document.write( "Question 243566: The perimeter of a rectangle is 800 feet, and length of it is 50 feet less than half the width. What are it's dimensions? \n" ); document.write( "

Algebra.Com's Answer #178424 by oberobic(2304) $\"\"$ $\"About$

You can put this solution on YOUR website!
CORRECTION! I read the problem too fast the first time. (sorry)
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\n" ); document.write( "The perimeter of a rectangle = 2L + 2W, where L=length and W=width.
\n" ); document.write( "We are told L is 50 ft less than HALF the W, so L = .5W -50.
\n" ); document.write( "We are told the perimeter = 800.
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\n" ); document.write( "Substituting we have:
\n" ); document.write( "800 = 2L + 2W
\n" ); document.write( "800 = 2(.5W-50) + 2W
\n" ); document.write( "Multiplying the factor
\n" ); document.write( "800 = W - 100 + 2W
\n" ); document.write( "Collecting like terms
\n" ); document.write( "800 = 3W - 100
\n" ); document.write( "Adding 100 to both sides
\n" ); document.write( "900 = 3W
\n" ); document.write( "3W = 900
\n" ); document.write( "Dividing both sides by 3
\n" ); document.write( "W = 900/3 = 300
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\n" ); document.write( "Looking back, we know L = .5W-50, so L = 150 - 50 = 100.
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\n" ); document.write( "Check the perimeter.
\n" ); document.write( "2W + 2L = 2(300) + 2(100) = 800.
\n" ); document.write( "This is what it's supposed to be.
\n" ); document.write( "Done. \n" ); document.write( "

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