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"
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...
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
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
L + W = 6
L * W = 8
L = 4
W = 2