Quick Help .............

Question

The perimeter of a rectangular field is 80 meters and its area is 320 square meters. Find the dimensions of the field, correct to the nearest tenth of a meter.

Answer 1

There are 2 equations that can be made from what is described in this problem.

Let x = the width of the field

Let y = the length of the field

Perimeter equation: 80=2x+2y

Area equation: 320=xy

Can you solve the system of equations by substitution or elimination?

Answer 2

perimeter is L + W + L + W or 2L + 2W

And the area is L•W

So, since perimeter is given as 80 and perimeter is 2L + 2W, then 2L + 2W = 80

And, since the area is given as 320, then L•W = 320

If L•W = 320 then L = 320/W. That means, in the first equation, we can plug in "320/W" wherever there we see "L" so:

2L + 2W = 80

2(320/W) + 2W = 80

640/W + 2W = 80 [multiply both sides by W to get:

640 + 2W2 = 80W [This is a quadratic equation. To solve a quadratic, we need to set it equal to zero. So subtract 80W from both sides to get:

2W2 - 80W + 640 = 0 {divide both sides by 2 to get:

W2 - 40W + 320 = 0 [factor it to get:

w=11.06

So, when W = 11.06, we can plug 11.06 in for W in the equation L = 320/W to get L = 28.94

So the rectangle is 11.06 meters by 28.94 meters