegiventriangles.

6. Findtheremaininganglesinthegivenrightangledtriangle.

10.2 Congruent and Similar Figures

10.2.1 IdentificationofCongruentandSimilarFigures• Congruent FiguresTwogeometricalfiguresaresaidtobecongruent,iftheyhavesameshapeandsize.Lookatthefollowingfigures.

Thesidesandanglesoftwogivenfigurescanbematchedasgivenbelow.

Fromtheabove,itcanbeseenthatthetwofigureshaveexactlythesameshapeandsize.Therefore,wecansaythatthesetwofiguresarecongruent.


10 11



10.2.2 Recognizing the Symbol of Congruency

Wehavelearntthattwogeometricalfiguresarecongruent iftheyhave thesameshapeandsamesize.Thecongruencyof twofiguresisdenotedbyasymbolbwhichisreadas“iscongruentto”.Thesymbolbismadeupoftwoparts.i.e.• ; meansthesameshape(similar).•=meansthesamesize (equal).

ThesymbolforcongruencewasdevelopedbyGottfriedLeibniz.Hewasbornin1646anddiedin1716.GottfriedLeibniz made very important contributions to thenotationofMathematics.

• Similar Figures Thefigureswith thesameshapebutnotnecessarily thesamesizearecalledsimilarfigures.Thesimilarityofthegeometricalfiguresisrepresentedbythesymbol“;”.Forexample,allcirclesaresimilartoeachotherandallsquaresarealsosimilar.

Figure 1Butthesearenotcongruenttoeachotherasthesizeofeachcircleisdifferent.

10.2.3 Applying the Properties for two Figures to the Congruent or Similar

Thedifferencebetweensimilarandcongruentfiguresisthat:• Congruentfigureshavethesameshapeandsamesize.• Similarfigureshavethesameshape,butmaybedifferentinsizes. Letususethesamepropertiesfortwofigurestofindwhether theyarecongruentorsimilar.

Example 1: Decide whether the figures in each pair arecongruentoronlysimilar.Solution:

Example 2: Howarethefollowingshapesrelated?Solution:


12 13



Forexample,if

b) Twosidesofonetriangleandtheirincludedanglearecongruent tothetwocorrespondingsidesandangleoftheothertriangle, i.e.SASbSAS.Forexample,

c) Two angles of one triangle and their included side are congruent to the two corresponding angles and side of the othertriangle,i.e.ASAbASA.Forexample,

d) Thehypotenuseandoneside (baseorattitude)ofatriangle arecongruenttothecorrespondinghypotenuseandoneside oftheothertrianglei.e.RHSbRHS.Forexample,

EXERCISE 10.2

1. Definesimilargeometricalfigureswithexamples.2. Aresimilarfigurescongruent?Giveexamples.3. Arecongruentfiguressimilar?Provethiswithexamples.4. Identifycongruentandsimilarpairsoffigures.

10.3 Congruent Triangles

Whileconsideringtriangles,twotriangleswillbecongruentif:

a) All the threesidesofone trianglearecongruent toall three correspondingsidesoftheothertriangle,i.e.SSSbSSS.


14 15



EXERCISE 10.3

1. If themeasuresof twoanglesofa triangleare35°and80°, thenfindthemeasureofitsthirdangle.2. Inthegivenfigure,∠1b∠3and∠2b∠4. Thenprovethat,3ABD b 3BDC

3. If the bisector of an angle of a triangle bisects its opposite side,thenprovethatthetriangleisanisoscelestriangle.

4. Inthegivenfigure, and LN MP LP MN≅ ≅ , thenprovethat and P N LMN MLP∠ ≅ ∠ ∠ ≅ ∠

5. Inthegiventriangle3ABC, and CD AB CA CB⊥ ⊥ , thenprove

that and AD BD ACD BCD≅ ∠ ≅ ∠

6. Lookatthefiguretoshowthat3ABPb3DCP

10.4 Circle

A circle is themost familiar shape of the geometry thatweoftenobservearoundus,awheel,theSun,fullMoon,coinsofone,twoandfiverupeesaresomeexamplesofacircle.So,wecandefineitas:“A circle is a set of points in a plane which areequidistantfromafixedpoint,calledcenterofthecircle”.Let“P”beanypointwhichismovingsothatitremainsatequaldistancefromafixedpoint“O”.Thispointwill traceacirclewhosecenterwillbe“O”asshowninthefigure.

• Radius Thedistancebetweenthecenterandanypointonthecircleiscalledradius.Herethedistancebetween“O”and“P”iscalledradius.InthisfigureOP istheradialsegment.

• DiameterAlinesegmentthatpassesthroughthecenterofacircleandtouchestwopointsonitsedgeiscalledthediameterofthecircle.Inthegivenfigure, AB isthediameterofthecircle.


16 17



• Chord Alinesegmentjoiningtwopointsonacircleiscalledthechord.Thefigureshowsthat AB isthechordofthecircle.

• Arc Ifwe cut a circle itwill giveus the curvedshapeasshown in thefigure.Thiswholefigure

iscalledsectorwhereasBC iscalledarcof the

circlewithradii and OB OC

Anarc consistsof twoendpoints andall thepointson the circlebetweentheseendpoints.Whenwecutacircleinsuchawaythatasectorofthecircleissmallerthantheother,wegettwotypesofarcs,i.e.minorarcandmajorarc.• Minor Arc Anarcwhichissmallerthanhalfofthecircle iscalledminorarc.Itisnamedbyusingtwoendpointsofthearc.• Major Arc Anarcwhichismorethanhalfofthecircleiscalledmajorarc.It isnamedbythreepoints.Thefirstandthirdarcendpointsandthemiddlepointisanypointonthearcbetweentheendpoints.Forexample,IfwecutacircleatanypointsAandB.Itwillprovide

ustwoarcsAB andACB .ThearcAB isminorarc

whereasawaythroughACB makesamajorarc.

10.4.2 Semicircle

Nowconsiderthecase,ifwecutacirclesuchthatboththearcsareequal.Thenitcanhappenonlywhenitiscutalongitsdiameter.Thisprocessgeneratestwosemicirclesortwohalfcircles.

On joining these semicircles, we get the same circle again. Nowletusdemonstratethepropertyofasemicirclethattheangleinasemicircleisarightangle.Step 1: Draw a circle, mark its center and draw a diameter throughthecenter.Usethediametertoformonesideof atriangle.Theothertwosidesshouldmeetatavertex somewhereonthecircumference.Step 2: Dividethetriangle intwobydrawingaradiusfromthe centertothevertexonthecircumference.Step 3: Recognizethateachsmalltrianglehastwosidesthatare radii. All radii are the same in a particular circle. This meansthateachsmall trianglehastwosidesthesame length.Theymustthereforebothbeisoscelestriangles.Step 4: Becauseeachsmalltriangleisanisoscelestriangle,they musteachhavetwoequalangles.Step 5: The sum of internal angles in any triangle is 180°. By comparisonwiththediagraminstep5,wenoticethatthe threeanglesinthebigtrianglearea,banda+b.So,we canwriteanequationas:

a+b + (a+b)=1800

2a+2b=1800

00180 90

2a b+ = =

Henceproved,theangleinasemicircleisarightangle.


18 19



10.4.3 Segment of a Circle

Segmentofacircleisapartofcircle,cutalonganychord.Lookatthefollowingfigures.

ThearcACB correspondingtothechordAB iscalledsegmentofthecircle.InsecondfigureAOBisthesectorofthecircle.Now cut the circle along a chord other thendiameter. Youhave two segments.Nowdrawtwo inscribed angles of the smaller segment∠ACB and∠ADB. Nowmeasure them,wewillseethatbothanglesareequalinmeasurement.i.e.m∠ACB=m∠ADB.Again draw the two angles in the major segments of the circle,∠APBand∠AQB.Measurethemagain,haveyounoticedthatagainm∠APB=m∠AQB.Therefore,wecandrawaconclusion;theanglesinthesamesegmentofacircleareequal.

EXERCISE 10.4

1. Drawacirclewithradius 5OA cm= andfinditsdiameter.2. Inthegivenfigurelocatemajorandminorarcs.

3. Howmanydiametersofacirclecanbedrawn.Drawacircle andtraceatleast‘5’differentdiameters.

4. Drawasemicircleofaradiusof8cm.5. Draw a circle and cut it into two segments. Construct two inscribedanglesineachofthesegmentandmeasurethem.

