document.write( "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. \n" ); document.write( "

Algebra.Com's Answer #399241 by stanbon(75887) $\"\"$ $\"About$

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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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\n" ); document.write( "Let the height be \"x\".
\n" ); document.write( "Then width = \"x+17\".
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\n" ); document.write( "Equation:
\n" ); document.write( "Pythagoras:
\n" ); document.write( "x^2 + (x+17)^2 = 53^2
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\n" ); document.write( "x^2 + x^2 + 34x + 289 = 2809
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\n" ); document.write( "2x^2 + 34x - 2520 = 0
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\n" ); document.write( "x^2 + 17x - 1260 = 0
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\n" ); document.write( "x = [-17 +- sqrt(17^2-4*-1260)]/2
\n" ); document.write( "--------
\n" ); document.write( "x = [-17 +- sqrt(5329)]/2
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\n" ); document.write( "Positive solution:
\n" ); document.write( "x = [-17+73]/2
\n" ); document.write( "---
\n" ); document.write( "x = 28 (height)
\n" ); document.write( "x+17 = 45 (width)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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