document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=124499>Question 124499</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A square of side length 10cm is inscribed in a circle. Calculate the diameter of the circle </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #91244 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=91244\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It might help you to visualize this problem if you made a rough sketch of it. Draw a circle and<BR>\n" );
document.write( "then, on the inside of the circle draw a square that has each of its corners on the circle.<BR>\n" );
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document.write( "Once you do that, draw a diagonal line that connects one corner of the square with the other<BR>\n" );
document.write( "corner on the opposite side.  This diagonal line will be a diameter of the circle.<BR>\n" );
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document.write( "Notice that this diagonal line also divides the square into two right triangles. You can see<BR>\n" );
document.write( "that the two legs of each triangle are sides of the square. Therefore, each leg is 10 cm in length.<BR>\n" );
document.write( "You then need to find the hypotenuse (long side) of the triangle and when you do, you will have<BR>\n" );
document.write( "the length of the diameter of the circle.<BR>\n" );
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document.write( "You can find the length of the hypotenuse by applying the Pythagorean theorem which says that<BR>\n" );
document.write( "the sum of the squares of the two legs of a right triangle is equal to the square of the<BR>\n" );
document.write( "hypotenuse. In equation form this is:<BR>\n" );
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document.write( "A^2 + B^2 = C^2<BR>\n" );
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document.write( "where A and B are the two legs and C is the hypotenuse.<BR>\n" );
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document.write( "For this problem, A and B are both 10 cm. So, substituting 10 for A and 10 for B makes the<BR>\n" );
document.write( "equation:<BR>\n" );
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document.write( "10^2 + 10^2 = C^2<BR>\n" );
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document.write( "and since 10^2 is equal to 100 you can reduce the equation to:<BR>\n" );
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document.write( "100 + 100 = C^2<BR>\n" );
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document.write( "Add the two numbers on the left side and you have:<BR>\n" );
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document.write( "200 = C^2<BR>\n" );
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document.write( "You can solve for C by taking the square root of both sides to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28200%29+=+C ALIGN=MIDDLE  ALT=\"sqrt%28200%29+=+C\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Just to get this in a conventional form, let's reverse the sides of the equation to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+sqrt%28200%29 ALIGN=MIDDLE  ALT=\"C+=+sqrt%28200%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "But 200 can be replaced by 100*2 to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+sqrt%28100%2A2%29 ALIGN=MIDDLE  ALT=\"C+=+sqrt%28100%2A2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "and by the rules of square roots this becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+sqrt%28100%2A2%29+=+sqrt%28100%29%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"C+=+sqrt%28100%2A2%29+=+sqrt%28100%29%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "and since the square root of 100 is 10, this further reduces to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+10%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"C+=+10%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The units being used are cm, so the answer is that C, the diagonal of the<BR>\n" );
document.write( "circle, is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=C+=+10%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"C+=+10%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > cm<BR>\n" );
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document.write( "Hope this helps you to understand the problem a little better.<BR>\n" );
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