document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=231878>Question 231878</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In class we had to figure out how to solve problems bu using solving the sub-problem of that problem. Which is basically just steps to get to figuring out the big picture. In the following problem I wasn't quite sure where to start with that. Here is the problem:\r<BR>\n" );
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document.write( "\"In right triangle RED, angle R is the right angle. Point O is on segment ER and point A is on segment ED. Segment OA is perpendicular to segment ED. EA = 6, AD = 14, and ER = 16. Find the area of quadrilateral ROAD.\"\r<BR>\n" );
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document.write( "So with the picture you draw with that, do I have to cut the right triangle in half and then solve? I was so confused nor do I know where to start with finding the sub problems. If anyone has a suggestion it would be very helpful. Much appreciated in advance!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #171494 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=171494\">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( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=drawing%28%0D%0A500%2C+500%2C+-5%2C+5%2C+-5%2C+5%2C%0D%0Atriangle%28-3%2C4%2C-3%2C-4%2C3%2C-4%29%2C%0D%0Ared%28circle%28-3%2C1.6%2C.05%29%2C%0D%0Acircle%28-1.2%2C1.6%2C.05%29%2C%0D%0Aline%28-3%2C1.6%2C-1.2%2C1.6%29%2C%0D%0Alocate%28-3.2%2C1.6%2CO%29%2C%0D%0Alocate%28-1%2C1.6%2CA%29%29%2C%0D%0Ablue%28locate%28-3.2%2C4.2%2CE%29%2C%0D%0Alocate%28-3.2%2C-4.2%2CR%29%2C%0D%0Alocate%283.2%2C-4.2%2CD%29%29%0D%0A%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Notice a few things right off:\r<BR>\n" );
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document.write( "1.  EA = 6 and AD = 14 means that ED = 20.\r<BR>\n" );
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document.write( "2.  OA perpendicular to RE means OA parallel to RD, hence:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \">*  Triangle OEA similar to triangle RED\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \">*  Quadrilateral ROAD is a trapeziod \r<BR>\n" );
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document.write( "3.  20 is 4 times 5 and 16 is 4 times 4, hence RD has to be 12: 4 times 3 -- 3-4-5 right triangle.<BR>\n" );
document.write( "(verify by using Pythagoras and calculating <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large RD\ =\ \sqrt{20^2\ -\ 16^2}\"> if you want)\r<BR>\n" );
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document.write( "The hypotenuse of OEA is 6, so <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large \ \ \frac{6}{5}\ =\ \frac{OE}{4}\ =\ \frac{OA}{3}\ \ \ \"> because of similar triangles.  Hence OE = 4.8 and OA = 3.6.  Verification left as an exercise for the student.\r<BR>\n" );
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document.write( "Since OE = 4.8, RO = 16 - 4.8 = 11.2.  And because of right angle R, RO is an altitude of the trapezoid\r<BR>\n" );
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document.write( "The formula for the area of a trapezoid is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A_t\ = a\left(\frac{b_1\ +\ b_2}{2}\right)\">\r<BR>\n" );
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document.write( "Plug in the values:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A_{ROAD}\ = 11.2\left(\frac{3.6\ +\ 12}{2}\right)\">\r<BR>\n" );
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document.write( "You can do your own arithmetic.\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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