document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1166972>Question 1166972</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram, triangle ABC is inscribed with diameter AC. Semicircles are constructed having AB and BC as diameters. The total area represented by regions a and b, in square units, is:<BR>\n" );
document.write( "A) 105.125 π<BR>\n" );
document.write( "B) 210<BR>\n" );
document.write( "C) 420<BR>\n" );
document.write( "D) 210.25 π<BR>\n" );
document.write( "E) 105\r<BR>\n" );
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document.write( "https://imageshack.com/i/pmYzXW65j\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #791531 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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/cgi-bin/show-face.mpl?user=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=791531\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "The first things you need are the three simplest pieces of this puzzle, the radius of the large circle, and the area of the two smaller semi-circles.\r<BR>\n" );
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document.write( "The radius is just half of the diameter that you can find using Pythagoras on the two chord lengths which are the legs of a right triangle with the necessary diameter as the hypotenuse.  The area of either of the semicircles is half of the associated chord length squared times <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \pi\"> and then divide by 2.\r<BR>\n" );
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document.write( "Now that you have the two semicircle areas, you need to find the area of the associated segments of the large circle. For each, you will need the <b><i>radian</i></b> measure of the subtended central angle.  Use:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ 2\arcsin\(\frac{c}{2R}\)\">\r<BR>\n" );
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document.write( "where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large c\"> is the measure of the chord and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large R\"> is the radius.\r<BR>\n" );
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document.write( "Once you have the central angle measure, you can calculate the area of the segment using:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{R^2}{2}\(\theta\ -\ \sin\theta\)\">\r<BR>\n" );
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document.write( "Once you have calculated the two segment areas, you can subtract those areas from the corresponding semicircle areas to get the two desired areas which you can then sum to find your final answer.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi}\ +\ 1\ =\ 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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document.write( "I > Ø<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \\r\"> </SPAN>\n" );
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