How to use algebra to find length and width of rectangle
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How to use algebra to find length and width of rectangle

[From: ] [author: ] [Date: 11-10-08] [Hit: ]
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The perimeter of a rectangle is 12 inches, and its area is 8 square inches. Find the length and width of the rectangle. Please help with the problem and thank you!

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LET THE LENGTH BE X AND WIDTH BE Y

2( X+Y) = 12

X +Y = 6

XY = 8

X = 8/Y

8/Y + Y = 6

Y^2 -6Y + 8 = 0
( Y -4) ( Y -2) = 0

Y = 4 OR 2

X = 2 OR 4

answer LENGTH = 4 " WIDTH = 2"

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formula for perimeter is P=2W + 2L
area is A=WL

equate 12=2W +2L
8=WL
W=L/8
12=2(L/8) + 2L
12=L/4 + 2L
12= 9L/4
multiply it by 4 to cancell the denominator
4(12=9L/4)
36=9L
9L=36
L=4

in the perimter 12=2W +2L
use L=4
12=2W+2(4)
12=2W+8
2w=12-8
2w=4
w=2


so the Length is 4 and the width is 2...

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2l + 2w = 12
2(l + w) = 12
l + w = 6
l = 6 - w

And

lw = 8
l = (8 / w)

So

6 - w = 8 / w
6w - w^2 = 8
-w^2 + 6w - 8 = 0
w^2 - 6w + 8 = 0
(w-4)(w-2) = 0
w = 4 or 2

so l = 4 or 2

check:
2(4) + 2(2) = 12
8 + 4 = 12
12 = 12

(4)(2) = 8
8 = 8

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2L + 2W = 12

L + W = 6

L * W = 8

L = 4

W = 2
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