SOLUTION: A rectangle is drawn so the width is 17 inches longer than the height. If the rectangle's diagonal measurement is 53 inches, find the height.

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Question 633877: A rectangle is drawn so the width is 17 inches longer than the height. If the rectangle's diagonal measurement is 53 inches, find the height.
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A rectangle is drawn so the width is 17 inches longer than the height. If the rectangle's diagonal measurement is 53 inches, find the height.
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Let the height be "x".
Then width = "x+17".
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Equation:
Pythagoras:
x^2 + (x+17)^2 = 53^2
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x^2 + x^2 + 34x + 289 = 2809
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2x^2 + 34x - 2520 = 0
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x^2 + 17x - 1260 = 0
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x = [-17 +- sqrt(17^2-4*-1260)]/2
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x = [-17 +- sqrt(5329)]/2
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Positive solution:
x = [-17+73]/2
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x = 28 (height)
x+17 = 45 (width)
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Cheers,
Stan H.
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Answer by solver91311(24713) (Show Source):
You can put this solution on YOUR website!

Let represent the width. Then must represent the height. Our good friend and constant companion, Pythagoras, says that the square of the diagonal is equal to the square of the width plus the square of the height:

Collect terms and simplify:

Solve the factorable quadratic for , then calculate

John

My calculator said it, I believe it, that settles it