document.write( "Question 1056236: A rectangle is drawn so the width is 31 inches longer than the height. If the rectangle's diagonal measurement is 41 inches, find the height.\r
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Algebra.Com's Answer #671486 by ankor@dixie-net.com(22740)\"\" \"About 
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A rectangle is drawn so the width is 31 inches longer than the height.
\n" ); document.write( " If the rectangle's diagonal measurement is 41 inches, find the height.
\n" ); document.write( "Using pythag
\n" ); document.write( "h^2 + (h+31)^2 = 41^2
\n" ); document.write( "FOIL (h+31)(h+31)
\n" ); document.write( "h^2 + h^2 + 31h + 31h + 961 = 1681
\n" ); document.write( "combine like terms
\n" ); document.write( "2h^2 + 62h + 961 - 1681 = 0
\n" ); document.write( "2h^2 + 62h - 720 = 0
\n" ); document.write( "simplify, divide by 2
\n" ); document.write( "h^2 + 31h - 360 = 0
\n" ); document.write( "Factors to
\n" ); document.write( "(h + 40)(h - 9) = 0
\n" ); document.write( "The positive solution is all we want here
\n" ); document.write( "h = 9 inches is the height
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\n" ); document.write( "Check; the other side: 9+31
\n" ); document.write( "9^2 + 40^2 = 41^2 \n" ); document.write( "
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