REVIEW EXERCISE 10

1. Answerthefollowingquestions. (i) Whatismeantbytheadjacentangles? (ii) What is thedifferencebetween complementary angles andsupplementaryangles? (iii) Definetheverticallyoppositeangles. (iv) Whatisthesymbolofcongruency? (v) Whatisacircle? (vi) Differentiatebetweenmajorandminorarcs.

2. Fillintheblanks. (i) From the _______ angleswemean, angles next to each other. (ii) If the sumof twoangles is _______ then theanglesare calledcomplementaryangles. (iii) The non-adjacent angles which are formed from two intersectinglinesarecalled_______angles. (iv) Twofiguresarecongruentiftheyaresamein_______and ________. (v) A circle is a setofpointswhichareequidistant froma fixedpoint,calledits_______. (vi) Twotrianglesarecongruent,ifthreesidesofonetriangle are_______tothethreesidesofothertriangle. (vii) The figures with same shape but not necessarily the samesizearecalled_______figures. (viii) Verticallyoppositeanglesarealways_______inmeasure.

3. Tick(p)thecorrectoption.


20 21



4. Find unknown measures of the sides and angles for these congruentshapes.

5. Ifaandbarecomplementaryangles,thenfindthevalueof‘b’ ifmeasureof‘a’is40°.6. Ifxandyaretwosupplementaryangleswhereasm∠x=600. Thenfindthemeasureofy.

SUMMARY

• Two angles with a common vertex, one common arm anduncommon arms on opposite sides of the common arm, arecalledadjacentangles.

• If the sum of two angles is 90°, then the angles are calledcomplementaryangles.

• If the sum of two angles is 180°, then the angles will besupplementarytoeachother.

• If two lines intersect each other, the non adjacent angles, soformedarecalledverticalangles.

• Twogeometricalfiguresaresimilariftheyaresameinshape.• Figuresarecongruentiftheyaresameinshapeandsize.• Twotriangleswillbecongruent ifanyof thefollowingproperty

holds.(i) SSSb SSS (ii) SASb SAS (iii) ASAb ASA (iv) RHSb RHS

• A path traced by a point remaining equidistant from the fixedpoint,generatescircle.

• Alinesegmentjoiningtwopointsonacircleiscalledchordofacircle.

• Asegmentofacirclecutacrossdiameteriscalledsemicircle.• Anangleinasemicircleisarightangle.• Theanglesinthesamesegmentofacircleareequal.

CHAPTER

11 Practical Geometry

Version: 1.1

Animation 11.1: Parallelogram AreaSource & Credit: .wikipedia

https://en.wikipedia.org/wiki/File:Parallelogram_area_animated.gif

https://en.wikipedia.org/wiki/Parallelogram


2 3

11. Practical Geometry 11. Practical GeometryeLearn.Punjab eLearn.Punjab



After studying this unit, students will be able to:• Dividealinesegmentintoagivennumberofequalsegments.• Dividealinesegmentinternallyinagivenratio.• Constructatrianglewhenperimeterandratioamongthelengths

ofsidesaregiven.• Constructanequilateraltrianglewhen

• baseisgiven• altitudeisgiven

• Constructanisoscelestrianglewhen• baseandabaseanglearegiven,• vertexangleandaltitudearegiven,• altitudeandabaseanglearegiven.

• Constructaparallelogramwhen• twoadjacentsidesandtheirincludedanglearegiven,• twoadjacentsidesandadiagonalaregiven.

• Verifypracticallythatthesumof.• measuresofanglesofatriangleis180°• measuresofangleofaquadrilateralis360°

11.1 Line Segment

Weknowthatwecancomparetwolinesegmentsbymeasuringtheirlengths.

Infigure(i),wecanseethatthelinesegmentAB isshorterthanCDbecausethelengthof AB islessthanthatofCD i.e.mAB mCD<

In figure (ii), we can notice that the line segment AB is longer

thanCD becausethelengthof AB isgreaterthanthatofCD i.e.

mAB mCD> .

In figure (iii), we can check the third and final possibility of thecomparison of two line segments. Herewe can see that two line

segmentsareequalinlength,i.e.mAB mCD=Suchlinesegmentswhichhaveequallengthsarecalledcongruentlinesegments.

11.1.1 Division of a Line Segment into Number of Equal Segments

Inourpreviousclass,wehavelearntthatafinesegmentcanbe divided into an even number of line segments by successivebisectionofitspartsasshownbelow:

Nowwe learn amethod for dividing a line segment into an oddnumberofcongruentparts,i.e.3,5,7,........andsoon.Weshalllearnthismethodwiththehelpofanexample.

Example 1: Dividealinesegmentoflength14cmin7equallinesegments.Solution:Steps of construction: (i)Drawa14cmlonglinesegmentPQ.(Usearuler)


4 5



(ii) DrawarayPRmakinganacuteanglewithalinesegmentPQ. (Usearuler)

(iii) DrawanotherrayQSmakingthesameacuteanglewithPQ (iv) Draw7arcs(accordingtotherequiredpartsofalinesegment) ofsuitableradius, intersectingtherayPRatpointsA, B, C, D, E, F andOrespectively.(StartfromAandforeacharcconsider thepreviouspointasthestartingpoint).(v) Similarly,draw7arcsofsameradius,intersectingtherayQSat pointsF’, E’, D’, C’, B’, A’andO’respectively.(vi) DrawlinesegmentsPO’, AA’, CC’, DD’, EE’,FF’ andOQ.Theseline segmentsintersectthelinesegmentPQatpointsU, V, W, X, Y andZ respectively.

(vii) Hence and PU ,UV ,VW ,WX ,XY ,YZ ZQ are the required 7

congruentpartsoflinesegmentPQ

Note: Resultcanbecheckedbymeasuringtheeachpartwitha divider.

11.1.2 Division of a Line Segment in a given Ratio Inpreviousexample,wecannoticethattheintersectingpoints

U, V, W, X, YandZarealsodividingthePQ inaspecificratio.• ThepointUisdividingthelinesegmentPQinratio1:6.• ThepointVisdividingthelinesegmentPQinratio2:5.

• ThepointWisdividingthelinesegmentPQinratio3:4.• ThepointXisdividingthelinesegmentPQinratio4:3.• ThepointYisdividingthelinesegmentPQinratio5:2.• ThepointZisdividingthelinesegmentPQinratio6:1.Nowwelearnthedivisionofalinesegmentinagivenratio.Supposethatthegivenratioisa: b: c.So,Step 1: DrawalinesegmentPQ.Step 2: Draw two rays PR and QS making acute angles with linePQ.Step 3: Draw a + b + c number of arcs at equal distance on

and PR QS .

Step 4: Jointhepointsof and PR QS correspondingtotheratio a : b : c.TheintersectingpointsdividethelinesegmentPQ inagivenratioa : b : c.

Example 2: Divide a long line segment of length 12cm in theratio1:3:4.Solution:Steps of construction: (i) Drawa12cmlinesegmentPQ.(Usearuler)(ii) DrawtworaysPRandQSmakingsameacuteanglewithline segmentPQ.(iii) Draw1+3+4=8arcsofsuitableradius,intersecttherayPRat pointsA, B, C, D, E, F, GandOandintersecttherayQSatpoint G’, F’, E’, D’, C’, B’, A’andO’.(iv) DrawlinesegmentsPO’,AA’,DD’andOQ’,theselinesegments intersectthelinesegmentPQatpointsXandY.


6 7



(v) The line segments PX, XY and YQ are three parts of a line segmentPQwhicharedividingitintheratio1:3:4.

EXERCISE 11.1

1. Dividealinesegmentoflength6cminto3congruentparts.2. Dividealinesegmentoflength7.5cminto5congruentparts.3. Draw a line segment of length 10.8cm and divide it into 6 congruentparts.4. Dividealinesegmentoflength10cminto5congruentparts.5. Draw a line segment of length 9.8cm and divide it into 7 congruentparts.6. Dividethelinesegment: a. AB oflength4cmintheratio1:2.

b. PQ oflength7.5cmintheratio2:3. c. XY oflength9cmintheratio2:4. d. DE oflength6cmintheratio1:2:3. e. DE oflength6cmintheratio1:1:2. f. LM oflength13.5cmintheratio2:3:4.

g. UV oflength11.2cmintheratio1:2:4.

11.2 Triangles

Wearealreadyfamiliarwiththedifferentmethodsoftriangleconstruction.Hereweshalllearnmoremethods.

11.2.1 Construction of a Triangle when its Perimeter and Ratio among the Lengths of Sides are Given

Atrianglecanalsobeconstructedifwehavetheperimeterofatriangleandtheratioamongthelengthsofitssides.

