The perimeter of a rectangular field is 60 meters. Its area is 200 square meters. Find the dimensions

Answer 1

Let the rectangle have a length (L) and a width (W) like this:

Then the perimeter is L + W + L + W or 2L + 2W

And the area is L•W

So, since perimeter is given as 60 and perimeter is 2L + 2W, then 2L + 2W = 60

And, since the area is given as 200, then L•W = 200

If L•W = 200 then L = 200/W. That means, in the first equation, we can plug in "200/W" wherever there we see "L" so:

2L + 2W = 60

2(200/W) + 2W = 60

400/W + 2W = 60 [multiply both sides by W to get:

400 + 2W² = 60W [This is a quadratic equation. To solve a quadratic, we need to set it equal to zero. So subtract 60W from both sides to get:

2W² - 60W + 400 = 0 {divide both sides by 2 to get:

W² - 30W + 200 = 0 [factor it to get:

(W - 10)(W - 20) = 0 [now, set each piece equal to zero

W - 10 = 0 or W = 10 and W - 20 = 0 or W = 20

So, when W = 10, we can plug 10 in for W in the equation L = 200/W to get L = 200/10 or L = 20

And, when W = 20, we can plug 10 in for W in the equation L = 200/W to get L = 200/20 or L = 10

So the rectangle is either 10 meters by 20 meters, or it is 20 meters by 10 meters