document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=574820>Question 574820</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the side of an equilateral triangle is x . its area is ___________ its perimeter is ________________.<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 #369375 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=369375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Several ways to find the area. You can construct an altitude from one vertex to the opposite side, use 30-60-90 triangles, et cetera. For simplicity, I would use the area formula\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE Area = \frac{1}{2}ab \sin C\"> where C is the included angle between sides a and b. This becomes\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE Area = \frac{1}{2}x^2 \sin 60^{\circ}\">\r<BR>\n" );
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document.write( "sin 60 = sqrt(3)/2, so\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE Area = \frac{\sqrt{3}}{4}x^2\">\r<BR>\n" );
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document.write( "Perimeter of the same triangle is just x+x+x, or 3x. </SPAN>\n" );
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