document.write( "Question 192009This question is from textbook Elementary and intermediate algebra
\n" ); document.write( ": The ancient Greeks thought that the most pleasing shape for a rectangle was one for which has the ratio of the length to the width was approximately 8 to 5, the golden ratio. If the length of a rectangular painting is 2 ft longer than its width, then for what dimensions would the length and width have the golden ratio?\r
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Algebra.Com's Answer #144401 by jojo14344(1513)\"\" \"About 
You can put this solution on YOUR website!

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\n" ); document.write( "Golden Ratio: \"L%2FW=8%2F5\"\r
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\n" ); document.write( "\n" ); document.write( "And the painting, \"L=+W%2B2ft\", Length is 2 ft longer than Width.
\n" ); document.write( "Substitute this \"L\" to our Golden Ratio:\r
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\n" ); document.write( "\n" ); document.write( "\"highlight%28W%2B2ft%29%2FW=8%2F5\"\r
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\n" ); document.write( "\n" ); document.write( "Cross multiply,
\n" ); document.write( "\"%285%29%28W%2B2%29=%288%29%28W%29\"
\n" ); document.write( "\"5W%2B10=8W\"
\n" ); document.write( "\"10=8W-5W\"
\n" ); document.write( "\"10=3W\" -----> \"10%2F3=cross%283%29W%2Fcross%283%29\"
\n" ); document.write( "\"red%28W=%2810%2F3%29ft%29\"}Width\r
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\n" ); document.write( "\n" ); document.write( "So, \"L=W%2B2=10%2F3%2B2=%2810%2B6%29%2F3\"
\n" ); document.write( "\"red%28L=%2816%2F3%29ft%29\"} Length\r
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\n" ); document.write( "\n" ); document.write( "Let us check on our Golden Ratio:
\n" ); document.write( "\"8%2F5=L%2FW=%2816%2F3%29%2F%2810%2F3%29\"
\n" ); document.write( "\"8%2F5=%2816%2Fcross%283%29%29%28cross%283%29%2F10%29\"
\n" ); document.write( "\"8%2F5=16%2F10\"
\n" ); document.write( "Reduce left term by dividing both num.&den. by \"2\":
\n" ); document.write( "\"8%2F5=cross%2816%298%2Fcross%2810%295\"
\n" ); document.write( "\"highlight%288%2F5=8%2F5%29\"\r
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\n" ); document.write( "\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo

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