geometry / math? - XP Math - Forums

 XP Math - Forums geometry / math?

 07-30-2007 #1 marianne d Guest   Posts: n/a geometry / math? The length of a rectangular playing field is 5 ft. less than twice its width. If the perimeter of the playing field is 230 ft. find the length and width of the field.
 07-30-2007 #2 M Jith Guest   Posts: n/a Let, width = xlength = 2x - 5Perimeter = 2( L + w)P = 2(x + 2x - 5)230 = 6x - 10x = 40 ft.Width = 40 ft.length = 2x-5 = 75 ft.there u go.
 07-30-2007 #3 dadi Guest   Posts: n/a L=2W-5 P=2(L+W)230=2(L+W) divide both side by 2115=L+W115-W=Lsince L=2W-5115-W=2W-5115+5=2W+W120=3WW=40ftsubstitute to L=2W-5L=2(40)-5L=75ft
 07-30-2007 #4 russell m Guest   Posts: n/a perimeter = 2(L+W)L= (2W-5)Perimeter = 2((2W-5)+W)or 230' = 2((2W-5)+W)230= 2((2W-5)+W)--------------------------- 2115 = 2W-5+W115=3W-5115(+5) = 3W -5 (+5)120=3W120=3W------------ 340= W
 07-30-2007 #5 Amethyst Cherry W Guest   Posts: n/a easy.let the breadth of the field be x.2(2x-5)+2x=2304x-10+2x=2306x=240x=40breadth=40lenght=40X2-5 =75

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