Areas and Mensuration

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AreasandMensuration

Areasandmensurationisatopicdependsentirelyonapplicationofformulas.Ifyouremembertheformula,solvingproblemsinthisareaisacakewalk.Foreasylearningandtoremembertheformulasweprovidedthemintoasimpletable.Beforeyoumoveontohavealookatthesolvedexamples,studythetablesandtrytounderstandtherelevantformula

2DimensionalFigures:

Twodimensionalfigureshaveonlyanytwooflength,breadth,andheight.Theyhaveonlyareasbutnotvolumes.Perimeterisaunidimensionalmeasure.Ifyouobservecarefully,thepowerofthetermsintheformulasofperimeteris1,andofthetermsintheareasis2.

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SolvedExamples

1.Ifsidesofatriangleare8cm,15cmand17cmrespectively.Finditsarea.

Areaofthetrianglewhenallthethreesidesaregiven=s(s a)(s b)(s c)wheres =a + b + c

2

s=8 + 15 + 17

2=20cm

Therefore,Area=20 (20 8) (20 15) (20 17)

=20 12 5 3=4 5 3 4 5 3=4x5x3=60cm2Trick:Thetriangleisrightangletriangleas172 = 152 + 82

Therefore,Areaofrightangletriangle=12x8x15=60cm2

2.Twoparallelsidesofatrapeziumare4cmand5cmrespectively.Theperpendiculardistancebetweentheparallelsidesis6cm.Findtheareaofthetrapezium.

Areaoftrapeziumwhenheightandtwoparallelsidesaregiven=12

h (a + b)=12x6x(4+5)=27cm2

3.Ifperimeterandareaofasquareareequal.Sideofthesquare(incm)is:Given,(Side)2=4x(Side)Therefore,Side=4cm

4.Thediameterofthewheelofavehicleis5metre.Itmakes7revolutionsper9seconds.Whatisspeedofthevehicleinkm/h?

Radiusofthewheel=52metre

Distancecoveredin1revolution=Circumferenceofthewheel=2r=2x227

52metre

Therefore,Distancecoveredinonesecond=2x227

52

79metre

Therefore,Speedperhour=2x227

52

79

185=44km/h

5.Theperimeterofarhombusis60cmandoneofitsdiagonalis24cm.Findtheotherdiagonaloftherhombus.LetABCDistherhombuswhosediagonalsBDandACintersectingatpoint'O'.

Sideofrhombus=14xPerimeter=

14x60=15cm.

LetBD=24cm

Then,BO=12BD=12cm

Now,AO=AB2BO2 = 152122=9Therefore,AC=2AO=2x9=18cm

6.Afieldis40metrelongand35metrewide.Thefieldissurroundedbyapathofuniformwidthof2.5metrerunsrounditontheoutside.Findtheareaofthepath.RemembertheformulafortheAreaofpath

=2xWidthx[Length+Breadth+(2xWidth)]=2x25x(40+35+2x2.5)=5x(75+5)=400m2

7.Findareaofuniformpathofwidth2metrerunningfromcentreofeachsideoftheoppositesideofarectanglefieldmeasuring17metreby12metre.RemembertheformulafortheAreaofpath=Widthofpathx(Lengthoffield+Breadthoffield)(Widthofpath)2

=2x(17+12)(2)2

=584=54m2

8.Asquareandarectangleeachhasperimeter60metre.Differencebetweentheareasofthetwofiguresis16squaremetre.Findlengthoftherectangle.

Sideofthesquare=604=15metre

Differenceinareasofsquareandrectangle=16m2

Therefore,Increaseanddecreaseindimensions=16=4metreTherefore,Sidesofrectangleare(15+4)and(154)metreTherefore,Lengthofrectangle=19metre

9.Findtheradiusofacirclewhoseareaisequaltothesumofareasofthreecircleswithradii8cm,9cm,12cmrespectively.Letradiusofnewcircleis'R'cm.ThenR2=82 + 92 + 122 = 64 + 81 + 144 = 289Therefore,R2=289=>R=17cm.

