Gery wants to fence a rectangular field whose area is 1200 sq.m. He has only 100 meters of fencing so he decided to fence only the three side of the rectangle letting the wall be the fourth side. How wide the rectangle should be?
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With two short sides (x) and one long side y,
2x+y = 100
The area is
a = xy = x(100-2x)
So, you just need to solve
x(100-2x) = 1200
2x^2-100x+1200 = 0
2(x-20)(x-30) = 0
now can you figure it out?
To find the width of the rectangle, we can use the information given about the area and the amount of fencing available.
Let's assume the length of the rectangle is L, and the width is W.
We know that the area of a rectangle is given by the formula:
Area = Length * Width
In this case, the area of the rectangular field is given as 1200 sq.m. So we have:
1200 = L * W
We also know that Gery wants to fence only three sides of the rectangle, using 100 meters of fencing. The three sides that will be fenced are two lengths (L) and one width (W). We can write this information as an equation:
2L + W = 100
Now we have a system of two equations:
1) 1200 = L * W
2) 2L + W = 100
To solve this system, we can rearrange equation 2) to solve for W:
W = 100 - 2L
Substitute this value of W in equation 1):
1200 = L * (100 - 2L)
Now we have a quadratic equation. Simplify and rearrange:
1200 = 100L - 2L^2
Rearrange the equation to form a standard quadratic equation:
2L^2 - 100L + 1200 = 0
We can solve this equation using factoring, completing the square, or the quadratic formula. Factoring is the easiest method in this case:
2L^2 - 60L - 40L + 1200 = 0
2L(L - 60) - 40(L - 60) = 0
(2L - 40)(L - 60) = 0
This gives us two possible solutions:
1) 2L - 40 = 0, which gives L = 20
2) L - 60 = 0, which gives L = 60
Since a length of 20 meters would leave only 60 meters for the two widths, which is not enough, we can discard that solution.
Therefore, the width of the rectangle must be 60 meters.