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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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