document.write( "Question 206775This question is from textbook Elementary and intermediate algebra
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document.write( ":  The Golden Ratio. The ancient Greeks thought that the most pleasing shape for a rectangel was one for which the ratio of the length to the width was approximately 8 to 5, the golden ratio. If the length of a rectangle painting is 2 feet longer than its width, then for what dimensions would the length and the width have to the golden ratio? Ok, the ratio is L=8 to W=5. However, the length is 2 feet more, so is the L=8+2 and W=5. Would the golden ratio be 10 to 5? I am lost.   \n" );
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| Algebra.Com's Answer #156306 by stanbon(75887)      You can put this solution on YOUR website! The ancient Greeks thought that the most pleasing shape for a rectangel was one for which the ratio of the length to the width was approximately 8 to 5, the golden ratio. \n" ); document.write( "If the length of a rectangle painting is 2 feet longer than its width, then for what dimensions would the length and the width have to the golden ratio? \n" ); document.write( "Ok, the ratio is L=8 to W=5. However, the length is 2 feet more, so is the L=8+2 and W=5. Would the golden ratio be 10 to 5? \n" ); document.write( "--- \n" ); document.write( "Let the width be \"x\" ft. \n" ); document.write( "Then the length is \"x+2\" ft \n" ); document.write( "------------------- \n" ); document.write( "Proportion: \n" ); document.write( "(x+2)/x = 8/5 \n" ); document.write( "8x = 5(x+2) \n" ); document.write( "8x = 5x + 10 \n" ); document.write( "3x = 10 \n" ); document.write( "x = 3 1/3 ft (this is the width) \n" ); document.write( "x+2 = 5 1/3 ft (this is the length) \n" ); document.write( "======================================= \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " \n" ); document.write( " |