The length of a rectangle is 9 in greater than the width the area is 36 in squared find the dimensions?

Question

The length of a rectangle is 9 in greater than the width the area is 36 in squared find the dimensions?

Answer 1

do you need to know the formula or anything?

cause thats a pretty easy question...

the dimensions are clearly 3 and 12

and that was guess and test.

its summer time so i can't really solve it at the moment using a formula or anything fancy...

Answer 2

L = W + 9

Area = LW

A = (W + 9)(W)

36 = W^2 + 9W

0 = W^2 + 9W - 36

0 = (W - 3)(W + 12)

W = 3, W = -12

W = 3 because width can never be negative

L = 3 + 9

L = 12

The dimensions are 12 in by 3 in.

Answer 3

The area of a rectangle is given by length * width. Therefore, we find the lengths and widths of the sides, set them equal to the area and solve for one to find the other.

Let w = width. Then length = w + 9

l*w = a

(w + 9)*w = 36

w^2 + 9w = 36

w^2 + 9w - 36 = 0

Now lets factor this last expression so that we may use the zero product property to solve for the width of the rectangle.

(w + 12)(w - 3) = 0

Therefore,

w + 12 = 0

w = -12 in, which is not plausible since we cannot have a negative length, or

w - 3 = 0

w = 3 in, which must be the answer since there is no other from which to choose.

Since w = 3 in, we get that the length is (w + 9), or 12, which completes the problem.

Answer 4

L = W + 9

A = (W + 9) (W)

W² + 9W = 36

W² + 9W - 36 = 0

(W + 12)(W - 3) = 0

W = 3 in

L = 12 in

Answer 5

L=B+9

L*B=36

B^2+9B-36=0

B=-12 OR +3(12 IS REJECTED), SO L=3+9=12

SO L=12 IN, B=3 IN