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albebra required for this

let a = side 1 and side 3 of a rectangle

let b = side 2 and side 4 of a rectangle

then the perimeter = a + b + a + b = 2a + 2b+ 28

and the area = a x b = 18

taking 2a + 2b = 28 and making a the subject of the formula you get a = 28-b

taking a= 28-b into the axb = 18 formula gives

b(28-b)=18

so b2-28b+18 = 0

by quadratic formula . i will say b = 14 +/- the root of 178

which you can work out. sorry i have run out of time

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Q: Is it possible to have a perimeter of 28 with 18 square units?
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