# A rectangular field has a length 10 feet more than it is width. If the area of the field is 264, what are the dimensions of the rectangular field? Let L be the length of the field.  Let W be the width of the field.  We know from your question that

L = W + 10

We also know that W*L = 264

This is because the area of a rectangle is equal to the length times the width.

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Let L be the length of the field.  Let W be the width of the field.  We know from your question that

L = W + 10

We also know that W*L = 264

This is because the area of a rectangle is equal to the length times the width.

So now we subsitute in

(W + 10)*W = 264

W^2 + 10W = 264

W^2 + 10W - 264 = 0

Now we have to find the factors.  We can factor this out to

(w + 22) (w - 12) = 0

So the width could be -22 or 12.  Of course it cannot be negative so

W = 12

and therefore

L = 22.

Approved by eNotes Editorial Team Let us assume that the  length is (L) and the width is (W)

We know that the area of a rectangle is a=L*W

==> LW = 264 ...(1)

But we know that L=W+10

Substitute with (1):

(W+10)W =264

w^2 + 10w = 264

w^2+10w -264=0

Factorize:

(w-12)(w+22)=0

W= 12

L= 12+10 = 22

Approved by eNotes Editorial Team