Example 1: Constructatrianglewhoseperimeteris12cmand2:3:4istheratioamongthelengthsofitssides.Solution:Steps of construction:(i) DrawalinesegmentPQoflength12cm.(usearuler)(ii) DividethelinesegmentPQinthegivenratio2:3:4.(iii) Consider thepointLascenteranddrawanarcbyusingthe lengthofPLasradius.(iv) AgainconsiderthepointMascenteranddrawanotherarcby usingthelengthofMQasradius.

(v) LabelthepointofintersectionoftwoarcsasN.(vi) JointhepointNtoLandMrespectively. ∆LMNistherequiredtriangle.

11.2.2 Construction of Equilateral Triangles

Anequilateraltriangleisatriangleinwhichallthreesidesareequalandallthreeanglesarecongruent.Itcanbeconstructedusingagivenlengthofalinesegment(baseandaltitude).Letusconstructanequilateraltrianglewhen:


8 9



• Base is GivenHereitbeginswiththegivenbasewhichisthelengthofeachsideoftherequiredequilateraltriangle.Letusmakeitclearwithanexample.

Example 2: Constructanequilateraltriangle∆ABCwhosebaseis4.5cmlong.Solution:Steps of construction:(i) Drawa linesegmentof length4.5cmusinga ruler. Label its endpointsAandB.(ii) Placetheneedleofthecompassesat pointAandopenitsothatthetipof thepenciltouchesthepointB.(iii) Draw an arc of radius AB with centreA.(iv) Drawanotherarcofradius AB with centre B. This arc will intersect the first arc at one point. Name the meetingpointoftwoarcsasC.(v) FinallyjointhepointCwiththepointAandwiththepointB. Thetriangle∆ABCistherequiredequilateraltriangle.• Altitude is Given Anequilateraltrianglecanalsobeconstructedifitsaltitudeisgiven.Letuslearnthismethodwithanexample.

Example 3: Construct an equilateral triangle ∆XYZ whosealtitudeis ofmeasure5cm.Solution:Steps of construction:(i) DrawalineABusingarulerandmarkanypointPonit.

(ii) DrawaperpendicularPQ

onlineAB,i.e.PQ AB⊥

.(iii) FrompointPdrawanarcofmeasure5cm.Thisarcwillcutthe perpendicularPQatthepointXasshown.

(iv) Construct the anglesof 30° at pointX i.e.m∠PXY = 30° and m∠PXZ=30°.

∆XYZistherequiredequilateraltriangle.

11.2.3 Construction of Isosceles Triangles

Asisoscelestriangleisatriangleinwhichtwosidesareequalinlength.Thesetwosidesarecalledlegsandthirdsideiscalledthebase.Theanglesrelatedtothebasearealsocongruent.Anisoscelestrianglecanbeconstructedwhen:• Base and a Base Angle are Given Weknowthatthebaseanglesofanisoscelestrianglearealwaysequal.So,wecanconstructanisoscelestrianglewiththemeasureofitsbaseandbaseanglearegiven.

Example 4: Constructanisoscelestriangle∆ LMNwhosebaseisofmeasure6cmandmeasureofbaseangleis30°.Solution:Steps of construction:(i) DrawalinesegmentLM oflength6cm.(Byusingaruler)(ii) Constructananglem∠MLN=30°atthepointL.(iii) ConstructanotherangleLMNof30°atpointM.TheproducedarmsoftheseanglesintersectatpointN.


10 11



∆LMN istherequiredisoscelestriangle.• Vertex Angle and Altitude are Given Inanisoscelestriangle,theangleformedbythetwosidesofequallengthandoppositetothebaseiscalledvertexangle.Whenthealtitudeisdrawntothebaseofanisoscelestriangle,itbisectsthevertexangle.Thispropertycanbeusedtoconstructanisoscelestrianglewithitsvertexangleandmeasureofaltitudearegiven.

Example 5: Constructanisoscelestrianglewithaltitude=3.5cmandvertexangle=50o

Solution:Steps of construction:(i) DrawalineXYandchooseanypointDonit.

(ii) DrawaperpendicularDG

abovethebaseline.(iii) From pointD, draw an arc of radius 3.5cm to cut this perpendicularatpointC.(iv) Since the vertex angle is 50° and an altitude bisects it. So, perpendicular will show the angle of 25° on both sides,

i.e.0

050 252

=

(v) Drawtwoarmsmakinganangle25°withperpendicularCD, onbothsidesandletthesearmscutthebaselineatpoints A andB.

∆ABC istherequiredisoscelestriangle.

• Altitude and Base Angle are Given Weknowthatinanisoscelestriangle,baseanglesarecongruentandsumofthreeangles is180°. Itmeans, if thebaseangleofanisosceles triangle is given, we can find its vertex angle as shownbelow.Letthebaseanglebe40°andvertexanglebex.Thenaccordingtotheisoscelestriangle:40°+40°+x=180°

80°+x=180°x=180°-80°x=100°

Thus,thevertexangleisofmeasure100°.

Example 6: Constructanisoscelestrianglewithaltitude=4cmandbaseangle=50°.Solution:Steps of construction:(i) Weknowthat: 50°+50°+vertexangle=180° 100°+vertexangle=180° vertexangle=180°-100°=80°(ii) DrawalinePQandchooseanypointOonit.(iii) DrawtheperpendicularODabovethebaseline.(iv) From point O, draw an arc of radius 4cm and cut the perpendicularatpointC.

(v) Draw twoarmsmakinganangleof0

050 252

= at pointC on

bothsidesoftheperpendicularlineOD .(vi) LetthearmstouchthebaselineatpointsAandB.

∆ ABC isrequiredisoscelestriangle.


12 13



EXERCISE 11.2

1. Constructtheequilateraltrianglesofgivenmeasures. (i) base=4cm (ii) altitude=6cm (iii) altitude=5.5cm (iv) base=3.5cm

2. Constructanisoscelestrianglewhose: (i) base=3cmandbaseangle=45° (ii) altitude=4.8cmandvertexangle=100° (iii) base=5cmandbaseangle=65° (iv) altitude=4.2cmandbaseangle=35°

3. Constructatriangle∆LMNwhoseratioamongthelengthsofits sidesis2:3:4andperimeteris10cm.

4. Constructatriangle∆XYZwhoseperimeteris13cmand3:4:5is theratioamongthelengthofitssides.

5. Theperimeterofa∆XYZis12cmandratioamongthelengths ofitssidesis4:2:3.Constructthetriangle∆XYZ.

11.3 Parallelogram

Aparallelogramisafour-sidedclosedfigurewithtwoparallelandcongruent(equal in measurement)oppositesides.Theoppositeanglesofaparallelogramarealsocongruentanditsdiagonalsbisecteachotherasshowninthefigure(a).

figure(a)

Intheabovefigure(a),wecanseethat,

(i) and AB CD AD BC

(ii) m∠AB=m∠CDandm∠AD=m∠BC(iii) m∠DAB=m∠DCB.i.e.∠DAB = ∠DCBand m∠CDA=m∠CBA.i.e.∠CDA = ∠CBA

11.3.1 Construction of Parallelogram when two Adjacent Sides and their Included Angle are Given

Wecanconstructaparallelogramwhenthemeasurementsoftwoadjacent sides and the relatedanglewhich is formedby twosides,aregiven.

Example: ConstructaparallelogramABCDif,06 3 5 60mAB cm mAD . cm m A= = ∠ =

Solution:Steps of construction:(i) Draw AB 6cmlong.(ii) Constructanangleof60°atpointAi.e.m∠A=600.(iii) Drawanarcofradius3.5cm.(iv) NowtakethepointBascenteranddrawanotherarcofradius 3.5cm.(v) NowagainconsiderthepointDascenteranddrawanarcof radius 6cm. This arc will intersect the previous arc on the pointC.(vi) Join the point C and pointD and also join the point C and pointB.

Result: ABCDistherequiredparallelogram.


14 15



EXERCISE 11.3

1. ConstructtheparallelogramABCDwhere

07 4 60mAB cm mBC cm m ABC= = ∠ =2. ConstructtheparallelogramPQRSwhere

08 4 75mPQ cm mQR cm m PQR= = ∠ =3. ConstructtheparallelogramLMNOwhere

06 5 4 5 45mLM . cm mMN . cm m LMN= = ∠ =4. ConstructtheparallelogramBSTUwhere

07 7 4 4 30mBS . cm mST . cm m BST= = ∠ =5. ConstructtheparallelogramOABCwhere

06 3 3 1 70mOA . cm mAB . cm m OAB= = ∠ =6. ConstructtheparallelogramDBASwhere

09 2 8 40mBA cm mAS . cm m DBA= = ∠ =

• Construction of Parallelogram when two Adjacent sides and a Diagonal are given:

Aparallelogramcanbeconstructedwhenwehavetwoadjacentsidesandonediagonalasgivenintheexample.

