document.write( "Question 83589: Help please.\r
\n" ); document.write( "\n" ); document.write( "If the sides of a square are decreased by 2cm, the area is decreased by 36cm^2. What were the dimensions of the original square?
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Algebra.Com's Answer #60105 by Earlsdon(6294)\"\" \"About 
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Let x be the original dimension (side) of the square.
\n" ); document.write( "The original area would then be: \"A+=+x%5E2\"
\n" ); document.write( "Now if you decrease the original side by 2cm, you will have (x-2) and the new area can be found by \"%28x-2%29%5E2\", which is given in the problem as \"x%5E2-36\".
\n" ); document.write( "Now you can set up the equation to solve for x:
\n" ); document.write( "\"%28x-2%29%5E2+=+x%5E2-36\" Simplify and solve for x.
\n" ); document.write( "\"x%5E2-4x%2B4+=+x%5E2-36\" Subtract \"x%5E2\"from both sides.
\n" ); document.write( "\"-4x%2B4+=+-36\"Subtract 4 from both sides.
\n" ); document.write( "\"-4x+=+-40\" Divide both sides by -4.
\n" ); document.write( "\"x+=+10\" The dimensions of the original square were 10cm by 10cm.\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "The original area was:
\n" ); document.write( "\"A%5B0%5D+=+%2810%29%2810%29\"
\n" ); document.write( "\"A%5B0%5D+=+100\" sq.cm.
\n" ); document.write( "The new area is:
\n" ); document.write( "\"A%5B1%5D+=+%2810-2%29%2A%2810-2%29\"
\n" ); document.write( "\"A%5B1%5D+=+8+X+8\"
\n" ); document.write( "\"A%5B1%5D+=+64\"sq.cm.
\n" ); document.write( "\"A%5B0%5D-A%5B1%5D+=+100-64\" = 36sq.cm.
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