document.write( "Question 769978: The width of a rectangle is 1cm more than half the length. When both the length and the width are increased by 1cm, the area increases by 20cm^2. Find the original dimensions of the rectangle. \n" ); document.write( "
Algebra.Com's Answer #469173 by stanbon(75887)\"\" \"About 
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The width of a rectangle is 1cm more than half the length. When both the length and the width are increased by 1cm, the area increases by 20cm^2. Find the original dimensions of the rectangle.
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\n" ); document.write( "Original DATA:
\n" ); document.write( "length = L
\n" ); document.write( "width = (1/2)L+1
\n" ); document.write( "Area = (1/2)L^2 + L
\n" ); document.write( "------------------------
\n" ); document.write( "New DATA:
\n" ); document.write( "Length = L+1
\n" ); document.write( "width = (1/2)L+2
\n" ); document.write( "Area = (L+1)[(1/2)L+2]
\n" ); document.write( "----
\n" ); document.write( "Equation:
\n" ); document.write( "New area - Old area = 20 cm^2
\n" ); document.write( "(L+1)((1/2)L+2) - [(1/2)L^2+L] = 20
\n" ); document.write( "-----
\n" ); document.write( "(1/2)L^2 + (5/2)L +2 - (1/2)L^2 - L = 20
\n" ); document.write( "-------
\n" ); document.write( "(7/2)L = 18
\n" ); document.write( "--
\n" ); document.write( "L = 36/7 cm (original Length)
\n" ); document.write( "Original width = (1/2)L + 1 = (18/7)+(7/7) = 25/7 (original width)
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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