document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=203382>Question 203382</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the area of an inscribed square whose radius is 8 inches? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #153453 by </B><A HREF=/tutors/aboutme.mpl?userid=jsmallt9><B>jsmallt9(3758)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jsmallt9><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=153453\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Better wording would be: What is the area of a square inscribed in a circle whose radius is 8 inches? If this is your problem then , the picture looks like:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28400%2C+400%2C+-2%2C+2%2C+-2%2C+2%2C+circle%280%2C+0%2C+sqrt%282%29%29%2C+line%28-1%2C+-1%2C+1%2C+-1%29%2C+line%281%2C+-1%2C+1%2C+1%29%2C+line%281%2C+1%2C+-1%2C+1%29%2C+line%28-1%2C+1%2C+-1%2C+-1%29%2C+line%280%2C+0%2C+1%2C+1%29%2C+line%280%2C+0%2C+1%2C+0%29%2C+locate%28.4%2C+.7%2C+8%29%2C+locate%28.2%2C+.2%2C+45%29%2C+locate%281.1%2C+.5%2C+x%29%2C+locate%28-.9%2C+0%2C+2x%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Since the area of a square is the square (hence the term) of the side, we need to find the side of the square. As I hope you can see, finding the leg of the right triangle will help us find the side of the square.<br><BR>\n" );
document.write( "In all 45-45-90 triangles the ratio of the hypotenuse to a leg is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Sqrt%282%29 ALIGN=MIDDLE  ALT=\"Sqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >. In the 45-45-90 right triangle shown, the hypotenuse is 8.  Substituting this into the ratio we can find the leg (using x for leg because \"l\" can be confused with \"1\"): <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%2Fx+=+sqrt%282%29 ALIGN=MIDDLE  ALT=\"8%2Fx+=+sqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Multiplying both sides by x<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8+=+x%2Asqrt%282%29 ALIGN=MIDDLE  ALT=\"8+=+x%2Asqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Divide both sides by sqrt(2):<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8%2Fsqrt%282%29+=+x ALIGN=MIDDLE  ALT=\"8%2Fsqrt%282%29+=+x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Now the leg, x, is 1/2 of the side of the square, s. So the side of the square is twice as much:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s+=+2%288%2Fsqrt%282%29%29+=+16%2Fsqrt%282%29 ALIGN=MIDDLE  ALT=\"s+=+2%288%2Fsqrt%282%29%29+=+16%2Fsqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+s%5E2+=+%2816%2Fsqrt%282%29%29%5E2+=+256%2F2+=+128 ALIGN=MIDDLE  ALT=\"A+=+s%5E2+=+%2816%2Fsqrt%282%29%29%5E2+=+256%2F2+=+128\"   DISABLED_style= \"max-width:90%; height:auto;\" > square inches. </SPAN>\n" );
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