SOLUTION: Please help me solve this problem: The length of a rectangle is 5 in. longer than its width. The diagonal is 5 in. shorter than twice the width. Find the length, width, and dia

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Question 1006429: Please help me solve this problem:
The length of a rectangle is 5 in. longer than its width. The diagonal is 5 in. shorter than twice the width. Find the length, width, and diagonal measures of the rectangle.

Thank you so much!
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let the length be L, the width W, and the diagonal d
if the length of a rectangle is 5in longer than its width, we have
L=W%2B5
if the diagonal is 5 in. shorter than twice the width, we have
d=2W-5
we know that d%5E2=L%5E2%2BW%5E2...substitute d and L from above
%282W-5%29%5E2=%28W%2B5%29%5E2%2BW%5E2........solve for W
4W%5E2-20W%2B25=W%5E2%2B10W%2B25%2BW%5E2
4W%5E2-20W%2B25=2W%5E2%2B10W%2B25
4W%5E2-20W%2B25-2W%5E2-10W-25=0
2W%5E2-30W=0....simplify, divide by 2
W%5E2-15W=0
%28W-15%29W=0
solutions:
W=0....since W represents the width, disregard this solution
highlight%28W=15%29=> your solution
now find the length
L=15%2B5
highlight%28L=20%29
and diagonal
d=2%2A15-5
d=30-5
highlight%28d=25%29
check the result:

d%5E2=L%5E2%2BW%5E2
25%5E2=20%5E2%2B15%5E2
625=400%2B225
625=625