document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=221444>Question 221444</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the area of an equilateral triangle in which each side has a length of 2cm. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #166123 by </B><A HREF=/tutors/aboutme.mpl?userid=drj><B>drj(1380)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drj><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=drj><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=166123\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Find the area of an equilateral triangle in which each side has a length of 2cm.\r<BR>\n" );
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document.write( "Step 1.  Area of equilateral triangle is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=bh%2F2 ALIGN=MIDDLE  ALT=\"A=bh%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > where b is the base and h is the height.  Draw a perpendicular line from one angle to the midpoint of the opposite side.  Doing this forms a right triangle and bisected in half the opposite side.\r<BR>\n" );
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document.write( "Step 2.  The base is 1/2 the side of the equilateral triangle or 2cm/2=1 cm.\r<BR>\n" );
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document.write( "Step 3.  The height can be calculated using the Pythagorean Theorem which states that the sum of the squares of the legs of a right triangle (h and b) is equal to the hypotenuse squared or side s=2 cm.\r<BR>\n" );
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document.write( "Then, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%5E2=h%5E2%2Bb%5E2 ALIGN=MIDDLE  ALT=\"s%5E2=h%5E2%2Bb%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=2%5E2=h%5E2%2B1%5E2 ALIGN=MIDDLE  ALT=\"2%5E2=h%5E2%2B1%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=4=h%5E2%2B1 ALIGN=MIDDLE  ALT=\"4=h%5E2%2B1\"   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=h%5E2=3 ALIGN=MIDDLE  ALT=\"h%5E2=3\"   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=h=sqrt%283%29 ALIGN=MIDDLE  ALT=\"h=sqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Step 4.  ANSWER:  Area <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=bh%2F2=1%2Asqrt%283%29%2F2=sqrt%283%29%2F2 ALIGN=MIDDLE  ALT=\"A=bh%2F2=1%2Asqrt%283%29%2F2=sqrt%283%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" > square centimeters.\r<BR>\n" );
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document.write( "I hope the above steps were helpful.\r<BR>\n" );
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document.write( "For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.\r<BR>\n" );
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document.write( "Good luck in your studies!\r<BR>\n" );
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document.write( "Respectfully,<BR>\n" );
document.write( "Dr J </SPAN>\n" );
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