10.Theareoftheringbetweentwoconcentriccircles,whosecircumferencesare88cmand132cmis:

AreaofRing=14

1322 882 =14

x(132+88)x(13288)

=14

722

x220x44=770cm2

11.Twopoleswhoseheightsare11metreand5metrestanduprightinafield.Ifthedistancebetweentheirfeetis8metre,whatisthedistancebetweentheirtops?

( )

Distancebetweenfeetofbigpoleandthatofsmallpole=8metreDifferencebetweenheightsoftwopoles=115=6metreDistancebetweenthetopsofpoles=62 + 82=10metre

12.Findtheareofthelargestcirclethatcanbeinscribedinasquareof14cm,aside.Sideofthesquare=Diameterofthecircle=14cmTherefore,Radiusofthecircle=7cm.

Areaofthecircle=r2 =227x7x7=154cm2.

13.Ahorseistiedtooneofthecornersofasquarefieldwhosesideis20metre.Iflengthoftheropeis14metre,findtheareaofungrazedfield.Areaofsquarefield=(20m)2=400m2

Areaofthefieldgrazedbythehorse=14r2

=14

227x14x14=154m2

Therefore,Fieldungrazed=(400154)=246m2

14.Ahorsesaretiedonetoeachcornerofasquarewith14maside,andlengthoftheropeis7m.Findtheareaofungrazedfield.Ungrazedarea=AreaofsquareAreaofcircle

=142 227x7x7=196154=42m2

ShortCutMethod:Radiusofcircle=7m.

Therefore,Ungragedfield=67

72=42m2

MCQ's

1.Arectangularcarpethasanareaof60sq.m.Itsdiagonalandlongersidetogetherequal5timestheshorterside.Thelengthofthecarpetis:a.5mb.12m

c.13md.14.5mCorrectOption:BExplanation:Letthelength=pmetresandbreadth=qmetresThenpq=60

p2 + q2 + p = 5q p2 + q2 = (5q p)2=25q2 + p2 10pq

Aspq=60, 25q2 10 60 = 0 q2 = 25orq=5p=60/5=12mLengthofthecarpet=12m

2.Arectangularcarpethasanareaof120sq.mandperimeterof46m.Thelengthofitsdiagonalis:a.15mb.16mc.17md.20mCorrectOption:CExplanation:Letlength=ametresandbreadth=bmetresThen,2(a+b)=46or(a+b)=23andab=120Diagonal=a2 + b2=(a + b)2 ab = (23)2 2 120=289 = 17m

3.Aparallelogramhassides60mand40mandoneofitsdiagonalsis80mlong.Then,itsareais:a.480sq.mb.320sq.mc.60015sq.md.45015sq.mCorrectOption:C

Explanation:AB=60m,BC=40morAC=80m

s=12(60 + 40 + 80)m = 90m

(sa)=30m,(sb)=50mand(sc)=10mAreaofABC=s(s a)(s b)(s c)=90 30 50 10m2 = 30015m2Areaof//gmABCD=60015m2

4.Ifasquareandaparallelogramstandonthesamebase,thentheratiooftheareaofthesquareandtheparallelogramis:a.greaterthan1b.equalto1

c.equalto12

d.equalto14

CorrectOption:BExplanation:

LetABCDbethesquareandABEFbetheparallelogram.Then,inrighttrianglesADFandBCE,wehaveAD=BC(sidesofasquare)andAF=BE(sidesofparallelogram).Therefore,DF=CE[DF2 = AF2 AD2 = BE2 BC2 = CE2]

Thus,ADF=BCE ADF + ABCF=BCE + ABCF AreaofparallelogramABEF=AreaofABCD

5.Inarhombus,whoseareais144sq.cim,oneofitsdiagonalsistwiceaslongastheother.Thelengthofitsdiagonalsare:a.24cm,48cmb.12cm,24cmc.62cm,122cmd.6cm,12cmCorrectOption:BExplanation:12

x 2x = 144 x2 = 144orx=12

Lengthofdiagonals=12cm,24cm

6.Thelengthofaropebywhichacowmustbetetheredinorderthatshemaybeabletograzeanareaof9856sq.m.is:a.56mb.64mc.88md.168mCorrectOption:A

Explanation:

Grazingareaisequaltotheareaofacirclewithradiusr.

