SOLUTION: if the area is 40 square meters, and the perimeter is 26 meters, what is the length and the width? (this was a question on my son's grade 4 math test that is stumping me...)

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Question 290059: if the area is 40 square meters, and the perimeter is 26 meters, what is the length and the width? (this was a question on my son's grade 4 math test that is stumping me...)
Found 2 solutions by scott8148, dabanfield:
Answer by scott8148(6628) (Show Source):
You can put this solution on YOUR website!
P = 2L + 2W, so half the perimeter is L + W

this means you are looking for two numbers that add to 13 (26/2), and multiply to 40

like, eight and five ...

the equation mechanics:

L * W = 40 ___ and ___ L + W = 13 ___ L = 13 - W

substituting ___ (13 - W) * W = 40 ___ 0 = W^2 - 13W + 40

0 = (W - 8) (W - 5)

W - 8 = 0 ___ W = 8

W - 5 = 0 ___ W = 5

so W equals 8 or 5 and L equals the other one

Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website!
if the area is 40 square meters, and the perimeter is 26 meters, what is the length and the width? (this was a question on my son's grade 4 math test that is stumping me...)
Wow. 4th grade??? !!!
Let W be the length and L be the length of the rectangle?
Then the area is W*L and the perimeter is 2*W + 2*L. So we have:
1.) W*L = 40
2.) 2*W + 2*L = 26
From 1.) we know that W = 40/L so we can substitute 40/L for W in equation 2.):
2*(40/L) + 2*L = 26
80/L + 2*L = 26
Multiply both sides of this equation by L:
80 + 2L*L = 26*L
80 + 2L^2 = 26*L
2L^2 - 26L + 80 = 0
Divide both sides by 2:
L^2 - 13L + 40 = 0
(L-8)*(L-5) = 40
So L-8 = 0 or L-5 = 0.
L = 8 or L = 5
If L = 8 then W = 40/8 = 5
If L = 5 then W = 40/5 = 8
I don't see an easier way to do this unless it's just to "see" the answer or experiment until you get the right combination for L and W.