document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1201695>Question 1201695</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi<BR>\n" );
document.write( "A square of area 70cm^2 is inscribed inside a circle . What is the area of the circle<BR>\n" );
document.write( "Thank you  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #836182 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=836182\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "The square has area 70 square cm.<BR>\n" );
document.write( "The side length is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2870%29 ALIGN=MIDDLE  ALT=\"sqrt%2870%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > cm.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%0D%0Adrawing%28400%2C400%2C-4%2C4%2C-4%2C4%2C%0D%0Acircle%280%2C0%2C3%29%2C%0D%0Aline%28-2.12%2C2.12%2C2.12%2C2.12%29%2Cline%282.12%2C2.12%2C2.12%2C-2.12%29%2Cline%282.12%2C-2.12%2C-2.12%2C-2.12%29%2Cline%28-2.12%2C-2.12%2C-2.12%2C2.12%29%2Cline%28-2.12%2C-2.12%2C2.12%2C2.12%29%2C%0D%0Acircle%280%2C0%2C0.0412%29%2Ccircle%280%2C0%2C0.0824%29%2Ccircle%280%2C0%2C0.1236%29%2C%0D%0Alocate%280.86%2C1.36%2C%22r%22%29%2Clocate%28-1.26%2C-0.56%2C%22r%22%29%2C%0D%0Alocate%28-2.74%2C0.2%2Csqrt%2870%29%29%2Clocate%28-0.36%2C2.68%2Csqrt%2870%29%29%0D%0A%29%0D%0A ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "For each right triangle, the hypotenuse is the diameter of the circle.<BR>\n" );
document.write( "r = radius<BR>\n" );
document.write( "2r = diameter = hypotenuse<BR>\n" );
document.write( "Use the pythagorean theorem to find that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=hypotenuse+=+2r+=+sqrt%28140%29 ALIGN=MIDDLE  ALT=\"hypotenuse+=+2r+=+sqrt%28140%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Or you could use the 45-45-90 triangle template to connect the leg length to the hypotenuse.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=matrix%281%2C3%2Chypotenuse=%28matrix%281%2C2%2Cleg%2Clength%29%29%2Asqrt%282%29%2C+%22%22%2C+matrix%281%2C4%2Conly%2Cfor%2C45-45-90%2Ctriangles%29%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Then it leads to<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r+=+sqrt%28140%29%2F2 ALIGN=MIDDLE  ALT=\"r+=+sqrt%28140%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The area of the circle is then...<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2Ar%5E2 ALIGN=MIDDLE  ALT=\"A+=+pi%2Ar%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2A%28sqrt%28140%29%2F2%29%5E2 ALIGN=MIDDLE  ALT=\"A+=+pi%2A%28sqrt%28140%29%2F2%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2A%28%28%28sqrt%28140%29%29%5E2%29%2F%282%5E2%29%29 ALIGN=MIDDLE  ALT=\"A+=+pi%2A%28%28%28sqrt%28140%29%29%5E2%29%2F%282%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2A%28140%2F4%29 ALIGN=MIDDLE  ALT=\"A+=+pi%2A%28140%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+pi%2A35 ALIGN=MIDDLE  ALT=\"A+=+pi%2A35\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+35pi ALIGN=MIDDLE  ALT=\"A+=+35pi\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "The exact area of the circle is 35pi square cm.\r<BR>\n" );
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document.write( "If you use the approximation pi = 3.14, then the approximate area of the circle would be 35*pi = 35*3.14 = 109.9 square cm.<BR>\n" );
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