Answer:
Width = 10ft
Length = 12ft
Step-by-step explanation:
For a rectangle of length L and width W, the area is calculated as:
A = L*W
In this case, we know that:
The area is 120 ft^2.
The length is 2 ft more than the width.
Then we can write:
L = W + 2ft.
If we use these two equations and replace them in the equation for the area of a rectangle, we get:
A = 120ft^2 = W*L = W*(W + 2ft)
120ft^2 = W*(W + 2ft)
Now we can solve this for W
120 ft^2 = W^2 + 2ft*W
Now we can rewrite this as:
W^2 + 2ft*W - 120ft^2 = 0
The solutions of this equation can be found by the Bhaskara's formula.
We know that for an equation of the form:
a*x^2 + b*x + c = 0
The two solutions are given by:
[tex]x = \frac{-b +\sqrt{b^2 - 4*a*c} }{2*a}[/tex]
in our case, we have:
a = 1
b = 2ft
c = -120ft^2
Then the solutions for the equation W^2 + 2ft*W - 120ft^2 = 0 are:
[tex]W = \frac{-2ft + -\sqrt{(2ft)^2 - 4*1*(-120ft^2)} }{2*1} = \frac{-2ft +- 22ft}{2}[/tex]
Then the two solutions are:
W = (-2ft - 22ft)/2 = -12ft (this is a negative measure, does not have a real meaning in this case, so this solution can be discarded)
The other solution is:
W = (-2ft + 22ft)/2 = 10ft (this is the correct solution)
Then the width is 10ft, and we know that the length is 2 ft longer than the width, then the length is:
L = W + 2ft = 10ft + 2ft = 12ft
