document.write( "Question 58533: A rectangle is twice as long as it is wide. If the length and width are each decreased by 4cm, the area is reduced by224cm squared. Find the dimensions of the original rectangle. \n" ); document.write( "
Algebra.Com's Answer #40075 by ankor@dixie-net.com(22740)\"\" \"About 
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A rectangle is twice as long as it is wide.
\n" ); document.write( "x = width; 2x = length
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\n" ); document.write( "If the length and width are each decreased by 4cm,the area is reduced by 224 cm squared.
\n" ); document.write( "Original area = 2x^2
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\n" ); document.write( "New dimensions: (x-4; (2x-4); new area given as 2x^2 - 224
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\n" ); document.write( "2x^2 - 224 = (x-4)*(2x-4)
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\n" ); document.write( "2x^2 - 224 = 2x^2 - 12x + 16
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\n" ); document.write( "2x^2 - 2x^2 + 12x = 16 + 224
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\n" ); document.write( "12x = 240
\n" ); document.write( "x = 240/12
\n" ); document.write( "x = 20
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\n" ); document.write( " Find the dimensions of the original rectangle.
\n" ); document.write( "x = 20cm is the width; 40 cm is the length
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\n" ); document.write( "Check:
\n" ); document.write( "20*40 = 800 sq cm
\n" ); document.write( "16*36 = 576 sq cm
\n" ); document.write( "---------------------subtract
\n" ); document.write( ">>>>>>>>>224 sq cm\r
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