Question

The length of a rectangle is 5 less than twice its width. Find the perimeter of this rectangle if its area is 273 yd2

Answer 1

Let the width be $w$.

Therefore we know that the length $2w - 5$.

We know that area is $A = w*l$.

Therefore:

$273 = w * (2w-5)$

Now we can solve for the width.

$273 = 2w^2 - 5w$

$2W^2 - 5w - 273 = 0$
We can use the quadratic formula from here where a = 2, b = -5 and c = 273.
$w = \frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

$w = \frac{-(-5) \pm \sqrt{-5^2 -4(2)(-273} }{2(2)}$
Solving this gives us the possible answers of 13 and $- \frac{21}{2}$. 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.
$273 = 13*l$
$21 = l$
We now know the width and length. Finally, the last thing to do is calculate the perimeter using the formula $2w + l2 = P$

$2(13) + 2(21) = P$
$26 + 42 = P$
$68 = P$
So, the perimeter of the rectangle is 68 yards.