The length of a rectangular garden exceeds the width by 8 feet. If the perimeter of the rectangle is 28 feet, what are its dimensions?
width -- x
length -- x+8
solve for x ...
2(x + x+8) = 28
To find the dimensions of the rectangular garden, we can set up and solve a system of equations using the given information. Let's denote the width as "w" and the length as "l".
Given that the length exceeds the width by 8 feet, we can write the equation:
l = w + 8 (Equation 1)
We also know that the perimeter of a rectangle is given by the formula:
Perimeter = 2(length + width)
Substituting the given perimeter of 28 feet into the formula, we get:
28 = 2(l + w)
Since we have two variables (l and w) and two equations (Equation 1 and the modified Perimeter equation), we can solve this system of equations.
Let's begin by rearranging Equation 1 to solve for "l":
l = w + 8
Now, substitute this expression for "l" in the modified Perimeter equation:
28 = 2((w + 8) + w)
Simplifying, we have:
28 = 2(2w + 8)
28 = 4w + 16
12 = 4w
w = 3
Now, substitute the value of "w" back into Equation 1 to find "l":
l = w + 8
l = 3 + 8
l = 11
Hence, the dimensions of the rectangular garden are 3 feet by 11 feet.