227

r2 = 9856

Thenr2 = 9856 722

r=56m

7.Thecircumferencesoftwoconcentriccirclesare176mand132mrespectively.Whatisthedifferencebetweentheirradii?a.5metresb.7metresc.8metresd.44metresCorrectOption:BExplanation:

2R 2r = (176 132)

2(R r) = 44

(R r) =44 72 22

= 7m

8.Thediameterofacircleis105cmlessthanthecircumference.Whatisthediameterofthecircle?a.44cmb.46cm

( )

c.48cmd.49cmCorrectOption:DExplanation:d d = 105 ( 1)d = 105

227

1 d=105

d=715

105 cm=49cm

9.Acircleandasquarehavesamearea.Theratioofthesideofthesquareandtheradiousofthecircleis:a.:1b.1:rc.1:rd.r:1CorrectOption:BExplanation:

x2 = r2 xr

= = : 1

10.Thenumberofroundsthatawheelofdiameter711

mwillmakeingoing4km,is:

a.1000b.1500c.1700d.2000CorrectOption:DExplanation:

Numberofrounds=4 1000227

711

= 2000

11.Acircularwireofradius42cmiscutandbentintheformofarectanglewhosesidesareintheratioof6:5.Thesmallersideoftherectangleis:

( )( )

a.30cmb.60cmc.72cmd.132cmCorrectOption:BExplanation:

Circumference= 2 227

42 cm=264cm

Lettherectanglesidesare6x,5x.Thencircumferenceis2 (6x + 5x) = 264orx=12Smallersideofrecatngle=5x=60cm

12.Ifthediameterofacircleisincreasedby100%,itsareaisincreasedby:a.100%b.200%c.300%d.400%CorrectOption:CExplanation:

Originalarea= d2

2=

d2

4

Newarea= 2d2

2= d2

Increaseinarea= d2 d2

4=

3d2

4

Increasepercent=3d2

4

4

d2 100 %=300%

Shortcut:

Youcanuse A + B +AB100

%formula.SubstituteA=B=100

( )

( )( )

( )( )

( )

13.Iftheradiusofacircleisreducedby50%,itsareaisreducedby:a.25%b.50%c.75%d.100%CorrectOption:CExplanation:Originalarea= r2

Neware= r2

2=

r2

4

Reductioninarea= r2 r2

4=

3r2

4

Reductionpercent=3r2

4

1

r2 100 %=75%

Shortcut:

Youcanuse A + B +AB100

%formula.SubstituteA=B=50.

14.Ifthecircumferenceofacircleis352metres,thenitsareainm2is:a.9856b.8956c.6589d.5986CorrectOption:AExplanation:

2 227

r = 352 r = 352 722

12

= 56m

Area=227

56 56 m2 = 9856m2

15.Areaofsquarewithsidexisequaltotheareaofatrianglewithbasex.Thealtitudeofthetriangleis:

( )( )( )

( )

( )( )

a.x2

b.xc.2xd.4xCorrectOption:CExplanation:

x2 =12

x horh=2x2

x= 2x

16.Theperimeterofanisoscelestriangleisequalto14cmthelateralsideistothebaseintheratio5:4.Theareaofthetriangleis:

a.1221cm

2

b.3221cm

2

c.21cm2d.221cm2CorrectOption:DExplanation:Letlateralside=(5x)cmandbase=(4x)cm5x+5x+4x=14orx=1So,thesidesare5cm,5cmand4cm

Semiperimeter,S=12(5+5+4)cm=7cm.

(sa)=2cm,(sb)=2cmand(sc)=3cmArea=7 2 2 3cm2 = 221cm2

17.Ifthediagonalofasquareisdoubled,howdoestheareaofthesquarechange?a.Becomesfourfoldb.Becomesthreefoldc.Becomestwofoldd.NoneoftheseCorrectOption:AExplanation:

12

d2

12

(2d)2=

14

Newareabecomes4fold.

