Download Area of Common Shapes

AreaofCommonShapes
AreaofRectangle
Figure1:A4-by-9rectanglewith36unitsquaresinside
Ashape'sareameansthenumberof1-by-1squareunittheshapecovers.InFigure1,thisrectangle's
areais 9 ⋅ 4 = 36 squareunits,astherectanglecovers361-by-1unitsquares.It'sfairlyeasyto
understandthatarectangle'sareaformulais:
rectanglearea=base·height(orlength·width)
Areaoftriangle
Figure2:Atriangle'sareaishalfasbigasarectanglewiththesamebaseandheight
Bythisgraph,it'sfairlyeasytoseewhyatriangle'sareaishalfasbigastheareaofarectanglewiththe
samebaseandsameheight.Thusatriangle'sareaformulais:
1
triangle area = (base)(height) 2
[Example1]Atriangle'sbaseis4meters,anditsheightis2.5meters.Finditsarea.
[Solution]Bythetriangleareaformula,theareais:
A=
1
1
(base)(height ) = ⋅ 4 ⋅ 2.5 = 5squaremeters
2
2
Noticethattheunitofareaisdifferentfromtheunitofperimeter.Ifarectangle'sperimeterissome
meters,thenitsareamustbesomesquaremeters.
Forsimplicity,wecanalsowrite"5squaremeters"as"5m2".Theletter"m"represents"meters".
Similarly,"cm"representscentimeters,"in"representsinches,etc.
Noticethat"5m2"isdifferentfrom"52m":
•
•
"5m2"meansanareaof5squaremeters.Thesquarehasnothingtodowith5.
"52m"meansalengthof25meters.Thesquarehasnothingtodowith"m".
Arighttriangle'sheightisactuallyoneofitslegs,andanobtusetriangle'sheightliesoutsidethetriangle.
Seethefollowingfigures:
Figure3:righttriangle'sheightandobtusetriangle'sheight
Inthenextexample,youneedtosolveanequationbasedonatriangle'sareaformula.
[Example2]Atrianglecovers20squaremillimeters.Itsbaseis5millimeters.Finditsheight.
[Solution]Letthetriangle'sheightbehmillimeters.Plugthegivennumbersintothetrianglearea
formula,wehave:
1
(base )(height )
2
1
20 = ⋅ 5h
2
1
2 ⋅ 20 = 2 ⋅ ⋅ 5h
2
40 = 5h
A=
40 5h
=
5
5
8=h
Solution:Thetriangle'sheightis8millimeters.Notethatinthethirdrow,wemultipledbothsidesofthe
equationby2togetridofthefraction
1
.
2
AreaofCirlces
Next,let'slearnthefamouscircleareaformula: A = πr 2 .Theonlythingweneedtobecarefulaboutis
thatsometimesthediameterisgiven,andweneedtofindtheradiusfirst.
[Example3]Acircle'sdiameteris8yards.Findthiscircle'sareaintermsofπ,andthenroundtheareato
thehundredthplace.
[Solution]Wefirstfindthecircle'sradius,whichishalfofitsdiameter―4yardsinthisproblems.Next,
weuseacircle'sareaformula:
A = πr 2 = π ⋅ 4 2 = 16π yd2
Next,weuseascientificcalculatortochangetheresulttodecimal,andthenroundtothehundredth
place:
A = 16π = 16 ⋅ 3.1415026... ≈ 50.27 yd2
Solution:Thecircle'sareais16πyd2(accuratevalue),orapproximately50.27yd2.
Inthenextexample,theareaisgiven,andyouareaskedtofindthecircle'sradius.Weneedtoreview
theconceptof"squareroot"first.
02 = 0
0 =0
12 = 1
1 =1
22 = 4
4 =2
2
3 =9
9 =3
2
4 = 16
16 = 4
...
...
10 2 = 100
100 = 10
Squarerootdoestheoppositeofsquare.Ifweknow r 2 = 100 ,weusesquareroottofindr'svalue:
r 2 = 100
r = 100 r = 10
[Example4]Acircle'sareais100squaremeters.Findthisradiusdiameter.Roundyouranswerto
hundredthplace.
[Solution]Tofindacircle'sdiameter,weneedtofinditsradius.Plug A = 100 intoacircle'sarea
formula,wehave:
A = πr 2
100 = πr 2
100
π
100
π
100
π
=
πr 2
π
= r2
=r
5.64 ≈ r
Sinceacircle'sdiameteristwiceitsradius,wehave:
d = 2r = 2 ⋅ 5.64 = 11.28 meters
Solution:Thecircle'sdiameterisapproximately11.28meters.