document.write( "Question 55097: Can someone help me on this problem. I have gotten two different results and not sure which was is right.\r
\n" ); document.write( "\n" ); document.write( " The length of a rectangle is 2 cm more than twice its width. f the perimeter of the rectangle is 52 cm, find the dimensions of the rectangle.
\n" ); document.write( " Thanks,
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Algebra.Com's Answer #37331 by rchill(405) $\"\"$ $\"About$

You can put this solution on YOUR website!
Let w represent the width, then (2w+2) is the length (\"2 more than twice its width\"). Because a rectangle's perimeter is 2(l+w), your rectangle's perimeter becomes $\"2%28%282w%2B2%29%2Bw%29=52\"$ . Expanding the equation gives us $\"2%283w%2B2%29=52\"$ which further expands to $\"6w%2B4=52\"$ . Solving for w gives us $\"w=8\"$ , which means the length is $\"2%2A8%2B2\"$ which is 18. To prove that the rectangle is 8 by 18, it's perimeter should be 52: 8+8+18+18=52. Yeah! \n" ); document.write( "

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