6. The practice court at the park is a rectangle. Its width is exactly twice its length. The area of the practice field is 450 ft2. What are the length and width of the rectangle?
W = 2L
W * L = 450
2L * L = 450
Solve for L, then W.
w=30
L=15
To find the length and width of the rectangle, we can set up a system of equations based on the given information:
Let's say the length of the rectangle is represented by 'L' and the width is represented by 'W'.
From the problem statement, we know that the width is exactly twice the length:
W = 2L
Also, we know the area of the rectangle is 450 square feet:
Area = Length * Width
450 = L * W
Now we can substitute the value of W from the first equation into the area equation:
450 = L * (2L)
Simplifying the equation:
450 = 2L^2
Dividing both sides by 2:
225 = L^2
Taking the square root of both sides:
L = ±√225
L = ± 15
Since the length of a rectangle cannot be negative, we can conclude that the length is 15 feet.
Now we can substitute the length back into the first equation to find the width:
W = 2L = 2 * 15 = 30 feet
Therefore, the length of the rectangle is 15 feet and the width is 30 feet.