Example: ConstructtheparallelogramABCDif,

4 3 6mAB cm mBC cm mCA cm= = =

Solution: We can examine that and AB BC are two sides becauseboth have a common point B and diagonal AC is also biggerinmeasurement.Steps of construction:(i) Drawa AC ofmeasure6cm.(ii) ConsiderthepointAascentreanddrawanarcofradius4cm

ontheuppersideoflinesegment AC anddrawanotherarc radiusof3cmonthelowersideoflinesegmentAC.

(iii) NowconsiderthepointCascentreanddrawanarcofradius 3cmontheuppersideofsegmentACanddrawanotherarcof radius 4cm on the lower side of AC. (These points of arcs intersectatpointBandD).(iv) FinallyjointhepointsBandDwiththepointAandthenwith thepointC.

Thus,ABCDistherequiredparallelogram.

11.3.2 Sum of measure of Angles of a Triangle and a Quadrilateral

• The sum of Measures of Angles of a Triangle is 180° Inanytriangle,thesumofmeasuresofitsanglesis180°.Itcanbeverifiedasgivenbelow.Verification: Let∆ABC be a triangle, then according to the givenstatement,wehavetoverifythat. m∠ACB + m∠ABC + m∠BAC= 180° Step 1: Draw a line ED parallel to the line segmentBCasshowninthefigure(a).Step 2: SincelineEDandlinesegmentBCare parallel. So, according to the propertiesofparallellines, m∠ACB = m∠CAD m∠ABC = m∠BAEStep 3: Labeltwoequalangles(∠ACBand ∠CAD)asxandlabel other two equal angles (∠ABC and ∠BAE) as y. Finally, labeltheangle∠BACasz.


16 17



Step 4: It can be seen that the sum ofmeasurement of three anglesx,yandzis180°becausetheseanglesareonastraight linei.e.m∠x + m∠y + m∠z = 180°

HenceVerifiedm∠ACB + m∠ABC + m∠BAC = 180°

• The sum of Measures of Angles in a Quadrilateral is 360°Wehavelearntthatthesumofthreeanglesinatriangleis180°.Let us use the same fact to verify that the sum of angles in aquadrilateralis360°.Verification:LetABCDbeaquadrilateral,thenwehavetoverifythat, m∠A + m∠B + m∠C + m∠D = 360°Step 1: Join thepointBwith pointD as showninthefigure (b).Thiswill dividethequadrilateral intotwo triangles,i.e. ∆ABDand∆BCD.Step 2: Thesumofanglesinatriangleis180°. So,Intriangle∆ABD,wehave m∠a + m∠b + m∠c = 180° Intriangle∆BCD,wehave m∠d + m∠e + m∠f = 180°Step3: Findthesumofalltheanglesofthequadrilateralas, m∠a + m∠b + m∠c + m∠d + m∠e + m∠f = 180°+180° m∠a + (m∠b + m∠d) + m∠e + (m∠c + m∠f ) = 360°Henceverified,m∠A + m∠B + m∠C + m∠D = 360°

EXERCISE 11.4

1. ConstructtheparallelogramMNARwhere

5 2 8 7mMN cm mMA . cm mNA cm= = =2. ConstructtheparallelogramDGRPwhere

5 5 1 9 6 8mDG . cm mGP . cm mDP . cm= = =

3. ConstructtheparallelogramABCDwhere

3 1 6 5 8mAD . cm mGP . cm mDP cm= = =4. ConstructtheparallelogramVTSEwhere

1 5 3 6 4 8mSR . cm mRT . cm mTS . cm= = =5. ConstructtheparallelogramDBCOwhere

4 4 6 6 7 7mBC . cm mBO . cm mCO . cm= = =6. ConstructtheparallelogramMASKwhere

3 1 6 4 5 2mMA . cm mAS . cm mMS . cm= = =

REVIEw EXERCISE 11

1. Answerthefollowingquestions. (i) Whichlinesegmentsarecalledcongruentlinesegments? (ii) Writethesumofinterioranglesofatriangle. (iii) Defineanequilateraltriangle. (iv) Nametheequalsidesofanisoscelestriangle. (v) Whatismeantbythevertexangleinanisoscelestriangle?

2. Fillintheblanks. (i) An________trianglecanbeconstructedifthelengthofits onesideisgiven. (ii) We compare two line segments by measuring their __________. (iii) Two line segments of an ________ length are called congruentlinesegments. (iv) A polygonwith three sides and three vertices is called a_________. (v) Theoppositeanglesofaparallelogramarealso________. (vi) Two equal sides of an isosceles triangles are called ________and3rdsideiscalledthe________.



18

11. Practical Geometry eLearn.Punjab

version: 1.1

4. Dividealinesegmentoflength9.8cminto7congruentparts.5. DividealinesegmentLM oflength13.5cmintheratio2:3:4.6. Constructanequilateraltrianglewhosealtitudeis3.8cm.7. Constructanisoscelestrianglewhosealtitudeis5cmandbase

anglesare01672

8. ConstructaparallelogramABCD,if:

5 4 2 4 6 6mAB . cm mBC . cm mAC . cm= = =

SuMMARy

• The linesegmentshavinganequal lengtharecalledcongruentlinesegments.

• Atriangleisapolygonwiththreesidesandthreeverticesanditssumofinterioranglesis180°.

• A triangle canalsobe constructedwith its perimeter and ratioamonglengthsofitssides.

• An equilateral triangle is a triangle inwhich all three sides areequalandallthreeanglesarecongruent.

• Anisoscelestriangleisatriangleinwhichtwosidesareequalandbaseanglesarecongruent.

• Aparallelogramisafoursidedclosedfigurewithtwoparallelandcongruentoppositesides.

CHAPTER

12 CIRCUMFERENCE, AREAAND VOLUME

Version: 1.1


2 3

12. Circumference, Area and Volume 12. Circumference, Area and VolumeeLearn.Punjab eLearn.Punjab



After studying this unit, students will be able to:• Expresspastheratiobetweenthecircumferenceandthediameter

ofacircle.• Findthecircumferenceofacircleusingformula.• Findtheareaofacircularregionusingformula.• Findthesurfaceareaofacylinderusingformula.• Findthevolumeofcylindricalregionusingformula.• Solvereallifeproblemsinvolving;• circumferenceandareaofacircularregion.• surfaceareaandvolumeofacylinder.

12.1 Circumference, Area and Volume

12.1.1 Expressing p as the Ratio between Circumference of a Circle and diameter

Thecircumferenceofacircleisthedistancearoundtheedgeof thecircle. Itcouldbecalled theperimeterof thecircle.Tofindthecircumferenceofanycircularthinglikeacoin,simplywrapanyadhesivetapearounditsuchthattheendpointoftapemustmeetthestartingpoint.Nowunfoldthetapeandagainpasteonanyflatsurfacethenmeasurethelengthofthetapetofindthecircumferenceofthatcircularthing.Weobservesomefollowingfiguresofthecircleswhosecircumferenceanddiametershavebeenfoundbythemethodsgivenabove.

Here, we can also calculate the ratio between circumferenceand diameter of the circles given above by finding

thevaluecdwhere“c”isthecircumferenceand“d”isthediameter

ofthecirclesA, B andCasgiveninthefollowingtable.

CirclesCircumference

(c)Diameter

(d)Ratio(c/d)

A 7.6 2.4 3.1666B 12.57 4 3.1425C 19.48 6.2 3.1419

We can see that the ratio between circumference and diameteris approximately the same. We denote this constant value by aGreeksymbolpthevalueofwhichistakenapproximatelyequalto

227

or3.14.

So,wecanwritetheabovestatementas,circumference ( )

diameter ( )c

dp=

Orsimplywecanwriteitas,cd

p=Therefore,c=dpButweknowthatd=2rSo,c=2prHence,c=dpor2prwhere‘c’isthecircumference,‘d’isthediameterand‘r’istheradius.

