document.write( "Question 983972: Rectangle of perimeter 32 cm. if its length decreases 1 cm & its width increases 3 cm, it will be square. find area of square
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Algebra.Com's Answer #604764 by KMST(5347) $\"\"$ $\"About$

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For a rectangle, $\"perimeter=2%28length%2Bwidth%29\"$ ,
\n" ); document.write( "and for a square both measurements are the same $\"side\"$ .
\n" ); document.write( "So, in this case, for the original rectangle
\n" ); document.write( " $\"2%28length%2Bwidth%29=32cm\"$ <---> $\"length%2Bwidth=16cm\"$ .
\n" ); document.write( "If $\"length\"$ decreases by $\"1cm\"$ , and
\n" ); document.write( "if $\"width\"$ increases by $\"3cm\"$ , we get
\n" ); document.write( " $\"length-1cm=side\"$ of a square, and
\n" ); document.write( " $\"width%2B3cm=side\"$ of the same square.
\n" ); document.write( "That means that for the rectangle
\n" ); document.write( " $\"length-1cm=width%2B3cm\"$ <---> $\"length=width%2B4cm\"$ .
\n" ); document.write( "From $\"system%28length%2Bwidth=16cm%2Clength=width%2B4cm%29\"$ , we get
\n" ); document.write( " $\"%28width%2B4cm%29%2Bwidth=16cm\"$ <---> $\"2%28width%29%2B4cm=16cm\"$ <---> $\"2%28width%29=16cm-4cm\"$ <---> $\"2%28width%29=12cm\"$ <---> $\"width=12cm%2F2\"$ <---> $\"width=6cm\"$ .
\n" ); document.write( "I could now find the $\"length\"$ for the rectangle, but I don't need that.
\n" ); document.write( "I know that if $\"width=6cm\"$ for the rectangle, then for the square,
\n" ); document.write( " $\"side=6cm%2B3cm\"$ ---> $\"side=9cm\"$ ,
\n" ); document.write( "and that makes the area of the square
\n" ); document.write( " $\"%289cm%29%289cm%29=81\"$ $\"cm%5E2\"$ .
\n" ); document.write( "The area of the square is $\"highlight%2881%29\"$ square centimeters. \n" ); document.write( "

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