The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 82 cm^2, what is the length of the diagonal? Note: Your answer must be a number. It may not contain any arithmetic operations.
w(w+20) = 8000
w^2+20w = 8000
w^2+20w-8000 = 0
(w-80)(w+100) = 0
w-80 = 0
w=80
or
w+100 = 0
w = -100
SORRY I WROTE THE WRONG QUESTION PERVIOUSLY
I think that x=80 makes more sense. However what would the length be if the width is equal to 80 in this problem ? please help Im stuck and am not sure what the final answer for this question would be.
Length = L units.
Width = (2L-2) units.
A = L(2L-2) = 82cm^2
2L^2 - 2L - 82 = 0
L^2 - L - 41 = 0
Use Quadratic formula:
L = 6.92 cm.
Width = 2L - 2 = 2*6.92 - 2 = 11.84 cm.
The diagonal divides the rectangle into
2 congruent rt. triangles.
D^2 = L^2 + W^2 = 6.92^2 + 11.84^2=188
Diag. = 13.71 cm.
To find the length of the rectangle when the width is 80, let's go back to the given information.
The width of a rectangle is 2 less than twice its length. So we can express the width of the rectangle as:
width = 2 * length - 2
In this case, the width is given as 80, so we can set up the equation:
80 = 2 * length - 2
To solve for the length, we'll isolate the variable by first adding 2 to both sides of the equation:
80 + 2 = 2 * length
Simplifying:
82 = 2 * length
Now, divide both sides of the equation by 2:
82 / 2 = length
Which gives us:
41 = length
So, if the width of the rectangle is 80, then the length of the rectangle is 41 cm.
Therefore, the final answer to the question, "what is the length of the diagonal?", would be 41 cm.