document.write( "Question 643879: The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
\n" ); document.write( "I set it up like this: 216cm^2= w+ 6cm(w)= 216cm^2= w^2+ 6cmw= 216cm^2- w^2-6cmw=0
\n" ); document.write( "and that's when i just got stuck I'm not sure if i was doing it right in the first place. Can I please have a bit of guidance?
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Algebra.Com's Answer #404702 by MathTherapy(10555)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "The length of a rectangle is 6cm more than its width. The area is 216 sq cm. What are the dimensions? I have w= width, l= w+ 6cm = length A= 216cm^2
\n" ); document.write( "I set it up like this: 216cm^2= w+ 6cm(w)= 216cm^2= w^2+ 6cmw= 216cm^2- w^2-6cmw=0
\n" ); document.write( "and that's when i just got stuck I'm not sure if i was doing it right in the first place. Can I please have a bit of guidance?\r
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\n" ); document.write( "\n" ); document.write( "Let width be W
\n" ); document.write( "Then length = W + 6
\n" ); document.write( "A, or area = LW
\n" ); document.write( "Since area = 216, then: 216 = W(W + 6)\r
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\n" ); document.write( "\n" ); document.write( "\"216+=+W%5E2+%2B+6W\"\r
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\n" ); document.write( "\n" ); document.write( "\"W%5E2+%2B+6W+-+216+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "(W - 12) (W + 18) = 0 \r
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\n" ); document.write( "\n" ); document.write( "W - 12 = 0, or W = - 18 (ignore as measurement CANNOT be negative)\r
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\n" ); document.write( "\n" ); document.write( "W, or width = \"highlight_green%2812%29\"\r
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\n" ); document.write( "\n" ); document.write( "Length = 12 + 6, or \"highlight_green%2818%29\" cm\r
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\n" ); document.write( "\n" ); document.write( "Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
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