the perimeter of a basketball court is 268 feet. the area is 4200 ft2. find the dimensions. using quadractics
2L + 2W = 268
L * W = 4200
W = 4200/L
2L + 2(4200/L) = 268
Solve for L, then W.
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To find the dimensions of the basketball court, we can set up a quadratic equation based on the given information.
Let's assume the length of the basketball court is "L" and the width is "W."
We know that the perimeter of a rectangle is given by the formula P = 2(L + W), where P is the perimeter. In this case, the perimeter is 268 feet.
So, we have the equation: 268 = 2(L + W)
Simplifying the equation, we get: 134 = L + W ----(Equation 1)
We also know that the area of a rectangle is given by the formula A = L * W, where A is the area. In this case, the area is 4200 square feet.
So, we have the equation: 4200 = L * W ----(Equation 2)
Now we have a system of two equations with two variables (L and W). To solve for L and W, we can use the method of substitution or elimination.
Let's solve this system of equations using the method of substitution:
From Equation 1, we can express L in terms of W by subtracting W from both sides:
L = 134 - W
Substitute this value of L into Equation 2:
4200 = (134 - W) * W
Expanding the equation, we get:
4200 = 134W - W^2
Rearranging the equation in standard quadratic form, we have:
W^2 - 134W + 4200 = 0
Now, we can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula.
Let's use the quadratic formula to solve for W:
W = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = -134, and c = 4200.
Substituting these values into the quadratic formula, we get:
W = (-(-134) ± √((-134)^2 - 4 * 1 * 4200)) / (2 * 1)
W = (134 ± √(17956 - 16800)) / 2
W = (134 ± √1156) / 2
W = (134 ± 34) / 2
Now, we have two possible values for W:
W1 = (134 + 34) / 2 = 84
W2 = (134 - 34) / 2 = 50
So, the possible widths of the basketball court are 84 feet or 50 feet.
To find the corresponding lengths, we can substitute the values of W into Equation 1:
For W = 84:
L = 134 - 84 = 50
So, the length is also 50 feet when the width is 84 feet.
For W = 50:
L = 134 - 50 = 84
So, the length is also 84 feet when the width is 50 feet.
Therefore, the dimensions of the basketball court can be either 50 feet by 84 feet or 84 feet by 50 feet.