The Browns installed 58 feet of fencing around a rectangular play yard. If the yard covers 204 square feet, what are its dimensions?
204 = 17*12
2(17+12) = 58
2 L + 2 w = 58
so L + w = 29
and L=29-w
w (29-w) = 204
29 w - w^2 = 204
w^2 - 29 w + 204 = 0
w = [ 29 +/- sqrt(841-816) ] / 2
w = [ 29 +/- 5 ]/2
w = 34/2 = 17
L = 12 that works
or
L =17 and w = 12
To find the dimensions of the rectangular play yard, we need to solve a system of two equations. Let's assume the length of the yard is L and the width is W.
From the given information, we know that the perimeter of the yard is 58 feet. The formula for the perimeter of a rectangle is given by P = 2L + 2W. In this case, the perimeter is 58 feet, so we can write the equation as:
2L + 2W = 58 -- Equation 1
We also know that the area of the yard is 204 square feet. The formula for the area of a rectangle is given by A = L * W. In this case, the area is 204 square feet, so we can write the equation as:
L * W = 204 -- Equation 2
Now, we have a system of two equations:
2L + 2W = 58 -- Equation 1
L * W = 204 -- Equation 2
We can solve this system of equations to find the values of L and W.
One way to solve this system is through substitution. Rearrange Equation 1 to solve for L:
2L = 58 - 2W
L = (58 - 2W) / 2
L = 29 - W -- Equation 3
Now, substitute Equation 3 into Equation 2:
(29 - W) * W = 204
29W - W^2 = 204
Rearrange this quadratic equation to standard form:
W^2 - 29W + 204 = 0
Now, we can either factor this equation or use the quadratic formula. Factoring, we find:
(W - 17)(W - 12) = 0
This gives us two possibilities for W:
W - 17 = 0 --> W = 17
W - 12 = 0 --> W = 12
Now we can substitute these values of W back into Equation 3 to find L:
For W = 17:
L = 29 - 17
L = 12
For W = 12:
L = 29 - 12
L = 17
Thus, we have two possible sets of dimensions for the rectangular play yard: 12 feet by 17 feet or 17 feet by 12 feet.