document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=251322>Question 251322</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the area of a right triangle which has a perimeter of length 16 meters and a hypotenuse with a length of 7 meters.\r<BR>\n" );
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document.write( "a) sqrt (56)m^2 b) 8m2   c) 10m2   d) 12m2   e) 16m2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #183001 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=183001\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "If the hypotenuse is 7 and the perimeter is 16, then the sum of the measures of the other two sides must be 16 - 7 = 9.  Therefore, letting <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a\"> represent the measure of one leg and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE b\"> represent the measure of the other leg, we can say:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ a\ +\ b\ =\ 9\">\r<BR>\n" );
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document.write( "which is to say:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 9\ -\ b\">\r<BR>\n" );
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document.write( "But since we know that the hypotenuse is 7, we can also say:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ a^2\ +\ b^2\ =\ 7^2\ =\ 49\">\r<BR>\n" );
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document.write( "Substituting:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \left(9\ -\ b\right)^2\ +\ b^2\ =\ 49\">\r<BR>\n" );
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document.write( "Simplifying (verification of this step is left as an exercise for the student):\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ b^2\ -\ 9b\ +\ 16\ =\ 0\">\r<BR>\n" );
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document.write( "This quadratic does not factor, so using the quadratic formula:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ b = \frac{-(-9)\ \pm\ \sqrt{9^2\ -\ 4(1)(16)}}{2(1)}\ =\ \frac{9\ \pm\sqrt{17}}{2}\">.\r<BR>\n" );
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document.write( "Since\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 9\ -\ \frac{9\ +\sqrt{17}}{2}\ =\ \frac{9\ -\sqrt{17}}{2}\">\r<BR>\n" );
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document.write( "We can say one leg of the triangle is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{9\ +\sqrt{17}}{2}\">\r<BR>\n" );
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document.write( "And the other is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{9\ -\sqrt{17}}{2}\">\r<BR>\n" );
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document.write( "Since the area of a triangle is given by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A\ =\ \frac{bh}{2}\">\r<BR>\n" );
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document.write( "And since the legs of a right triangle are perpendicular so either can be considered the base and the other the height, all you need to do now is multiply the two roots of the quadratic we just solved by each other and then divide by 2.  Hint:  The two roots are a conjugate pair -- think difference of two squares.\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" );
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