A rectangle has a perimeter of 98 feet. The lenght is 8 feet longer than double the width. Find each Dimension.
let the width be x
"The lenght is 8 feet longer than double the width" ----> length = 2x+8
Now isn't 2(x + 2x+8) = 98 ?
Solve this little equation.
To find the dimensions of the rectangle, let's assign variables to the given information. Let's use 'l' to represent the length and 'w' to represent the width.
According to the problem, the rectangle has a perimeter of 98 feet. The formula for the perimeter of a rectangle is P = 2(l + w), where P represents the perimeter, l represents the length, and w represents the width.
Therefore, we can write the equation as:
98 = 2(l + w)
The problem also states that the length is 8 feet longer than double the width. Mathematically, this can be represented by the equation:
l = 2w + 8
Now we have two equations:
Equation 1: 98 = 2(l + w)
Equation 2: l = 2w + 8
To solve this system of equations, we can substitute the value of 'l' from Equation 2 into Equation 1.
Substituting 'l' in Equation 1 gives us:
98 = 2((2w + 8) + w)
Simplifying the equation:
98 = 2(3w + 8)
98 = 6w + 16
82 = 6w
w = 82/6
w = 13.66 (rounded to two decimal places)
Now, we know the value of 'w', but we need to find the values of 'l' and 'w' (which should be whole numbers since we are dealing with the dimensions of a rectangle).
Using Equation 2 to find 'l':
l = 2w + 8
l = 2(13.66) + 8
l = 27.32 + 8
l = 35.32 (rounded to two decimal places)
Therefore, the dimensions of the rectangle are:
Length (l) ≈ 35.32 feet
Width (w) ≈ 13.66 feet
It's important to note that since we are dealing with measurements, we usually round the values to an appropriate decimal place to represent the level of precision required.