I need the answer.
The area of a rectangle is 252 square feet. If the perimeter is 66 feet, find the length and width of a rectangle?
Hello, Jesse --
You need to use your information about area and perimeter to set up two equations with two variables representing length and width of the rectangle. Then you will set up a quadratic equation. Let's use L for the length and W for the width of the rectangle:
For the perimeter:
2L + 2W = 66 or
L + W = 33 or
W = 33 - L
For the area:
LW = 252
Now substitute W = 33 - L from the first equation into the second one:
L(33 - L) = 252 or
33L - L2 = 252 or
L2 - 33L + 252 = 0
This can be solved by factoring, or by using the quadratic equation.
(L - 12)(L - 21) = 0 or
L = 12, or L = 21
In this case, let's use 21 because length is usually the longer dimension. Now we substitute L = 21 into the first equation to find W.
W = 33 - 21 or
W = 12
Your answers are: Length = 21 and Width = 12
L+W = P/2 = 66/2 = 33
LW = A = 252
So L and W are the two roots of x2 - 33x + 252 = 0
which factors as (x - 12)(x - 21) = 0
So it is a 12 by 21 rectangle.
A = 252 sq. ft = l*w
2l + 2w = 66, 2(l+w) = 66, so l+w = 33
w = 252/l
l+(252/l) = 33
l^2 + 252 = 33l
l^2 - 33l - 252 = 0
You're at a quadriatic equation, which you can solve. You will get two answers - toss out the unreasonable one (hint - negative answer, won't fit with the width, etc.).
Mental math approach:
Half of the perimeter = length + width = 33 ft
Factor out 252 into two numbers such that the sum is 33: 252 = 21*12.
Therefore, the length is 21 ft and the width is 12 ft.