SOLUTION: If the perimeter of a rectangle is p and its diagnoal is d, the difference between the length and width of the rectangle is ______?
I have worked several formulas, 2L Plus 2W =
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-> SOLUTION: If the perimeter of a rectangle is p and its diagnoal is d, the difference between the length and width of the rectangle is ______?
I have worked several formulas, 2L Plus 2W =
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Question 892582: If the perimeter of a rectangle is p and its diagnoal is d, the difference between the length and width of the rectangle is ______?
I have worked several formulas, 2L Plus 2W = p, solved for L, and tryig to solve for W, I have worked it several ways. I have also used the Pathagoreum theorm, L squared plus w squared = d squared. I have not been able to solve it, begining to wonder if it is a trick problem from instructor. Can you help. Answer by josgarithmetic(39799) (Show Source):
and ;
You want to find .
You assume that the constants are d and p.
;
substitute into the diagonal equation.
Use the general solution of quadratic equation. -----this is just for w.
What about the other dimension, L?
... w=(p-2L)/2; when substituted into the diagonal equation.
General Solution, -----formula for L.
YOU can finish this.
Pay attention to both forms for w and L, since each has a PLUS form and a MINUS form. Remember, you are looking for THE DIFFERENCE or absolute value of w and L.
Please excuse the lack of rendering in three of the steps. Tracing the parentheses will be difficult. The majority of readable steps should be plenty helpful.