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): You can put this solution on YOUR website!
This can only be done using variables because no values are given.

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.


AMP Parsing Error of [w=(4p+- 4sqrt(p^2-2(p^2-d^2))/16]: Invalid expression. Closing bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187. .

-----this is just for w.

What about the other dimension, L?
... w=(p-2L)/2;
when substituted into the diagonal equation.




General Solution,


AMP Parsing Error of [L=(4p+- 4sqrt(p^2-2(p^2-d^2))/16]: Invalid expression. Closing bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 187. .

-----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.

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