document.write( "Question 1011941: Find the side of a regular octagon inscribed in a circle of radius 19 cm.
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Algebra.Com's Answer #627773 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
Think TRIANGLES. Triangles are the key to solving many problems.
\n" ); document.write( "They are at the heart of geometry, trigonometry and engineering.
\n" ); document.write( "Or think TRIGONOMETRY, if you wish.
\n" ); document.write( "Use LAW OF COSINES, if that is what you were taught recently.
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\n" ); document.write( "Here is the circle, with its center, the inscribed regular octagon,
\n" ); document.write( "and some line segments that are at the same time
\n" ); document.write( "diameters of the circle and diagonals of the octagon.
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\n" ); document.write( "The diagonals connecting opposite octagon vertices go through the center of the circle,
\n" ); document.write( "and split the octagon into \"8\" isosceles triangles.
\n" ); document.write( "Each of those triangles has two sides measuring \"19cm\"
\n" ); document.write( "(the radius of the circle), and an angle measuring
\n" ); document.write( "\"360%5Eo%2F8=45%5Eo\" (or \"2pi%2F8=pi%2F4\" if you prefer radians)
\n" ); document.write( "at the center of the circle.
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\n" ); document.write( "USING TRIANGLES:
\n" ); document.write( "Let's look at one of those triangles, and use the Pythagorean theorem
\n" ); document.write( " \"x%5E2%2Bx%5E2=19%5E2\"--->\"2x%5E2=19%5E2\"--->\"x%5E2=19%5E2%2F2\"--->\"x=19%2Fsqrt%282%29\" for the larger right triangle.
\n" ); document.write( "For the smaller right triangle, \"x%5E2%2B%2819-x%29%5E2=side%5E2\" .
\n" ); document.write( "We can substitute \"19%2Fsqrt%282%29\" for \"x\" (and \"19%5E2%2F2\" for \"x%5E2\" if you wish)
\n" ); document.write( "sooner or later, as you wish.
\n" ); document.write( "\"x%5E2%2B%2819-x%29%5E2=side%5E2\"<-->\"x%5E2%2B19%5E2-2%2A19x%2Bx%5E2=side%5E2\"<-->\"2x%5E2%2B19%5E2-2%2A19x=side%5E2\"<-->\"2%28x%5E2-19x%29%2B19%5E2=side%5E2\"
\n" ); document.write( "\"2%2819%5E2%2F2-19%2A19%2Fsqrt%282%29%29%2B19%5E2=side%5E2\"
\n" ); document.write( "\"19%5E2-19%5E2sqrt%282%29%2B19%5E2=side%5E2\"
\n" ); document.write( "\"2%2A19%5E2-19%5E2sqrt%282%29=side%5E2\"
\n" ); document.write( "\"19%5E2%282-sqrt%282%29%29=side%5E2\"--->\"side=sqrt%2819%5E2%282-sqrt%282%29%29%29\"--->\"highlight%28side=19sqrt%282-sqrt%282%29%29%29\" or for an approximate measure \"highlight%28side=about14.54cm%29\"(rounded)
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\n" ); document.write( "USING TRIGONOMETRY:
\n" ); document.write( "We can put one of those triangles on a set of coordinates to find the distance between the vertices of the octagon.
\n" ); document.write( "Here are two similar triangles:
\n" ); document.write( " \"B%28cos%2845%5Eo%29%2Csin%2845%5Eo%29%29\"=\"B%28sqrt%282%29%2F2%2Csqrt%282%29%2F2%29\" , and \"A%281%2C0%29\" ,
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\n" ); document.write( "\n" ); document.write( " \"CD=highlight%2819sqrt%282-sqrt%282%29%29%29\" is the length of the side of the octagon, in cm.
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\n" ); document.write( "USING LAW OF COSINES:
\n" ); document.write( " Law of cosine says that \"a%5E2=b%5E2%2Bc%5E2-2bc%2Acos%28A%29\" , so with length measured in cm,
\n" ); document.write( "\"a%5E2=19%5E2%2B19%5E2-2%2A19%2A19%2Acos%2845%5Eo%29\"
\n" ); document.write( "\"a%5E2=2%2A19%5E2-2%2A19%5E2%2A%28sqrt%282%29%2F2%29\"
\n" ); document.write( "\"a%5E2=2%2A19%5E2-19%5E2%2Asqrt%282%29\"
\n" ); document.write( "\"a%5E2=19%5E2%282-sqrt%282%29%29\"
\n" ); document.write( "\"a=sqrt%2819%5E2%282-sqrt%282%29%29%29\" ,
\n" ); document.write( "and \"a=highlight%2819sqrt%282-sqrt%282%29%29%29\" is the length of the side of the octagon, in cm.
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