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
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.
Your equation doesn't even match your given data
length --- x
width ---- 2x - 2
area = x(2x-2) = 82
2x^2 - 2x - 82 = 0
x^2 - x - 41 = 0
use the formula ...
x = (1 ± √165)/2
= appr 6.92 or some negative
length = 6.92 , length = 11.85
check: 6.92x11.85 = appr 82
To find the length of the rectangle when the width is equal to 80, we can use the given information in the problem.
Let's start by substituting the value of the width (w) into the expression for the area of the rectangle. We know that the area (A) is 82 cm^2, so we have:
Length × Width = Area
Length × 80 = 82
Now we can solve for the length:
Length = 82 / 80
Calculating this out, we get:
Length ≈ 1.025
Therefore, when the width of the rectangle is 80 cm, the length is approximately 1.025 cm.
Now, let's move on to finding the length of the diagonal. To do this, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (diagonal) is equal to the sum of the squares of the other two sides.
In this case, the length and width of the rectangle can be considered as the two sides of the right triangle, and the diagonal is the hypotenuse. Let the length of the rectangle be L and the width be W (80 cm).
Using the Pythagorean theorem, we have:
L^2 + W^2 = Diagonal^2
(1.025)^2 + 80^2 = Diagonal^2
Let's compute this to find the value of the diagonal:
Diagonal ≈ 80.00124997917006 cm
Therefore, when the width of the rectangle is 80 cm, the length of the diagonal is approximately 80.00124997917006 cm.