Let the width be [tex]w[/tex].
Therefore we know that the length [tex]2w - 5[/tex].
We know that area is [tex]A = w*l[/tex].
Therefore:
[tex]273 = w * (2w-5)[/tex]
Now we can solve for the width.
[tex]273 = 2w^2 - 5w[/tex]
[tex]2W^2 - 5w - 273 = 0[/tex]
We can use the quadratic formula from here where a = 2, b = -5 and c = 273.
[tex]w = \frac{-b \pm \sqrt{b^2 -4ac} }{2a}[/tex]
[tex]w = \frac{-(-5) \pm \sqrt{-5^2 -4(2)(-273} }{2(2)}[/tex]
Solving this gives us the possible answers of 13 and [tex]- \frac{21}{2}[/tex]. Since our width can't be negative, we know the answer is 13.
Next, we can plug 13 into the formula for area to find the length.
[tex]273 = 13*l[/tex]
[tex]21 = l[/tex]
We now know the width and length. Finally, the last thing to do is calculate the perimeter using the formula [tex]2w + l2 = P[/tex]
[tex]2(13) + 2(21) = P[/tex]
[tex]26 + 42 = P[/tex]
[tex]68 = P[/tex]
So, the perimeter of the rectangle is 68 yards.
