# 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. 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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Posted on 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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