document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=731997>Question 731997</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the ratio of the area of the small square to the area of the large square\r<BR>\n" );
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document.write( "http://tinypic.com/r/c8ade/6 (the diagram)  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #447545 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=447545\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28300%2C300%2C-5%2C5%2C-5%2C5%2C%0D%0Arectangle%28-4%2C-4%2C4%2C4%29%2C%0D%0Acircle%280%2C0%2C4%29%2C%0D%0Arectangle%28-2sqrt%282%29%2C-2sqrt%282%29%2C2sqrt%282%29%2C2sqrt%282%29%29%2C%0D%0Aline%28-4%2C4%2C4%2C-4%29%2C%0D%0Aline%282sqrt%282%29%2C-2sqrt%282%29%2C-2sqrt%282%29%2C2sqrt%282%29%29%2C%0D%0Aline%28-4%2C0%2C4%2C0%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" > If the measure of the side of the small square is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a ALIGN=MIDDLE  ALT=\"a\"   DISABLED_style= \"max-width:90%; height:auto;\" >, its area is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2 ALIGN=MIDDLE  ALT=\"a%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > .<BR>\n" );
document.write( "The Pythagorean theorem says that the diagonal of the small square is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2Asqrt%282%29=sqrt%28a%5E2%2Ba%5E2%29=sqrt%282a%5E2%29 ALIGN=MIDDLE  ALT=\"a%2Asqrt%282%29=sqrt%28a%5E2%2Ba%5E2%29=sqrt%282a%5E2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "Since that is the diameter of the circle, and the side of the large square,<BR>\n" );
document.write( "the area of the large square is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28a%2Asqrt%282%29%29%5E2=2a%5E2 ALIGN=MIDDLE  ALT=\"%28a%2Asqrt%282%29%29%5E2=2a%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Then, the ratio of the areas is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2a%5E2%2Fa%5E2=highlight%282%29 ALIGN=MIDDLE  ALT=\"2a%5E2%2Fa%5E2=highlight%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > . </SPAN>\n" );
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