document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=820311>Question 820311</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.<BR>\n" );
document.write( "what is the area of the isosceles triangle 2 sides each measuring 3s with a base s ?\r<BR>\n" );
document.write( "\n" );
document.write( "2. solve = 5/(4a-7bi) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #493455 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=493455\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> It will help if you draw a diagram:<ol><li>Draw an isosceles triangle and label the vertices so that AB = AC.</li><li>Label side AB as 3s and side AC as 3s. Label BC as s.</li><li>Draw a perpendicular from A to BC. Label the point where it intersects with BC as D. And label its length as h (for height).</li><li>Because the two right triangles, ADB and ADC, have the same hypotenuse, 3s, and the same leg, AD, they are congruent. This makes BD = CD. Since BC is s and BD = CD, BD = CD = s/2.</li></ol>Using the Pythagorean Theorem on one of the right triangles we get:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28s%2F2%29%5E2%2Bh%5E2=%283s%29%5E2 ALIGN=MIDDLE  ALT=\"%28s%2F2%29%5E2%2Bh%5E2=%283s%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Simplifying:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%5E2%2F4%2Bh%5E2=9s%5E2 ALIGN=MIDDLE  ALT=\"s%5E2%2F4%2Bh%5E2=9s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Multiply both sides by 4 (to eliminate the fraction):<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%5E2%2B4h%5E2=36s%5E2 ALIGN=MIDDLE  ALT=\"s%5E2%2B4h%5E2=36s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Subtracting <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=s%5E2 ALIGN=MIDDLE  ALT=\"s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4h%5E2=35s%5E2 ALIGN=MIDDLE  ALT=\"4h%5E2=35s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Dividing by 4:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h%5E2=%2835s%5E2%29%2F4 ALIGN=MIDDLE  ALT=\"h%5E2=%2835s%5E2%29%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Square root of each side:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28h%5E2%29=sqrt%28%2835s%5E2%29%2F4%29 ALIGN=MIDDLE  ALT=\"sqrt%28h%5E2%29=sqrt%28%2835s%5E2%29%2F4%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "(We will not use the <u>+</u> because h is a height which should never be negative.) Simplifying...<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=%28sqrt%2835%29sqrt%28s%5E2%29%29%2Fsqrt%284%29%29 ALIGN=MIDDLE  ALT=\"h=%28sqrt%2835%29sqrt%28s%5E2%29%29%2Fsqrt%284%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=%28sqrt%2835%29%2As%29%2F2 ALIGN=MIDDLE  ALT=\"h=%28sqrt%2835%29%2As%29%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "which can be rewritten as:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h=%28sqrt%2835%29%2F2%29s ALIGN=MIDDLE  ALT=\"h=%28sqrt%2835%29%2F2%29s\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "The area of the isosceles triangle ABC is 1/2 times the base times the height. Using a base of s and the height we just found we get:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F2%29s%2A%28sqrt%2835%29%2F2%29s ALIGN=MIDDLE  ALT=\"%281%2F2%29s%2A%28sqrt%2835%29%2F2%29s\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "which simplifies to:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28sqrt%2835%29%2F4%29s%5E2 ALIGN=MIDDLE  ALT=\"%28sqrt%2835%29%2F4%29s%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "P.S. Problem 2 is not correct. Expressions are not \"solved\". Either the problem is incomplete or you did not include the proper instructions for the problem. Plus, it really doesn't belong as a posting in the radicals category. (It belongs under Complex Numbers. Please re-post the full, correct problem under Complex Numbers. </SPAN>\n" );
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