document.write( "Question 92180: Hello I am not sure how to go about this. If the sides of a square are decreased by 3cm, the area is decreased by 81cm^2. What were the dimensions of the original square. I am thinking (x-3)^2=a-81cm^3, but where do I go from here? \n" ); document.write( "
Algebra.Com's Answer #67019 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Well, your initial idea is right on!
\n" ); document.write( "\"%28x-3%29%5E2+=+A-81\" Where A = original area and x = the original length of the side of the square.
\n" ); document.write( "The next step is to recall that \"A+=+x%5E2\", that is; the original area (A) is equal to the original sides squared \"%28x%5E2%29\", so substitute the A with \"x%5E2\" in the first equation then solve for x.
\n" ); document.write( "\"%28x-3%29%5E2+=+x%5E2-81\" Expand the left side.
\n" ); document.write( "\"x%5E2-6x%2B9+=+x%5E2-81\" Subtract \"x%5E2\" from both sides.
\n" ); document.write( "\"-6x%2B9+=+-81\" Subtract 9 from both sides.
\n" ); document.write( "\"-6x+=+-90\" Finally, divide both sides by -6.
\n" ); document.write( "\"x+=+15\"cm.\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "\"%2815-3%292+=+15%5E2-81\"
\n" ); document.write( "\"12%5E2+=+225-81\"
\n" ); document.write( "\"144+=+144\"
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