document.write( "Question 123994: Here is the problem\r
\n" ); document.write( "\n" ); document.write( "The large rectangle shown is a gold rectangle. This means that when a square is cut off, the rectangle that remains is similar to the original rectabgle.\r
\n" ); document.write( "\n" ); document.write( "The ratio of the length to width ina golden rectangle is called the golden ratio. Write the golden ratio in simplified radical form. Then use a calculator to find an approximation to the newest hundreth.\r
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\n" ); document.write( " |.....|........| (Ignore the periods in the rectangle. I had to do
\n" ); document.write( " |.....|........| that so it would look like a box.)
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\n" ); document.write( " ^------1-------^ < The measure of that whole bottom side is 1\r
\n" ); document.write( "\n" ); document.write( "So far I have concluded that x=1. As of what to do next I am completely on sure. My teacher gave me a hint to use the quadratic formula... but I still dont know what to do with it. Please help! THANK YOU!!!
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Algebra.Com's Answer #90953 by solver91311(24713) $\"\"$ $\"About$

You can put this solution on YOUR website!


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document.write( "To construct a golden rectangle, begin with a square (ABCD in the diagram) with
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document.write( "sides that measure 1 unit.\r
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document.write( "Extend one side of the square (red line in the diagram)\r
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document.write( "Bisect the side of the square that was extended (point E)\r
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document.write( "Strike an arc with radius DE to intersect the extended side of the square
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document.write( "(point F)\r
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document.write( "Construct the perpendicular to the extended side at F.\r
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document.write( "Extend the opposite side of the square to intersect the perpendicular at F.
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document.write( "(point G).\r
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document.write( "The resulting quadrilateral (ABFG) is a golden rectangle.\r
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document.write( "The dimensions of the golden rectangle are the measures of AB and BF.  But
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document.write( "BF = one half of BC plus EF.  But BC = AB, and EF = DE\r
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document.write( "The measure of DE can be determined by using the Pythagorean Theorem because
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document.write( "DE is the hypotenuse of a right triangle with sides CD measuring 1 and EC
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document.write( "measuring 1/2.\r
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document.write( "The rectangle that was added to the square is, itself, a golden rectangle
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document.write( "consisting of a square and a rectangle that is a golden rectangle, ad infinitum.
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document.write( "Solution
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document.write( "Let c = the measure of DE, a = the measure of EC, and b = the measure of CD.\r
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document.write( "a = 1
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document.write( "b = 1/2\r
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document.write( "Since the measure of DE = the measure of EF and the measure of BE is 1/2,
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document.write( "the measure of BF (the length of the golden rectangle) is  or \r
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document.write( " is roughly 1.62, so the Golden Ratio is approximately 1.62:1
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