12.1.2 Finding the Circumference of a Circle Using Formula

Example 1: Find the circumference of a circle with diameter3.2cm.Solution:Diameter(d)=3.2cm Circumference(c)=?Usingtheformula,c=dp

223 2 10.06 ( )7

up to two decimal placesc . cm= × =


4 5



Example 2: Theradiusofacircleis4.7cm.Finditscircumference.Solution: Radius(r)=4.7cm Circumference(c)=? Usingtheformula,c=2pr

222 4 7 29.54 ( )7

two decimal placesc . cm= × × =

Example 3: The circumference of a circle is 418cm. Calculatethediameterandradiusofthecircle.Solution:Circumference(c)=418cm Radius(r)=? Diameter(d)=?(i) Usingtheformula,c=2pr

418 7 66.5

2 2 22cr cmp

×= = =

×

r=66.5cm(ii) Usingtheformula,c=pd

418 7 133

22cr cmp

×= = =

d=133cm

EXERCISE 12.1

1. Findtheunknownquantitywhen227

p =

2. Thediameterofacircleis11.6cm.Findthecircumferenceof thecircle.3. Theradiusofacircleis9.8cm.Findthecircumferenceofthe circle.4. Thecircumferenceofacircleis1.54cm.Findthediameterand

radiusofthecircle(when227

p = ).

5. The circumference of a circular region is 19.5cm, find its diameterandradius(whenp c3.14).

12.1.2 Area of a Circular Region

The area of a circular region is the number of square unitsinsidethecircumferenceofthecircle.Weknowthatifwehavethemeasurementsof the lengthandthewidthofarectangle,wecanfindtheareaoftherectanglebythefollowingformula.

Areaofarectangle=LengthxWidthHereweusethesameformulatocalculatetheareaofacircle.Tomakeitclear,consideracircleofanysuitableradiusasshowninthefigure(a).Nowwedividetheabovecircularregioninto8,16and32equalpartsand rearrange its radial segments after cutting them carefully, asgivenbelow.

(i) Ifwedividethiscircularregioninto8equalpartsandrearrange theirradialsegments,wegetthefollowingfigure(b).


6 7



(ii) If we divide this circular region into 16 equal parts and rearrangetheirradialsegments,wegetthefollowingfigure(c).

(iii) If we divide this circular region into 32 equal parts and rearrangetheirradialsegments,wegetthefollowingfigure(d).

On considering these figures we see that these figures are ofparallelogram in which the edges of half of the sectors areupwards and half of the sectors are downwards and which areadjacent to each other. Similarly, if we continue the divisionofcircularregionintomoresectors,thenthisfigurewillcompletelybe converted into rectangle. Otherwise if we divide the lastradial segment into two equal parts and placed them at theboth ends of rectangle, then we can get the same rectangle.

In this way we get the length of rectangle which is half

of the circumference2

2rp

of circle and width of therectangle is equal to the radius of the circle. The length of therectangle ishalfof thecircumferenceofthecirclebecausehalfoftheradialsegmentsareupwardsandhalfaredownwardsandthetotal lengthoftheseallsegmentsisequaltothecircumferenceofthecircle.

Lengthoftherectangle=12 circumferenceofthecircle

1 (2 )2

r rp p= =

Widthoftherectangle=Radiusofthecircle=r Areaofthecircularregion=Areaoftherectangle =LengthxWidth =prxr = pr2 Therefore, Areaofthecircularregion=pr2

Example 1: Theradiusofacircleis14.3cm.Findtheareaofthecircle.Solution: Radius(r)=14.3cm Areaofthecircle=? Usingtheformula, Areaofthecircle=pr2

222 14 3 14 37

. . cm = × ×

=642.68cm2

Example 2: The area of a circle is 172.1cm2 . Find thecircumferenceofthecircle.


8 9



Solution: Areaofthecircle=172.1cm2 Circumference(c)=?Weknowthatcircumference=2prandwecancalculatetheradiusofthecirclefromitsarea. Areaofthecircle=pr2

222172 17

. r=

2 2172 1 722.r cm×

=

r2 = 54.76cm2

r= 7.4 cmSo, c = 2pr

22Circumference ( ) 2 7 4 46.517

c . cm= × × =

EXERCISE 12.2

1. Findtheareaofeachofthefollowingcircles

2. Findtheareaofacirclewhosecircumferenceis31.43cm.3.The radius of a circle is 6.3cm. Calculate the area and circumferenceofthecircle.4. Thecircumferenceofacircle is26.4cm.Findtheareaof the circle.5. Findthecircumferenceofacirclewhoseareais38.5m2.

12.2 Cylinder

We are already familiar with the shape of a cylinder in oureverydaylife.Tinpackofsoftdrinks,pineapple’sslicejar,gheetins,oildrums,chemicaldrums,differenttypesofrodsandpipes,allareexamplesofacylinder.Forfurtherdetail,weexaminethefollowingfigure(a)ofacylinder.

Fromthegivenfigure(a),wecanobservethatacylinderisasolidwhichconsistsofthreesurfacesofwhichtwoarecirclesofthesameradiusandoneiscurvedsurface.Wecanalsoseethattwocircularregionofacylinderareparalleltoeachotherandthecircumferenceofthecirclesisthewidthofthecurvedsurfaces.Thelengthofthecurved surface is called the height of the cylinder which can bedenotedby“h”wherewealreadyknowthat“r”istheradiusofbothcirclesand“d”isthediameter,i.e. Radius=r Diameter=d Height=h

12.2.1 Surface Area of the Cylinder

Wehavealready learnt the formulaof finding the areaof arectangleandacirclewhicharegivenbelow,areaofarectangle=lengthxwidth areaofacircle=pr2


10 11



Hereweshallusethesameformulaforfindingthesurfaceareaofacylinder.Weknowthatacylinderisthesumofthreeflatsurfaces(two circles and one curved surface)thatcanbeshownbyunfoldingacylinderasgiveninthefollowingfigure(b).

Fromthefigure (b),wecanexaminethethreeflatsurfacesof thecylinder. InwhichcircleA isthetopandcircleB isthebaseofthecylinderwhererectangleGOAT isthecurvedsurfacethat ifwerollupandjoinitstwoedgesGTandOAwegetagainthesamecurvedsurface.NowwecancalculatethesurfaceareaofacylinderbyfindingthesumofareasoftwocirclesA&BandareaoftherectangleGOATasgivenbelow: Widthofarectangle=circumferenceofacircle=2pr Lengthofarectangle=h AreaofarectangleGOAT=LengthxWidth =hx2pr=2prh Weknowthat, Areaofacurvedsurface=areaofarectangleGOAT So,areaofacurvedsurface=2prh AreaofacircleA=pr2

AreaofacircleB=pr2

Areaoftwocircles=areaofcircleA+areaofcircleB =pr2+pr2=2pr2

Thus,Surfaceareaofacylinder=areaofacurvedsurface+areaof twocircles =2prh+2pr2

=2pr(h+r)

Example 1: Find the surface area of a cylinder with length18.5cmandradius3.2cm.Solution: Length(h)=18.5cm Radius(r)=3.2cm Surfaceareaofacylinder=? Usingtheformula, Surface area of a cylinder 2 ( )r h rp= +

222(2 3 2(18.5 3 2)7

. . cm= × × +

222Surface area of a cylinder (2 3 2 21 7) 436.487

. . cm= × × × =

EXERCISE 12.3

1. Findthesurfaceareaofthefollowingcylinders.


12 13



2. Theradiusofacylinderis1.4cmandlengthis5.2cm.Calculate thesurfaceareaofthecylinder.3. Findthesurfaceareaofa7.4cmlongironrodof3.1cmradius.4. A cylinder is 5m long and radius of the cylinder is 5.3cm. Calculatethesurfaceareaofthecurvedsurface.5. Thediameterofacylinderis18.5cmandlengthis6.1m.Find thesurfaceareaofthecurvedsurface.

12.2.2 Volume of a Cylinder

We know that to find the volumeof any object,we use themeasurementsofthreedimensionsandtheformulaforfindingthevolumeofanobjectis:

Volume=lengthxbreadthxheightNowbyusingtheaboveformulaofvolume,weshalldiscoveranotherformulaforfindingthevolumeofacylinder.Todiscovertheformulaforfindingthevolumeofacylinder,wecanusethefollowingtwomethods.

• Bymakingacuboid• Bystackingcoins

• By making a Cuboid Wehavealreadylearntunderthetopic“areaofacircle”thatwhenwedivideacircleintoseveralbutevennumberofpartsandrearrange them according to the instructions we get a rectanglewhoselengthishalfofthecircumference(2pr)ofthecircleandwidthisequaltotheradius(r)ofthecircle.Thenbyusingtheformulaforthe areaof a rectangle,wefind the areaof a circle. Similarly,wecanusethesameabovemethodforacylinderbutwhenwecutacylinderintoseveralbutevennumberofpartsandrearrangethemwegetacuboidasshowninfollowingfigure.

