document.write( "Question 286661: The length of a square is increased by 8 and the width is decreased by 4. The area of the resulting is 12 square units more than the area of the square. How do I find the dimensions of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #207903 by mananth(16946) $\"\"$ $\"About$

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Let the length of side be x\r
\n" ); document.write( "\n" ); document.write( "area of square = x^2\r
\n" ); document.write( "\n" ); document.write( "side increased by 8 = x+8\r
\n" ); document.write( "\n" ); document.write( "side decreased by 4 = x-4\r
\n" ); document.write( "\n" ); document.write( "Resulting area = ( x-4)(x+8)\r
\n" ); document.write( "\n" ); document.write( "x^2+12= (x-4)(x+8)\r
\n" ); document.write( "\n" ); document.write( "x^2+12=x^2+8x-4x-32\r
\n" ); document.write( "\n" ); document.write( "x^2+12= x^2+4x-32\r
\n" ); document.write( "\n" ); document.write( "x^2-x^2-4x=-32-12\r
\n" ); document.write( "\n" ); document.write( "-4x=-44\r
\n" ); document.write( "\n" ); document.write( "x=11\r
\n" ); document.write( "\n" ); document.write( "x+8 is the length = 11+8=19\r
\n" ); document.write( "\n" ); document.write( "x-4= width = 11-4=7\r
\n" ); document.write( "\n" ); document.write( "The dimensions are 11 and 7\r
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