document.write( "Question 170636This question is from textbook Algebra structure and method book one
\n" ); document.write( ": Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dimensions of the new rectangle. \n" ); document.write( "
Algebra.Com's Answer #125964 by stanbon(75887)\"\" \"About 
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Originally the dimensions of a rectangle were 20 cm by 23 cm. When both dimensions were decreased by the same amount, the area of the rectangle decreased by 120 cm^2. Find the dimensions of the new rectangle.
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\n" ); document.write( "Original DATA:
\n" ); document.write( "area = 20*23 = 460 cm^2
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\n" ); document.write( "New DATA:
\n" ); document.write( "area = (20-x)(23-x) cm^2
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\n" ); document.write( "EQUATION:
\n" ); document.write( "original area - new area = 120 cm^2
\n" ); document.write( "460 -[460-43x+x^2] = 120
\n" ); document.write( "460 - 460 +43x - x^2 = 120
\n" ); document.write( "x^2 - 43x + 120 = 0\r
\n" ); document.write( "\n" ); document.write( "x = [43 +- sqrt(43^2 -4*120)}/2\r
\n" ); document.write( "\n" ); document.write( "x = [43 +- sqrt(1369)]/2\r
\n" ); document.write( "\n" ); document.write( "x = [43 +- 37]/2
\n" ); document.write( "Positive solution:\r
\n" ); document.write( "\n" ); document.write( "x = 3 cm or x = 40 cm
\n" ); document.write( "Only x = 3 cm is realistic.
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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