A basketball court is in the shape of a rectangle. Scottie and Jamal found that the width of their court was 44 feet less than the length. If the perimeter was 288 feet, what were the length and the width of their court?
2007-12-08 05:46:42 · 5 answers · asked by jared r 1 in Science & Mathematics ➔ Mathematics
Let the length be l and w be the width.
w=l-44
perimeter=2(w+l)=2w+2l
2w+2l = 2(l-44)+2l = 2l-88+2l =4l-88
4l-88=288
4l=288+88= 376
l=94
w=50
check : perimeter = 2(94+50)=2(144)=288
2007-12-08 05:53:04 · answer #1 · answered by cidyah 7 · 0⤊ 0⤋
Perimeter: Length=x, Width=x-44
2x+2(x-44)=288
4x-88=288
4x=288+88
4x=376
x=94
Length=94, Width =94-44=50
2007-12-08 14:00:00 · answer #2 · answered by Grampedo 7 · 0⤊ 0⤋
Call the width W and length L.
We are told that W = L - 44 and that 2W + 2L = 288. Plug the first equation into the second and you get 2(L-44) + 2L = 288. Get rid of the parentheses and you get 2L - 88 + 2L = 288. Combine like terms and you get 4L -88 = 288. Add 88 to both sides and you get 4L = 376. Divide by 4 and you get L = 94.
Now plug that into the first equation to get W = 94 - 44 = 50.
So the court is 50 feet wide and 94 feet long.
2007-12-08 13:54:10 · answer #3 · answered by Amy F 5 · 0⤊ 0⤋
let
L = length
L - 44 = width
288 = perimeter
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Perimeter formula
P = 2L + 2W
288 = 2L + 2(L - 44)
The distributive property on the right side of the equation
288 = 2l + 2L - 88
Collect like terms
288 = 4L - 88
Transpose 88
288 + 88 = 4L - 88 + 88
Collect like terms
376 = 4L
Divide both sides of the equation by 4
376 / 4 = 4L / 4
376/4 = L
94 = L. . .â. .The Length
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The length is 94 feet
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The width = L - 44
94 - 44 = 50
The width is 50 feet
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2007-12-08 14:22:37 · answer #4 · answered by SAMUEL D 7 · 0⤊ 0⤋
94' by 50'
2007-12-08 13:56:33 · answer #5 · answered by Joe K 1 · 0⤊ 0⤋