Question 1171867
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35/4 = 8.75
27/2 = 13.5


The area is 8.75 square meters and the perimeter is 13.5 meters.


Let L and W be the length and width
If the area is 8.75 square meters, then
A = L*W
L*W = A
L*W = 8.75


If the perimeter is 13.5 meters, then
P = 2*(L+W)
2*(L+W) = 13.5
L+W = 13.5/2
L+W = 6.75
L = 6.75-W


Plug this into the first equation and get everything to one side
L*W = 8.75
(6.75-W)*W = 8.75
6.75W-W^2 = 8.75
6.75W-W^2-8.75 = 0
-W^2+6.75W-8.75 = 0
100W^2-675W+875 = 0
In the last step I multiplied both sides by -100 to clear out the decimals and to make the leading term positive.


From here we could divide every term by 25 going from
100W^2-675W+875 = 0
to
4W^2-27W+35 = 0


Solving that equation is the same as solving 
4x^2-27x+35 = 0


Let's use the quadratic formula to solve

{{{x = (-b+sqrt(b^2-4ac))/(2a)}}} or {{{x = (-b-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-(-27)+sqrt((-27)^2-4(4)(35)))/(2(4))}}} or {{{x = (-(-27)-sqrt((-27)^2-4(4)(35)))/(2(4))}}}


{{{x = (27+sqrt(169))/(8)}}} or {{{x = (27-sqrt(169))/(8)}}}


{{{x = (27+13)/(8)}}} or {{{x = (27-13)/(8)}}}


{{{x = (40)/(8)}}} or {{{x = (14)/(8)}}}


{{{x = 5}}} or {{{x = 1.75}}}


So the width is either W = 5 or W = 1.75


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If the width was W = 5, then,
L = 6.75-W
L = 6.75-5
L = 1.75


Then note how
L*W = 1.75*5 = 8.75
and
2*(L+W) = 2*(1.75+5) = 13.5
which matches with our area and perimeter requirements. 


So one possible set of dimensions is
width = 5 meters
length = 1.75 meters


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If the width was W = 1.75, then,
L = 6.75-W
L = 6.75-1.75
L = 5
We basically get the flip of the previous section. So the order of the length and width doesn't matter.


We could say either
length = 5 meters
width = 1.75 meters
OR
length = 1.75 meters
width = 5 meters


Regardless of the order, they will both satisfy the equations we set up earlier, so they fit the requirements your teacher has set.


In reality we only have one unique set of dimensions.

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Answer: The rectangle is 5 meters by 1.75 meters
Note: In fraction form, 1.75 = 7/4
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