document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=442432>Question 442432</A>: <I><SPAN STYLE=\"word-break: break-all;\"> area of an isosceles triangle is 60sq. cm. the same two side of is  13 cm, what is the other one leangth ?? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #305174 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=305174\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "Construct triangle ABC with the vertex B common to the two congruent sides upward and the unknown side as the base.  Construct a perpendicular to the base through the vertex at the top of the triangle. Label the point of intersection of the perpendicular and the base with D.\r<BR>\n" );
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document.write( "Your picture should look like one of the following:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28%0D%0A500%2C+600%2C+-1%2C11%2C-1%2C13%2C%0D%0Aline%280%2C0%2C5%2C12%29%2C%0D%0Aline%280%2C0%2C10%2C0%29%2C%0D%0Aline%285%2C12%2C10%2C0%29%2C%0D%0Agreen%28line%285%2C12%2C5%2C0%29%29%2C%0D%0Alocate%282.4%2C-.2%2C%22x%22%29%2C%0D%0Alocate%285.2%2C6%2C%22y%22%29%2C%0D%0Alocate%284.9%2C-.2%2C%22D%22%29%2C%0D%0Alocate%28-.2%2C-.2%2C%22A%22%29%2C%0D%0Alocate%284.9%2C12.5%2C%22B%22%29%2C%0D%0Alocate%289.9%2C-.2%2C%22C%22%29%0D%0A%0D%0A%29 ALIGN=MIDDLE   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=drawing%28%0D%0A1200%2C+300%2C+-1%2C25%2C-1%2C6%2C%0D%0Aline%280%2C0%2C12%2C5%29%2C%0D%0Aline%280%2C0%2C24%2C0%29%2C%0D%0Aline%2812%2C5%2C24%2C0%29%2C%0D%0Agreen%28line%2812%2C5%2C12%2C0%29%29%2C%0D%0Alocate%285.9%2C-.2%2C%22x%22%29%2C%0D%0Alocate%2812.2%2C2.5%2C%22y%22%29%2C%0D%0Alocate%2811.9%2C-.2%2C%22D%22%29%2C%0D%0Alocate%28-.2%2C-.2%2C%22A%22%29%2C%0D%0Alocate%2811.9%2C5.5%2C%22B%22%29%2C%0D%0Alocate%2823.9%2C-.2%2C%22C%22%29%0D%0A%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> represent the measure of segement AD.  Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\"> represent segment BD.  Since we know from a common geometric proof that the perpendicular from the apex to the base of an isosceles triangle bisects the base, the apex angle, and the triangle itself.  Therefore the area of triangle ABD is equal to the area of triangle BCD and the area of each is one-half of the area of the entire triangle ABC.\r<BR>\n" );
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document.write( "From the formula for the area of a triangle, we have:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A_{ABD}\ =\ \frac{xy}{2}\">\r<BR>\n" );
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document.write( "Hence,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A_{ABC}\ =\ xy\">\r<BR>\n" );
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document.write( "And further,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ xy\ =\ 60\">\r<BR>\n" );
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document.write( "From which we can derive:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{60}{x}\">\r<BR>\n" );
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document.write( "From the fact that segment BD is perpendicular to segment AD, we can deduce that triangle ABD is a right triangle, for which we are given the measure of the hypotenuse is 13 and we have identified the measure of the legs as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\">.  Using Pythagoras:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 13^2\">\r<BR>\n" );
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document.write( "And using what we learned earlier about the relationship between <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large y\">, we can write:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ \frac{3600}{x^2}\ =\ 169\">\r<BR>\n" );
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document.write( "Multiply both sides by <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x^2\">:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^4\ +\ 3600\ =\ 169x^2\">\r<BR>\n" );
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document.write( "Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large u\ =\ x^2\">, substitute, and put the resulting quadratic into standard form:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ u^2\ -\ 169u\ +\ 3600\ =\ 0\">\r<BR>\n" );
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document.write( "Factor:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ (u\ -\ 144)(u\ -\ 25)\ =\ 0\">\r<BR>\n" );
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document.write( "Then\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 144\ \Rightarrow\ x^2\ =\ 144\ \Rightarrow\ x\ =\ \pm12\">\r<BR>\n" );
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document.write( "or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ u\ =\ 25\ \Rightarrow\ x^2\ =\ 25\ \Rightarrow\ x\ =\ \pm5\">\r<BR>\n" );
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document.write( "Discarding the negative roots because we are attempting to calculate length, we have <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ =\ 12\"> or <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ =\ 5\">.\r<BR>\n" );
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document.write( "Having previously determined that segment  BD bisects segment AC, and by construction, AD plus DC is equal to AC, we can deduce that AC is equal to two times AD.  Therefore the base measurement that we seek as an answer to this problem is two times the value of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ AC\ =\ 2x\ =\ 24\">\r<BR>\n" );
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document.write( "or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ AC\ =\ 2x\ =\ 10\">\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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