document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1107835>Question 1107835</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Quadrilateral ABCD is an isosceles trapezoid with AB parallel to DC, AC=DC, and AD=BC. If the height h of the trapezoid is equal to AB, find the ratio AB:DC<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #722966 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13203)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=722966\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "I used coordinate geometry to solve the problem; there are certain to be other paths, possibly much easier....<br><BR>\n" );
document.write( "Let points C and D be C(0,0) and D(1,0).<br><BR>\n" );
document.write( "Consider the arc in the first quadrant of the circle with center C and radius CD.  To make AC=DC, point A must be somewhere on that arc.<br><BR>\n" );
document.write( "The requirement is that the height h of the trapezoid be equal to the length of base AB; so we need to find the coordinates of the point A on the arc such that the y coordinate (the height of the trapezoid) is equal to the length of base AB.<br><BR>\n" );
document.write( "Let h be the height of the trapezoid.<br><BR>\n" );
document.write( "By symmetry, the midpoint of base AB will have coordinates (0.5,h).<br><BR>\n" );
document.write( "Again by symmetry, the coordinates of point A will be (0.5+0.5h,h).<br><BR>\n" );
document.write( "Then since point A is on an arc of a circle with radius 1,<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%280.5%2B0.5h%29%5E2+%2B+h%5E2+=+1 ALIGN=MIDDLE  ALT=\"%280.5%2B0.5h%29%5E2+%2B+h%5E2+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2Bh%29%5E2%2B4h%5E2+=+4 ALIGN=MIDDLE  ALT=\"%281%2Bh%29%5E2%2B4h%5E2+=+4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2B2h%2Bh%5E2%2B4h%5E2+=+4 ALIGN=MIDDLE  ALT=\"1%2B2h%2Bh%5E2%2B4h%5E2+=+4\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5h%5E2%2B2h-3+=+0 ALIGN=MIDDLE  ALT=\"5h%5E2%2B2h-3+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%285h-3%29%28h%2B1%29+=+0 ALIGN=MIDDLE  ALT=\"%285h-3%29%28h%2B1%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h+=+3%2F5 ALIGN=MIDDLE  ALT=\"h+=+3%2F5\"   DISABLED_style= \"max-width:90%; height:auto;\" >  or  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h+=+-1 ALIGN=MIDDLE  ALT=\"h+=+-1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Clearly we need to choose the positive solution.<br><BR>\n" );
document.write( "So h, the length of base AB, is 3/5; then since the length of base CD is 1, the ratio of the lengths of the bases AB:CD is 3:5.<br><br><BR>\n" );
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document.write( "Having solved the problem by that method, I see that the right triangle formed by AC, DC, and the altitude from A to CD is a 3:4:5 right triangle; that suggests to me another possibly easier solution.<br><BR>\n" );
document.write( "Use the same figure as before; and let BE and AF be altitudes of the trapezoid.<br><BR>\n" );
document.write( "Let CE = DF = x and CD = y.  Then in right triangle ACF,<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=CF+=+y-x ALIGN=MIDDLE  ALT=\"CF+=+y-x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AF+=+EF+=+y-2x ALIGN=MIDDLE  ALT=\"AF+=+EF+=+y-2x\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=AC+=+y ALIGN=MIDDLE  ALT=\"AC+=+y\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Then<br><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y-x%29%5E2%2B%28y-2x%29%5E2+=+y%5E2 ALIGN=MIDDLE  ALT=\"%28y-x%29%5E2%2B%28y-2x%29%5E2+=+y%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2-2xy%2Bx%5E2%2By%5E2-4xy%2B4x%5E2+=+y%5E2 ALIGN=MIDDLE  ALT=\"y%5E2-2xy%2Bx%5E2%2By%5E2-4xy%2B4x%5E2+=+y%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y%5E2-6xy%2B5x%5E2+=+0 ALIGN=MIDDLE  ALT=\"y%5E2-6xy%2B5x%5E2+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28y-x%29%28y-5x%29+=+0 ALIGN=MIDDLE  ALT=\"%28y-x%29%28y-5x%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><br><BR>\n" );
document.write( "Clearly y=x does not make sense in the problem; so y = 5x.<br><BR>\n" );
document.write( "But that makes the lengths of the sides of right triangle ACF 3x, 4x, and 5x.<br><BR>\n" );
document.write( "And since AB is the height of the trapezoid, the ratio AB:CD of the lengths of bases of the trapezoid is 3x:5x, or 3:5.<br><BR>\n" );
document.write( "Fun problem....!<BR>\n" );
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