document.write( "Question 971976: find the area of the largest trapezoid that can be inscribed a semicircle of diameter \"d\" in terms of \"d\"?. what is the length of the bottom base \"x\" of the same trapezoid ?? \n" ); document.write( "
Algebra.Com's Answer #594442 by KMST(5328)\"\" \"About 
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In the American definition, a trapezoid has two parallel sides .
\n" ); document.write( "I am assuming that the drawing above shows what was meant: a trapezoid inscribed in a semicircle.
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\n" ); document.write( "Intuitively, we could assume that the largest polygon that can be inscribed in a circle is the regular polygon of its kind,
\n" ); document.write( "so that the largest inscribed triangle would be an equilateral triangle,
\n" ); document.write( "the largest quadrilateral would be a square,
\n" ); document.write( "the largest pentagon a regular pentagon,
\n" ); document.write( "the largest hexagon a regular hexagon, and so on.
\n" ); document.write( "If that is so, the largest trapezoid inscribed in a semicircle may be half of a hexagon.
\n" ); document.write( "That trapezoid would have base angles measuring \"60%5Eo\",
\n" ); document.write( "and could be thought of as made of three equilateral triangles: .
\n" ); document.write( "In a semicircle of radius \"r=d%2F2\" ,the area of that trapezoid is
\n" ); document.write( " .
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\n" ); document.write( "There may be a very simple proof supporting that maximum area value,
\n" ); document.write( "but I was taught a lot of math past the fifth grade, and that spoiled me forever,
\n" ); document.write( "so I can only offer a complicated proof or two.
\n" ); document.write( "I just cannot think of a proof that does not involve calculus.
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\n" ); document.write( "Using \"green%28x%29\" :
\n" ); document.write( "\"h=sqrt%28r%5E2-x%5E2%29\"
\n" ); document.write( "\"area=A=%282r%2B2x%29sqrt%28r%5E2-x%5E2%29%2F2=%28r%2Bx%29sqrt%28r%5E2-x%5E2%29\"
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\n" ); document.write( "\"dA%2Fdx=sqrt%28r%5E2-x%5E2%29%2B%28r%2Bx%29%2Ax%2Fsqrt%28r%5E2-x%5E2%29%29\"
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\n" ); document.write( "\"dA%2Fdx=%28r%5E2-x%5E2-rx-x%5E2%29%2Fsqrt%28r%5E2-x%5E2%29%29\"
\n" ); document.write( "\"dA%2Fdx=%28-2x%5E2-rx%2Br%5E2%29%2Fsqrt%28r%5E2-x%5E2%29%29\"
\n" ); document.write( "\"dA%2Fdx%3E0\" between both roots of \"-2x%5E2-rx%2Br%5E2=0\"<--->\"2x%5E2%2Brx-r%5E2=0\"
\n" ); document.write( "Solving that quadratic equation we realize that the solutions are
\n" ); document.write( "\"x=-r\" (which, making \"x%3C0\" does not make sense, and
\n" ); document.write( "\"x=r%2F2=d%2F4\" , which makes sense and makes \"dA%2Fdx=0\" ,
\n" ); document.write( "marking the maximum for \"A=%28r%2Bx%29sqrt%28r%5E2-x%5E2%29\" .
\n" ); document.write( "So, that maximum is
\n" ); document.write( "\"area%5Bmax%5D=%28r%2Br%2F2%29sqrt%28r%5E2-%28r%2F2%29%5E2%29\"
\n" ); document.write( "\"area%5Bmax%5D=%283r%2F2%29sqrt%28r%5E2-r%5E2%2F4%29\"
\n" ); document.write( "\"area%5Bmax%5D=%283r%2F2%29sqrt%283r%5E2%2F4%29\"
\n" ); document.write( "\"area%5Bmax%5D=%283r%2F2%29sqrt%283%29%28r%2F2%29%29\"
\n" ); document.write( "\"area%5Bmax%5D=3sqrt%283%29%28r%2F2%29%5E2%29\"
\n" ); document.write( "\"area%5Bmax%5D=3sqrt%283%29%28r%2F4%29%5E2%29\"
\n" ); document.write( "\"area%5Bmax%5D=highlight%283sqrt%283%29d%5E2%2F16%29\"
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\n" ); document.write( "Using \"red%28alpha%29\" :
\n" ); document.write( "The trapezoid is made of three isosceles triangles.
