document.write( "Question 724205: a rectangle has whole number sides.The length of the rectangle is tripled and the width is doubled.the new area is 204 cm squared.the new perimeter is 110 cm.what was the old perimeter? \n" ); document.write( "

Algebra.Com's Answer #443686 by KMST(5328) $\"\"$ $\"About$

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$\"x\"$ = the length of one of the sides of the new, enlarged rectangle
\n" ); document.write( " $\"y\"$ = the length of the other side of the new, enlarged rectangle
\n" ); document.write( "The new area is $\"xy=204\"$ .
\n" ); document.write( "The new perimeter is $\"2%28x%2By%29=100\"$ --> $\"x%2By=55\"$
\n" ); document.write( "I could set up to solve $\"system%28xy=204%2Cx%2By=55%29\"$ , end up with a quadratic equation, and solve it by factoring. The hint that the length of the sides were whole numbers tells me that the factoring approach would work.
\n" ); document.write( "Finding two numbers that add up to 55 and multiply to yield 204 is a shorter way to do that.
\n" ); document.write( "Since $\"51%2B4=55\"$ and $\"51%2A4=204\"$ the new rectangle sides measure $\"51cm\"$ and $\"4cm\"$
\n" ); document.write( "Since $\"51cm=3%2A17cm\"$ and $\"4cm=2%2A2cm\"$ , it is obvious that the original length and width were $\"17cm\"$ and $\"2cm\"$ .
\n" ); document.write( "(The original length and width could not have been $\"4cm%2F3=4%2F3\"$ $\"cm\"$ and $\"17cm%2F2=17%2F2\"$ $\"cm\"$ respectively, because that would gives us a length shorter than the width and no whole number side lengths).
\n" ); document.write( "With a length of $\"17cm\"$ and a width of $\"2cm\"$ , the perimeter of the original rectangle is
\n" ); document.write( " $\"2%2817cm%2B2cm%29=highlight%2838cm%29\"$ \n" ); document.write( "

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