The expression 2(l+w) may be used to find the perimeter of a rectangle. What are the length and width of a rectangle if the area is 13 1/2 square units and the length of one side is 1/5 the measure of the perimeter?
To solve this problem, we need to use the given information to set up equations and then solve for the variables.
Let's assume that the length of the rectangle is represented by 'l' and the width is represented by 'w'.
We know that the area of a rectangle can be calculated by multiplying its length and width:
Area = Length × Width
Given that the area is 13 1/2 square units, we can write the equation as:
13 1/2 = l × w
We are also given that the length of one side is 1/5 the measure of the perimeter. The perimeter of a rectangle can be calculated by summing the lengths of all four sides:
Perimeter = 2(l + w)
Since we know that the length of one side is 1/5 the measure of the perimeter, we can write the equation as:
l = (1/5) × Perimeter
Now we can substitute this value of 'l' into the area equation:
13 1/2 = ((1/5) × Perimeter) × w
Since we have two equations involving two variables (Perimeter and w), we can solve them simultaneously.
Now, let's solve the equation to find the value of 'w' and eventually 'l'.
First, let's simplify the area equation:
13 1/2 = (1/5) × Perimeter × w
27/2 = (1/5) × Perimeter × w (converting the mixed number to an improper fraction)
Simplifying the equation further, we can multiply both sides by 2/27 to isolate the product of Perimeter and w:
(27/2) × (2/27) = (27/2) × (1/5) × Perimeter × w
1 = (Perimeter/5) × w
Now we have an equation relating w and Perimeter.
Substituting the value of Perimeter found earlier, which is Perimeter = 2(l + w), we get:
1 = (2/5)(l + w) × w
We can simplify this equation further:
1 = (2/5)lw + (2/5)w^2
Since we are solving for the length and width, we need to eliminate 'l'. We can substitute l = (1/5)Perimeter into the equation:
1 = (2/5)((1/5)Perimeter)w + (2/5)w^2
1 = (2/25)Perimeter × w + (2/5)w^2
Now we have an equation only involving the variable 'w'.
By rearranging the equation and combining like terms, we get a quadratic equation:
(2/5)w^2 + (2/25)Perimeter × w - 1 = 0
By simplifying and rearranging further, the equation becomes:
10w^2 + 2Perimeter × w - 25 = 0
You can now solve this quadratic equation using methods like factoring, completing the square, or using the quadratic formula to find the value(s) of 'w'. Once you find 'w', substitute it back into the equation l = (1/5)Perimeter to find the value of 'l'.
"the length of one side is 1/5 the measure of the perimeter" --> l = (1/5(2(l+w))
5l = 2l + 2w
3l = 2w
l = 2w/3
lw = 13.5
(2w/3)(w) = 13.5
w^2 = 20.25
w = sqrt(20.25) = 4.5
then l = 3