Fromtheabovefigure,wecanexaminethatthelengthofthecuboidishalfofthecircumference(2pr)ofacircularregion,heightisequaltothelength(h)ofthecylinderandbreadthisequaltotheradiusofthecircularregion.So,ifwecalculatethevolumeofthecuboidthatwillbeequaltothevolumeofthecylinderasgivenbelow:• lengthofthecuboid=halfofthecircumference

1 length (2 )2

r rp p= =

• breadthofthecuboid=radiusofthecircularregion ∴ abreadth=r• heightofthecuboid=lengthofthecylinder ∴ aheight=h• volumeofthecylinder=volumeofthecuboid =lengthxbreadthxheight = pr x r x h ∴ volume= pr2 h

• By stacking coinsTake5coinsofRs.5andstackupapilewhichgivesusashapeofasmallcylinderAof1cmheightasgivenbelow.Wecancheckthismeasurementbystackingupapileof5coinsofRs.5Similarly,stackuptwomorepilesof10coinsand15coinsofRs.5thatgivesustwocylindersof2cmand3cmrespectively,labelthem


14 15



asBandCasshowninthefollowingfigures.

Supposethat theradiusof thecoin is r thentheareaofcircularregionwillbepr2,i.e.,Areaofthecircularregion =lengthxbreadth=pr2

Whereheight =1cm,2cmand3cmUsingtheaboveinformation,weonebyonecalculatethevolumeofcylinderA,BandC.VolumeofthecylinderA =(lengthxbreadth)xheight

= (pr2 x 1) cm3 = pr2 cm3

VolumeofthecylinderB =(lengthxbreadth)xheight= (pr2 x 2) cm3 = 2pr2 cm3

Volumeofthecylinder C =(lengthxbreadth)xheight= (pr2 x 3) cm3 = 3pr2 cm3

Weconsiderthatwestackupapileofsomecoinswhoseheightishthenthevolumeofcylinderis: Volumeofcylinder=(lengthxbreadth)xheight =(pr2 xh) Therefore,volume=pr2 h

Example 1: Find the volume of a cylinder whose height is18.5cmandradiusis4.2cm.Solution: Radius(r)=4.2cm Height(h)=18.5cm Volume(v)=? Usingtheformula, Volume(v)= pr2 h

Example 2: Find the height of a cylinder whose volume is3,168cm2andradiusis6cm.

Solution: Radius (r) = 6cm Volume(v)=3168cm3 Height(h) = ? Using the formula, Volume = pr2 h

2

volume 3168 722 6 6

h cmrp

× = = × ×

Height=28cm

Example 3: Findtheradiusofacylinderwhoseheightis14cmandvolumeis891cm3.

Solution: Volume(v)=891cm3 Height(h)=14cm Radius(r)=? Usingtheformula, Volume=pr2h

volumer

hp=

891 722 14

cm× = × =20.25cm2

20 25 4 5r . cm . cm= =

EXERCISE 12.4

1. Findthevolumeofthefollowingcylinders.

3 322( 4 2 4 2 18 5 1025 647

. . . cm . cm= × × × =


16 17



2. Findthevolumeofacylinderwhoseheightis9.8cmandradius of5.6cm.3. Thevolumeofacylinderis311.85cm3andheightis10cm.Find theradiusofthecircularregionofthecylinder.4. Theradiusofcylinderis7cmanditsvolumeis2,233cm3.Find theheightofthecylinder.5. Findtheradiusofacylinderwhenitsheight is9.2cmandits volumeis5,667.2cm3

12.2.3 Real Life Problems

• Solving Real Life Problems involving Circumference and Area of a Circle

Example 1: Theradiusof thewheelofacar is0.28m.Find inhowmanyrevolutionsthecarwillcoveradistanceof880meter.Solution:Radius(r) = 0.28m Circumference(c) = ? Using the formula, c = 2pr

222 0 287

c . m= × ×

c=1.76mThecarwillcoverthedistanceof1.76minacompleterevolutionofitswheels.So,1.76mdistancecoveredbycar=1revolution.

1880m distance will be covered by a car 880 500 revolutions.1 76.

= × =

Example 2: The diameter of the wheel of Ahmed’s bicycle is0.72m.Thebicyclewheelcompletes750revolutionswhenAhmedcomesfromschooltohouse.Findthedistancebetweenschoolandhouse.Solution: Diameter (d) = 0.72m Circumference (c) = ? Using the formula, c = pd

22 0 72 2 267

. . m= × =

In1revolutionthedistancecovered=2.26mIn750revolutionsthetravelis=2.26x750=1695m

Example 3: The length of theminute hand of a time clock is3.5cm.Findthedistancecoveredbythepointerofminutehandin3hours.Solution:Thelengthoftheminutehand(radius)=3.5cmCircumference=?Usingtheformula,c=2pr

22(2 3 5)cm 227

. cm= × × =


18 19



We know that in 1 hour, pointer of minute hand completes onerevolution.So,

In1hour,pointerofminutehandcoversthedistance =22cm

In3hours,pointerofminutehandcoversthedistance =3x22cm=66cm

Example 4: The circumference of a circular floor is 55m.CalculatetheareaofthefloorandalsofindthecostofflooringattherateofRs.90/m2.Solution: Circumference(c)=55m Areaofthefloor=? Weknowthat, c=2pr

2crp

=

55 7Radius of circular floor 8 75 2 22

. metre× = = × Nowwecalculatetheareaoffloor. Areaofthecircularregion= pr2

222 8 75 8 75 240 62

7. . . m = × × =

Thecostoflm2=90rupees Thecostof240.62m2=(90x240.62)=Rs.21655.8

EXERCISE 12.5

1. Thediameterof thewheelof Irfan’sbike is0.7m.Thewheel completes1800revolutionswhenhereacheshomefromthe office.FindthedistancebetweenIrfan’shouseandoffice.2. The radius of a truck wheel is 0.55m. Calculate how much distancethetruckwillcoverin1,500revolutionsofthewheel.

3. Thelengthoftheminutehandofawatchis1.75cm.Findinhow manyhours, thepointer ofminutehandwillmove to cover 165cm.4. The length of the hour hand of a watch is 1.2cm. Find the distance covered by the hour pointer of hand in 24 hours. (Hint: The hour hand completes one revolution in 12 hours)5. Theradiusofacirculargardenis24.5m.Findthecostoffencing thegardenattherateofRs.175permeter.6. Thediameterofacircularroomis4.2m.Findthecostofflooring attherateofRs.150/m2.7. Findthewagesofgrasscuttingofacircularparkattherateof Rs.5/m2,wheretheradiusoftheparkis105m.8. The radius of a circular pool is 10.5m. Calculate the cost of flooring tiles used on the floor of the pool at the rate ofRs.180/m2.9. The diameter of a circular playground is 21m. Calculate the cost of repairing the floor of the playground at the rate of Rs.230/m2andalsofindthecostoffencingtheplaygroundat therateofRs.75/m.• Solving Real Life Problems involving Surface Area and Volume

of a Cylinder

Example 1: Thelengthofasteelpipeis2.1mandtheradiusis8cm.Calculateitssurfaceareaifpipeisopenatboththeends.Solution:Thepipehasonlycurvedsurfaceso,Length(h) =2.1m=(2.1x100)=210cmRadius(r) =8cmAreaofacurvedsurface =?Usingtheformula,

Areaofacurvedsurface =2prh

Example 2: Findthesurfaceareaofanoildrumwhoselengthis1.6mandthediameteris63cm.

2222 8 210 105607

cm = × × × =


20 21



Solution: Length(h)=1.6m=(1.6x100)cm=160cm

( ) 63Radius ( ) 31 52 2

diameter dr cm . cm= = =

Surfaceareaofthedrum=? Usingtheformula, Surfaceareaofthedrum(cylinder)=2pr(h+r)

222 31 5 (160+31.5)7

.= × ×

222 31 5 191 57

. .= × × ×

Surfaceareaofthedrum=37917cm2

Example 3: Heightofanoildrumis250cmitsradiusis70cm.Findthevolumeorcapacityofcylinderinlitre.Solution: Height(h)=250cm Radius(r)=70cm Capacityofanoildruminlitres=volume=?

Volumeofcylinder = pr2 h

Volumeofcylinder =3,850,000cm3

Weknowthat:1,000cm3 =1liter

EXERCISE 12.6

1. A cylindrical wooden piece is 19.4cm long. Find the surface areaofthewoodenpieceifitsdiameteris14cm.2. Atinpackofasoftdrinkis10cmlongandtheradiusofthetin packis3.3cm.Findthesurfaceareaofthetinpack.3. A circular pillar of 22.5cm radius is 6.3m long. Calculate the surfaceareaofthepillar.

