document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=80754>Question 80754</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the surface area of a sphere with diameter 2 sqrt(3)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #57944 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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=57944\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> SPHERE TOTAL SURFACE AREA=4piR^2 GIVEN A DIAMETER OF 2SQRT3 THEN THE RADIUS IS<BR>\n" );
document.write( "1/2*2SQRT3=SQRT3=1.732 FOR THE RADIUS. THUS:<BR>\n" );
document.write( "TSA=4*3.14*1.732<BR>\n" );
document.write( "TSA=21.754   </SPAN>\n" );
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