\n" ); document.write( "They have legs of length \"r\" ,
\n" ); document.write( "and vertex angles measuring \"red%28alpha%29\" , \"red%28180%5Eo-2alpha%29\" , and \"red%28alpha%29\" .
\n" ); document.write( "The area of the trapezoid is the sum of the triangles' areas, so it is
\n" ); document.write( "\"area=A=r%5E2sin%28alpha%29%2F2%2Br%5E2sin%28180%5Eo-2alpha%29%2F2%2Br%5E2sin%28alpha%29%2F2\"
\n" ); document.write( "\"A=r%5E2sin%28alpha%29%2B%28r%5E2%2F2%29sin%28180%5Eo-2alpha%29\"
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\n" ); document.write( "\"dA%2Fd%28alpha%29=-r%5E2cos%28alpha%29%2B%28r%5E2%2F2%29cos%28180%5Eo-2alpha%29\"
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\n" ); document.write( "\"dA%2Fd%28alpha%29=-r%5E2cos%28alpha%29-%28r%5E2%2F2%29cos%282alpha%29\"
\n" ); document.write( "\"dA%2Fd%28alpha%29=-r%5E2cos%28alpha%29-%28r%5E2%2F2%29%282cos%5E2%28alpha%29-1%29\"
\n" ); document.write( "\"dA%2Fd%28alpha%29=-%28r%5E2%2F2%29%282cos%28alpha%29%2B2cos%5E2%28alpha%29-1%29\"
\n" ); document.write( "\"dA%2Fd%28alpha%29=0\" for \"2cos%28alpha%29%2B2cos%5E2%28alpha%29-1=0\"
\n" ); document.write( "and \"dA%2Fd%28alpha%29%3E=0\" between both roots,
\n" ); document.write( "so \"A%5Bmax%5D\" happens for the greatest root of \"2cos%28alpha%29%2B2cos%5E2%28alpha%29-1=0\" .
\n" ); document.write( "Making \"cos%28alpha%29=u\" for short, the equation is
\n" ); document.write( "\"2u%5E2%2B2u-1=0\" and the roots are
\n" ); document.write( "\"u+=+%28-1+%2B-+sqrt%281%5E2-4%2A2%2A%28-1%29%29%29%2F%282%2A2%29+\"
\n" ); document.write( "\"u+=%28-1+%2B-+sqrt%281%2B8%29%29%2F4=%28-1+%2B-+sqrt%289%29%29%2F4=%28-1+%2B-+3%29%2F4\"
\n" ); document.write( "The roots are \"u%5B1%5D=%28-1-3%29%2F4=%28-4%29%2F4=-1%3C%28-1%2B3%29%2F4=2%2F4=1%2F2\" ,
\n" ); document.write( "so \"area\" increases with \"cos%28alpha%29\" ,
\n" ); document.write( "between \"cos%28alpha%29=0\"<-->\"alpha=90%5Eo\" , and
\n" ); document.write( "\"cos%28alpha%29=1%2F2\"<-->\"alpha=60%5Eo\" ,
\n" ); document.write( "and the maximum is at \"cos%28alpha%29=1%2F2\"<-->\"alpha=60%5Eo\" , where
\n" ); document.write( "\"area%5Bmax%5D=r%5E2sin%28alpha%29%2Br%5E2sin%28180%5Eo-2alpha%29%2F2\"
\n" ); document.write( "\"area%5Bmax%5D=r%5E2sin%2860%5Eo%29%2Br%5E2sin%28180%5Eo-2%2A60%5Eo%29%2F2\"
\n" ); document.write( "\"area%5Bmax%5D=r%5E2sin%2860%5Eo%29%2Br%5E2sin%2860%5Eo%29%2F2\"
\n" ); document.write( "\"area%5Bmax%5D=%283r%5E2%2F2%29sin%2860%5Eo%29\"
\n" ); document.write( "\"area%5Bmax%5D=%283r%5E2%2F2%29%28sqrt%283%29%2F2%29\"
\n" ); document.write( "\"area%5Bmax%5D=%283%2F4%29r%5E2sqrt%283%29%7D%7D+%2C+and+since+%7B%7B%7Br=d%2F2\" ,
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