document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=261102>Question 261102</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a right triangle has hypotenuse of lenght 13 and legs A and B.  If the area of the triangle is 14, what is the sum A+B?  We do not think this is possible.  If the Square of the hypotenuse is 169, and it is an equilateral triangle, the sides would be about 9, and one half of 9*9 (the area of the triangle) is more like 40 than 14 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #192360 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=192360\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  a right triangle has hypotenuse of lenght 13 and legs A and B. If the area of the triangle is 14, what is the sum A+B? <BR>\n" );
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document.write( "a^2 + b^2 = 169<BR>\n" );
document.write( "ab/2 = 14 --> a = 28/b<BR>\n" );
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document.write( "(28/b)^2 + b^2 = 169<BR>\n" );
document.write( "784 + b^4 = 169b^2<BR>\n" );
document.write( "b^4 - 169b^2 + 784 = 0<BR>\n" );
document.write( "Sub x for b^2<BR>\n" );
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document.write( " <TR><TD  ALIGN=MIDDLE>Solved by <A  rel=nofollow HREF=http://wiki.algebra.com/index.php/Tutoring_-_pluggable_solvers>pluggable</A> solver: <A  rel=nofollow HREF=solve_quadratic_equation.solver>SOLVE quadratic equation (work shown, graph etc)</A></TD></TR>\n" );
document.write( "<TR><TD  ALIGN=LEFT>Quadratic equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E2%2Bbx%2Bc=0 ALIGN=MIDDLE  ALT=\"ax%5E2%2Bbx%2Bc=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (in our case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1x%5E2%2B-169x%2B784+=+0 ALIGN=MIDDLE  ALT=\"1x%5E2%2B-169x%2B784+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >) has the following solutons:<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  <BR>\n" );
document.write( "  For these solutions to exist, the <U>discriminant</U> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" > should not be a negative number.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  First, we need to compute the discriminant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" >: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac=%28-169%29%5E2-4%2A1%2A784=25425 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%28-169%29%5E2-4%2A1%2A784=25425\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Discriminant d=25425 is greater than zero. That means that there are two solutions: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5B12%5D+=+%28--169%2B-sqrt%28+25425+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28--169%2B-sqrt%28+25425+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D+=+%28-%28-169%29%2Bsqrt%28+25425+%29%29%2F2%5C1+=+164.22609359551 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%28-169%29%2Bsqrt%28+25425+%29%29%2F2%5C1+=+164.22609359551\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B2%5D+=+%28-%28-169%29-sqrt%28+25425+%29%29%2F2%5C1+=+4.77390640449013 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%28-169%29-sqrt%28+25425+%29%29%2F2%5C1+=+4.77390640449013\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <BR>\n" );
document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1x%5E2%2B-169x%2B784 ALIGN=MIDDLE  ALT=\"1x%5E2%2B-169x%2B784\"   DISABLED_style= \"max-width:90%; height:auto;\" > can be factored:<BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1x%5E2%2B-169x%2B784+=+%28x-164.22609359551%29%2A%28x-4.77390640449013%29 ALIGN=MIDDLE  ALT=\"1x%5E2%2B-169x%2B784+=+%28x-164.22609359551%29%2A%28x-4.77390640449013%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 164.22609359551, 4.77390640449013.\n" );
document.write( "Here's your graph:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-169%2Ax%2B784+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-169%2Ax%2B784+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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document.write( "b^2 =~ 164.226<BR>\n" );
document.write( "b^2 =~ 4.7739<BR>\n" );
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document.write( "b =~ 12.8166, a =~ 2.1849 or vice versa<BR>\n" );
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document.write( "a + b = 15<BR>\n" );
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document.write( "In any right triange with a specified lenght of the hypotenuse, the max area is<BR>\n" );
document.write( "c^2/2.<BR>\n" );
document.write( "For a hyp = 13, that's 169/2 = 84.5 sq units.  Any area less than that can be made with hyp = 13.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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