Question

find the area of a rectangle having a perimeter of 182yd and a width 20yd shorter than twice its length.

Answer 1

You have to use both the perimeter formula and the area formula for a rectangle. The perimeter is P = 2L + 2W, right? And they tell you that the perimeter's measurement is 182; they also tell you that the width is 20 yds shorter than twice the length. So we need to width in terms of length so we only have one variable, L, instead of 2, L and W. So...

Answer 2

W=2L-20 is "the width is 20 less than twice the length". So in our perimeter formula, we have a value for P and we also have width in terms of length...
P = 2L + 2W
182=2L + 2(2L - 20)

Answer 3

We need to solve that for the length. We will get:
182=2L + 4L - 40. Solving that for L, we get 182+40=2L+4L. 222=6L. L = 37. So the length is 37, and the width is 2(37)-20 which is 54. Now we have the length and the width of the rectangle, use those in the area formula to find the area of the rectangle.
A = L * W...A = 37 * 54 which is 1998

Answer 4

Thanks, got confused at 182+40=2L+4L but figured out what to do from there thanks @IMStuck