4. Acylindricalchemicaldrumis220.5cmlongandtheradiusof thedrumis42cm.Calculatethecostofpaintingthedrumat therateofRs.0.15/cm2.5. Radiusofaroundswimmingpool is17.5mandthedepthof thepoolis3m.Calculatethecostoftilesusedonthewallofthe poolattherateofRs.120/m2.6. Theinternaldiameterofaroundmosqueis31.5mandheight ofwallsis7m.Findthecostofcementingtheroundwallofthe mosqueattherateofRs.l9/m2.7. Acylindricalwatertankis7.7mhighanditsinnerradiusis5m. Calculatethepriceofmarbleusedintheinnersideofthetank attherateofRs.500/m2.8. Findtheheightofanoildrumwhosevolumeis12,474m3and radiusis6.3m.9. Acylindricaltincanis77cmhighanditsradiusis20cm.Find howmanyliterofoilmaybecontainedinthetincan.10. Find the capacity of a circularwater tank in literswhen the heightofthetankis420cmanditsdiameteris510cm.

REVIEW EXERCISE 12

1. Answerthefollowingquestions. (i) Definethecircumferenceofacircle. (ii) Whatisanareaofacircularregion? (iii) Writetheformulaforfindingthesurfaceareaandvolume ofacylinder. (iv) Writetheformulaforfindingthecircumferenceandarea ofacircle. (v) Whatistheapproximatevalueofp?

2. Fillintheblanks. (i) The_______ofacircle is themeasurementof itsclosed curve. (ii) Twocircularregions,ofacylinderare_______toeachother. (iii) Thelengthofthe_______iscalledtheheightofthecylinder.

322( 70 70 250)7

cm= × × ×

3 850 000Volume (litre) litres 3 850 litres1 000

, , ,,

= =


22 23



(iv) Theratiobetweencircumferenceanddiameterofacircle isdenotedbythesymbol_______. (v) Surfaceareaofacylinder=areaof thecurvedsurface +_______.


4. Find theareaandcircumferenceof thecircle, if227

p = and radiusis:

(i) 2.8cm (ii) 4.9cm (iii) 10.5cm

(iv)1102

cm (v)162

cm

5. Findthesurfaceareaandvolumeofthefollowingcylinders. (i) r=14cm,h=15cm (ii) r=3.5cm,h=100cm (iii) r=10cm,h=21cm (iv) r=4cm,h=12cm

6. Area of a round bed of roses is 7.065m . Find the cost of fencing around it at the rate of Rs.20 per meter (when pc3.14).

7. TheradiusofthewheelofAslam’scycleis35cm.Toreachschool from house, the wheel completes 1200 rounds. Find the

distancefromhousetoschool(when227

p ≈ ).

8. Find the surface area of 2m 99 long drumwhose radius of

baseis21cm(when227

p ≈ ).

9. Awell is20mdeepanditsdiameteris4m.Howmuchsoil is requiredtofillit.(whenpc3.14).

10. Find thecostofsprayingachemical inacircularfieldat the rateofRs.10/m2wheretheradiusofthecircularfieldis73.5m andalsocalculatethecostofmakinghurdlearoundthefield attherateofRs.25/m.

SUMMARY

• Thecircumferenceofacircleisthedistancearoundtheedgeofthecircle.

• The ratio between circumference and diameter of a circle isdenotedbyaGreeksymbolpwhoseapproximatevalueis3.14.

• Theareaofacircularregionisthenumberofsquareunitsinsidethecircle.

• Acylinderconsistsofthreesurfacesi.e.twocirclesofsameradiusandonecurvedsurface.

• C=dpor2pr,where ‘c’ isthecircumference, ‘d’ isthediameterand‘r’istheradius.

• Areaofthecircularregion=pr2

• Surfaceareaofacylinder=2pr(h+r)• Volumeofacylinder=pr2h

I

CHAPTER

13 INFORMATION HANDLING

Version: 1.1

Animation 13.1: Information HandlingSource & Credit: eLearn.Punjab



2 3

13. Information Handling 11. Information HandlingeLearn.Punjab eLearn.Punjab

2 version: 1.1version: 1.1


After studying this unit, students will be able to:• Demonstratedatapresentation.• Define frequency distribution (i.e. frequency, lower class limit,

upperclasslimit,classinterval).• Interpretanddrawpiegraph.

Introduction

In the world around us, there are a lot of questions andsituationsthatwewanttounderstand,describe,exploreandaccess.Forexample,• HowmanyhospitalsarethereindifferentcitiesofPakistan?• Howmanychildrenwerebornduringthelast10years?• Howmanydoctorswillberequiredinthenext5years?

Toknowabout such things,wecollect informationandpresent itinamanageablewaysothatusefulconclusionscanbedrawn.Thebranchofstatisticsthatdealswiththisprocessiscalledinformationhandling.

13.1 Data

Datameansfactsorgroupsofinformationthatarenormallythe results of measurements, observations and experiments.Theseresultshelpusinreviewingourpastperformanceandfutureplanning. For example, the government of a state prepares itsbudgets and development plans on the basis of a collected dataabouttheresourcesandpopulation.

13.1.1 Presentation of Data

After the collectionof adata, themost important step is itspresentationthatprovidesbasistodrawconclusions.Datacanbe

representedintheformoftablesanddifferentkindsofgraphs.We know that a data is collected in raw form and it provides usinformationaboutindividuals.Datainsuchformiscalledungroupeddata. After arranging thedata fordesired information, it is calledgrouped data. For example, a teacher collected the score of 20studentsinmathematicstest:11,52,40,95,65,45,35,30,88,56,75,90,81,82,28,49,67,98,64,92This is an ungrouped data. Now if we arrange it to representinformationintogroups,thenitiscalledgroupeddata.• Numberofstudentswhoscoredfrom11to40=5• Numberofstudentswhoscoredfrom41to70=7• Numberofstudentswhoscoredfrom71to100=8Itcanbeseenthat it iseasiertovisualizethegiven information ifdataispresentedingroupedform.Wecanalsorepresentagroupeddatausingatable.

Group Score Tally MarksNo. of

Students

11-40 11,40,35,30,28 5

41-70 52,65,56,45,49,67,64 7

71-100 98,88,75,90,81,82,98,92 8Themethodthatweusedtorecordtheresultsinthetableiscalledtallying inwhichwedraw tallymarksaccording to thenumberofindividualsofagroup.Wemakethesetoffivesbycrossingthefourmarkswiththefifthmark.Thismakeseasytocountthetallymarks.Forexample,toshow12individualsofagroupwedrawtallymarks

.Wecanalsocharacterizetheinformationpresentedintheexampleas;

Scores Characteristics71-100 Excellent41-70 Good11- 40 Poor


4 5



13.1.2 Frequency Distribution

Theconversionofungroupeddataintogroupeddatasothatthefrequenciesofdifferentgroupscanbevisualizediscalledfrequencydistribution

The table which shows the frequency of class intervals is called the frequency table.

• FrequencyThenumberofvaluesthatoccursinagroupofadataiscalleditsfrequency,e.g.intheabovegivenexample,

Thefrequencyof(11-40)is5.Thefrequencyof(41-70)is7.Thefrequencyof(71-100)is8.

• Class LimitsUpper Class Limit:Thegreatestvalueofaclassintervaliscalledtheupperclasslimit,e.g.intheclassinterval(41-70),70istheupperclasslimit.Lower Class Limit:Thesmallestvalueofaclassintervaliscalledthelowerclasslimit,e.g.intheclassinterval(71-100),71isthelowerclasslimit• Class IntervalsEachgroupofadataisalsoknownastheclassinterval.Forexample,(11-40),(41-70)and(71-100)areclassintervals.Eachintervalrepresentsallthevaluesofagroup.Size of the Class Interval: Thenumberofvaluesinaclassintervaliscalled itssizeor length.Forexample, thesizeor lengthofclassinterval(11-40)is30thatcanbecheckedbycounting.Itcanalsobecalculatedbysubtractingthelowestvalueofthedatafromgreatestvalueanddividetheresultbythenumberofclassintervalsasshownbelow:Lowestvalue=11Greatestvalue=100

Nowusetheformulatocalculatethesize.

greatest value lowest valueSize of interval = no. of intervals

-

100 11 89 29 6

3 3.-

= = =

Roundoff theanswer, this is the requiredsizeof the interval, i.e.29 6 30.

