A rectangle field has a perimeter of 118 feet. The width is 8 feet longer than half the length. Find the dimensions of the rectangle.
see my response to your almost identical question two postings below.
What is the width of the rectangle if the perimeter is 188 feet
To find the dimensions of the rectangle, we can set up a system of equations using the given information.
Let's start by assigning variables to the dimensions of the rectangle:
Let's say the length of the rectangle is 'L' feet.
And let's say the width of the rectangle is 'W' feet.
According to the given information, the perimeter of the rectangle is 118 feet. The perimeter of a rectangle is calculated by adding all four sides:
Perimeter = 2 * (Length + Width)
So, we can write the equation as:
2 * (L + W) = 118
Now, we are also given that the width is 8 feet longer than half the length.
The width is '8 feet longer than half the length' can be expressed as:
W = (1/2)L + 8
Now, we have a system of two equations that we can solve simultaneously to find the values of L and W.
Let's substitute the expression for the width in terms of length into the perimeter equation:
2 * (L + ((1/2)L + 8)) = 118
Now, simplify the equation:
2 * (L + (L/2) + 8) = 118
2 * (3/2)L + 16 = 118
3L + 16 = 118
3L = 118 - 16
3L = 102
L = 102 / 3
L ≈ 34
Now, substitute the value of L back into the expression for the width:
W = (1/2)L + 8
W = (1/2)(34) + 8
W = 17 + 8
W = 25
So, the dimensions of the rectangle are approximately: Length (L) ≈ 34 feet and Width (W) = 25 feet.