Question

Find the area of a rectangle having a perimeter of 182 yards and a width 20 yards shorter than twice its length

Answer 1

Let the length be $l$ in this equation.

We know that the perimeter of a rectangle is $w2 + l2 = P$

Width is defined in terms of length, so we know $w = l2 - 20$

We can substitute this value of w into the formula for perimeter to find the length.

$(l2 - 20)2 + l2 = 182$

$l4 - 40 + l2 = 182$

$l4 + l2 = 222$

$l6 = 222$

$l = 37$

So the length is 37. Now that we've found the length, we can go back to the Perimeter formula again and find width.

$w2 + (37)2 = 182$

$w2 + 74 = 182$

$w2 = 108$

$w = 54$

Width is 54. Now with both of these values found, we can calculate area with the formula $l * w = A$

$37 * 54 = A$

$1998 = A$

So the area of the rectangle is 1998 yards^2.