18.Ifthebaseofarectangleisincreasedby10%andtheareaisunchanged,thenthecorrespondingaltitudemustbedecreasedby:

a.9 111

%

b.10%c.11%

d.1119%

CorrectOption:AExplanation:Letbase=bandaltitude=h.thenarea=(bh)

Newbase= 110100

b =1110

b

Letnewaltitude=H

Then,1110

b H = bhorH= 1011

h

Decrease= h 1011

h =111

h

Decreasepercent= 111

h 1h

100 %=9 111

%

Shortcut:AssumeLengthis10unitsandAltitudeis11Units.NowArea=110Newlengthis11unitsandNewaltitudeishunits.NowArea=11hButgiven110=11h h=10

Sochangeinaltitude=1/11x100=9 111

%

19.Thelengthofarectangleistwiceitsbreadth.Ifitslengthisdecreasedby5cmandthebreadthisincreasedby

( ) ( )

( )( )

( )

5cm,theareaoftherectangleisincreasedby75cm2.Therefore,thelengthoftherectangleis:a.20cmb.30cmc.40cmd.50cmCorrectOption:CExplanation:Letbreadth=xcmandlength=(2x)cmThen.(2x5)(x+5)x 2x = 752x2 + 5x 25 2x2 = 75or5x=100orx=20Length=(2x)cm=40cm.

20.Arectanglehas15cmasitslengthand150cm2asitsarea.Itsareaisincreasedto113timestheoriginalarea

byincreasingonlyitslengthofitsnewperimeteris:a.50cmb.60cmc.70cmd.80cmCorrectOption:BExplanation:

Breadthoftherectangle=15015

cm=10cm

Newarea=43

150 cm2 = 200cm2

Newlength=20010

cm = 20cm

Newperimeter=2(20+10)cm=60cm

21.Thelengthofarectangularroomis4metres.Ifitcanbepartitionedintotwoequalsquarerooms,whatisthelengthofeachpartitioninmetres?a.1

( )( )( )

b.2c.4d.Datainadequate.CorrectOption:BExplanation:Letthesideofeachnewroom=ymetres.Then,y2 = 2xClearly,2xisacompletesquarewhenx=2y2 = 4ory=2m

22.Thelengthandbreadthofasquareareincreasedby40%and30%respectively.Theareaoftheresultingrectangleexceedstheareaofthesquareby:a.42%b.62%c.82%d.NoneoftheseCorrectOption:CExplanation:Letthesideofthesquare=100mNewlength=140m.newbreadth=130mIncreaseinarea=[(140 130) (100 100)]m2=8200m2

Increasepercent=8200

100 100 100 %=82%

23.Ahall20mlongand15mbroadissurroundedbyaverandahofuniformwidthof2.5m.ThecostofflooringtheverandahattherateofRs.3.50persq.metreis:a.Rs.500b.Rs.600c.Rs.700d.Rs.800CorrectOption:CExplanation:

( )

Areaofverandah=[(25 20) (20 15)]m2=200m2

Costofflooring=Rs.(200 3.50)=Rs.700

24.Iftheratiooftheareasoftwosquaresis9:1,theratiooftheirperimetersis:a.9:1b.3:1c.3:4d.1:3CorrectOption:BExplanation:Lettheareasofsquaresbe:(9x2)m2and(x2)m2

Then,theirsidesare9x2, x2or(3x)metres&xmetresrespectively.

25.Thelengthofarectangleisincreasedby60%.Bywhatpercentwouldthewidthhavetobedecreasedtomaintainthesamearea?

a.3712

b.60%c.75%d.120%CorrectOption:AExplanation:Initially,letlength=xandbreadth=y

Let,newbreadth=z.Thennewlength=160100

x =85x

85x z=xyorz=

5y8

Decreaseinbreadth= y 5y8

=38y

Decreasepercent=38y

1y

100 %=3712%

( )

( )

( )

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Areas and Mensuration