SOLUTION: A rectangle has a diagonal that is 3.6 feet longer than the length and 7.1 feet longer than the width. What are the dimensions of the rectangle? I would appreaciate it if someone c

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Question 144660: A rectangle has a diagonal that is 3.6 feet longer than the length and 7.1 feet longer than the width. What are the dimensions of the rectangle? I would appreaciate it if someone could help me solve this problem. This is what I've done so far but I know its not correct. Thank you for your time.
X^2 + (X+7.1)^2 = (X+3.6)^2
X^2 + X^2 + 14.2 + 50.41 = X^2 + 2X + 3.6
X^2+16.2X + 52.6=0
(X=13.15)(X=4.05)
Found 2 solutions by scott8148, solver91311:
Answer by scott8148(5895) (Show Source):
You can put this solution on YOUR website!
let x=diagonal, so x-3.6=length and x-7.1=width

x^2=(x-3.6)^2+(x-7.1)^2 __ x^2=x^2-7.2x+12.96+x^2-14.2x+50.41

subtracting x^2 __ 0=x^2-21.4x+63.37

use quadratic formula to find x

Answer by solver91311(12126) (Show Source):
You can put this solution on YOUR website!
It looks like you intended for to represent the length of the diagonal. If that is the case, and the diagonal is 3.6 feet longer than the length of the rectangle, then the length of the rectangle has to be 3.6 feet SHORTER than the diagonal. Hence, the length of the rectangle would be represented by . Similarly, the width of the rectangle would be . Since the diagonal is the hypotenuse of a right triangle, and the width and length of the rectangle are the legs of that triangle:

Now expand that and solve the quadratic to get your dimensions.

By the way, since your given dimensions are expressed to the nearest 1/10 of a foot, your answer should not be expressed to any greater precision.