Example 1: Thereare40studentsintheclassVIIwhogotthefollowingmarksinanEnglishtest.Makeafrequencytablebyusing5classesofanequalsize.35,9,26,41,27,15,18,60,46,33,24,15,52,39,28,89,74,68,56,38,92,49,28,82,19,21,34,23,43,77,65,64,21,59,15,33,66,29,33,65,Solution:Weknowthat,

greatest value lowest valueSize of class = no. of intervals

-

Wecanseefromtheaboveun-groupeddatathatof:Greatestvalue=92Lowestvalue=9No.ofclasses=5

92 9Size of class = 16 6 17 ( )5

. round up-= ≅

Class IntervalTally

MarksFrequency

9-25 10

26-42 13

43-59 6

60-76 7

77-93 4

EXERCISE 13.1

1. The telephone bills paid by 12 consumers are given below.


6 7



Makeafrequencytableof5classesofanequalsize. 510,700,356,603,422,674,481,545,718,592,685,5692. Inaboardexamination,20studentsoftheDawnPublicSchool gotthefollowingmarksoutof850marks.Constructafrequency tablebytaking100asaclassinterval. 551,786,678,725,788,580,720,690,750,651,599,609,719, 760,625,775,646,667,753,6753. The daily wages of 15 workers are given below. Make a frequencytableof4classesofanequalsize. 400,225,250,380,425,175,230,325,150,300,200,180,350, 375,2004. Acricketplayermadethelistofhislast18inningsscoreswhich isgivenbelow. 122,102,72,99,89,106,99,85,92,108,102,98,95,76,80,65, 101,96,Makeafrequencytableof6classesofanequalsize.5. Thefollowingdatashowsthedistanceinkmthatwastravelled byMr.Usmaninlast21days. 77,58,62,85,32,71,59,60,38,32,69,80,76,92,61,82,74,70, 99,44,53Makeafrequencytableof5classesofanequalsize.6. Thefollowingdataisshowingthesaleofabikecompanyduring lastmonths. 571,692,700,533,832,744,649,584,613,735,872,900,512, 864,654,782,777,555,632,880,628,529,680,756,567,548, 824,719,678,721 Makeafrequencytablebytaking100asaclassinterval.

13.2 Pie Graph

“Therepresentationofanumericaldataintheformofdisjointsectorsofacircleiscalledapiegraph.”Apiegraphisgenerallyusedforthecomparisonofsomenumericalfactsclassified indifferentclasses. In thisgraph, thecentralanglemeasures360°whichissubdividedintotheratioofthesizesofthegroupstobeshownthroughthisgraph.Followingexampleswillhelptounderstandtheconceptofapiegraph.

Example 1: Itiscompulsoryforeachstudenttotakepartinthedifferentgames.Outof1800studentsintheschool750playcricket,200 play badminton, 400 play hockey and 450 play football.In order to represent their comparison, draw a pie graph.Solution: Totalnumberofstudents=1800(i) Findtheangleforeachsectorbyusingthefollowingformula

0No. of students play a gameRequired angle = 360total students

×

0 0200Measure of angle associated with badminton = 360 401800

× =

0 0750Measure of angle associated with cricket = 360 1501800

× =

0 0400Measure of angle associated with hockey = 360 801800

× =

0 0450Measure of angle associated with football = 360 901800

× =

(ii) Inordertodrawapiegraph• Drawacircleofasuitableradius.• Drawanangleof40°representingthebadminton.• Drawanangle150°representingthecricket.• Drawanangle80°representingthehockey.• Remaininganglewillbeof90°representingthefootball.

(iii) Labeleachsectoraccordingtothefollowingfigure.

Have you noticed that students like cricket most?


8 9



Example 2: ThefollowingtableshowsthefavouritefoodofthestudentsofthegradeVII.Plotapiegraphtoshowthefavouritefoodofthestudents.

FoodFried

ChickenMutton Karahi

BiryaniMinced

MeatVegetables

No. of Students

40 20 10 6 4

Solution:(i) Findtheanglesforeachsectorbyusingthefollowingformula.

(a) 0No. of students like foodRequired angle = 360total students

×

(b) 0 040Angle for fried chicken = 360 18080

× =

(c) 0 020Angle for mutton karahi = 360 9080

× =

(d) 0 010Angle for biryani = 360 4580

× =

(e) 0 06Angle for minced meat = 360 2780

× =

(f) 0 04Angle for vegetables = 360 1880

× =

(ii) Drawacircleofanysuitableradius.(iii) Dividethecircleintothesectorsofcalculatedangles.(iv) Labeleachsectoraccordingtothefollowingfigure.

Did you see that fried chicken is the most favourite food of the students?

EXERCISE 13.2

1. Hinawentforshoppingandspent30%ofherpocketmoney forfood,35%onbuyingbooks,20%onschooldressandsaved 15%.Representthedataonpiegraph.

2. A media reporter conducted a survey of persons visiting marketduring the twohours.He foundthat therewere720 personsvisitedthemarketoutofwhich320werewomen,220 menand180children.Drawapiegraph.

3. In a class, the grades obtained by the students in the final examinationaregivenbelow.Drawthepiegraph.

Grade A+ A B C D E FNo. of students 2 6 10 30 6 4 2

4. Detailsofstudentsinfiveclassesofaschoolaregivenbelow. Drawapiegraphtoshowthecomparison.

Class I II III IV VNo. of students 300 270 225 150 135

5. Noreenhasthefollowingtypesofbooksinherlibrary.Draw piegraphshowingtheinformation.

Subject English Islamic Stories PoemsNo. of books 180 90 60 30

REVIEw EXERCISE 13

1. Answerthefollowingquestions. (i) Whatismeantbythegroupeddata? (ii) Defineaclassinterval. (iii) Defineapiegraph. (iv) Writetheformulaforfindingthesizeofclassinterval. (v) Whichmethodiscalledtallying


10 11



2. Fillintheblanks. (i) ________meansgroupsofinformationthatarenormally the results of measurements, observations and experiments. (ii) Each_________representallthevaluesofagroup. (iii) A data is collected in ________ form and it provides us informationaboutindividuals. (iv) The method which is used to record the result is called_______. (v) Thegreatestvalueofaclassintervaliscalledthe___limit. (vi) Thenumberofvaluesinaclassintervaliscalledits_______. (vii) The representation of a numerical data in the formof disjointsectorsofacircleiscalleda_______.


4. Theagesofpatients inyearsadmittedinahospitalduringa weekaregivenbelow.Groupthedatataking10asthesizeof aninterval. 25,50,49,47,26,10,2,1,15,17,18,19,27,28,30,35,40,37, 32,31,3,4,7,10,15,12,13,17,14,20,22,24,26,30,17,35,40, 36,32,31,37

5. The followingdata shows thedistance inkm thatMr.Ghani traveledinlastmonth. 90,44,15,19,28,9,92,17,8,84,50,60,77,69,24,89,63,74, 35,48,39,81,58,37,55,67,46,30,26,79. Constructthefrequencytableof6classesofanequalsize.

6. Aliandhisfriendseatbreadsinadayasshowninthetable.Meals Breakfast Lunch Dinner Supper

No. of Breads 12 24 16 8

Byusingthetable,drawapiegraph.

7. Inaparty,ahostservedtheguestbyfollowingfooditems.Food Items Cold Drink Sandwich Burger Cake

Quantity 180 124 330 86 Usethetabletodrawapiegraph.

SUMMARy

• Datameansfactsorgroupsofinformationthatarenormallytheresultsofmeasurements,observationsandexperiments.• Dataiscollectedinrawformanditprovidesusinformationabout

individualssuchformofthedataiscalledungroupeddata.• Inagroupeddata,eachgroupisalsoknownastheclassinterval.• Thegreatestvalueofaclassintervaliscalledtheupperclasslimit.• Thesmallestvalueofaclassintervaliscalledthelowerclasslimit.• Thenumberofvaluesthatoccursinaclassinterval iscalledits

frequency.• Thetablewhichshowsthefrequencyofclass intervals iscalled

frequencytables.• The representation of a numerical data in the formof disjoint

sectorsofacircleiscalledapiegraph.• Inpiegraph,thecentralanglemeasures360°whichissubdivided

intotheratioofthesizesofthegroups.

CHAPTER 1 SETS 7.pdf · 1. QUADRATIC E Q UATIONS eLearn.Punjab 1. QUADRATIC E Q UATIONS...

Documents

Transcript of CHAPTER 1 SETS 7.pdf · 1. QUADRATIC E Q UATIONS eLearn.Punjab 1. QUADRATIC E Q UATIONS...