The length of a rectangular field is 12m more than width.The value of its area= 4 times the value of its perimeter. Find dimensions of the field.
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calendarEducator since 2010
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We have to find the length of a rectangular given that the length is 12m more than the width and the numeric value of the area is four times that of the perimeter.
Let the length be L adn the width be W
The perimeter is 2*(L + W)
The area is L*W
The length is 12m more than the width, L = W + 12
Area = W*(W + 12) = 4* 2*(W + 12 + W)
=> W^2 + 12W = 8*(2W + 12)
=> W^2 + 12W = 16W + 96
=> W^2 - 4W - 96 = 0
=> W^2 - 12W + 8W - 96 = 0
=> W(W - 12) + 8(W - 12) = 0
=> W = 12 or -8
The width cannot be negative, so we take width = 12
The length = 12 + 12 = 24
The dimensions of the rectangle are 12m W by 24m L.
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
Let the length be L and the width be W
Given that the length is 12 more than the width.
==> L = 12+ W.............(1)
Also, given that the area is 4 times the perimeter.
==> A = 4*P
==> L*W = 4*(2L+2W)
(12+W)*W = 4(2(12+w) + 2w)
12W + w^2 = 4(24+4w)
12w + w^2 = 96 + 16w
==> w^2 - 4w - 96 = 0
Now we will factor.
==> (w- 12)(w+8) = 0
==> w = 12 ==> L = 12+12 = 24
Then the length of the rectangle is 24 m, and the width is 12 m.
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