document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=121821>Question 121821</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hello, \r<BR>\n" );
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document.write( "I have a question that I am having a lot of trouble with. I'll give you the question and try to explain (as best as I can) my working out so far and hopefully you'll be able to understand it and give me a detailed explanation of the answer. :^) \r<BR>\n" );
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document.write( "I think you might need a diagram to visualise the whole thing which I cannot actually draw, however, I'll do my best to describe it. (If you would like a similar picture I checked Google Images and did find an image that closely resembled what the question was. Except for the circle at the base of the larger circle, everything is correct. Insert the measurements and that's <BR>\n" );
document.write( "how it looks like. The link is: http://mimosa.cnice.mecd.es/~jcolon/tangen.gif\r<BR>\n" );
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document.write( "THE QUESTION<BR>\n" );
document.write( "Here it is: \r<BR>\n" );
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document.write( "Imagine a circle and a isosceles triangle is in it, each of its vertices touching the edges of the circle. The top vertice is labelled P and the two bottom vertices from left to right is labelled Q and R. (I hope you can either visualise it or sketch it on a piece of paper) QR = 18 and PQ = PR = 15. The questions asks me/you: What is the radius of the circle? \r<BR>\n" );
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document.write( "MY WORKING OUT<BR>\n" );
document.write( "I drew a line bisecting the triangle PQR in the way you told me to and extended it down to the diameter of the circle. Then I labelled the diameter PS and the intersection of PS and QR: T. I now have two right-angled triangles where I can label each side length a, b and c (the hypotenuse). \r<BR>\n" );
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document.write( "a = the side length I need to find<BR>\n" );
document.write( "b = T-R or T-Q (halve 18 [Q-R) = 9<BR>\n" );
document.write( "c = hypotenuse (15)\r<BR>\n" );
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document.write( "I then applied the Pythagorean Theorem to find (a) which is the side length I need to find, n amelly the segment PT. Using the Pythagorean Theorem PT = 12. All I needed to find now was the segment TS and in order to do that I needed to find the segment QS, and if I connect them I form another two triangles (QTS) and (PSQ) and I can use the sides of PTQ to work it out. But I'm stuck. I <BR>\n" );
document.write( "don't know where to go from here.\r<BR>\n" );
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document.write( "Please help, because Im not sure what to do next. Do I have to understand trigonometry, or geometry or more about triangles to solve it? I was told by my teacher there is a special relationship between the triangles PQS, PTQ and QTS. But I don't know what. \r<BR>\n" );
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document.write( "Thanks, \r<BR>\n" );
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document.write( "Jasmine (13)  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #89459 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=89459\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You are going to have to trust an old man's memory on this one.  I'm almost certain that we had a similar problem back when I took geometry in high school and we had to prove that your triangles PQS, PTQ, and QTS were all similar right triangles.  Of course, that was 45 years ago and what I certainly don't remember is how I went about proving it.\r<BR>\n" );
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document.write( "Presuming my memory is correct, and the three triangles are, in fact, similar, you can use a simple proportion to calculate the two missing sides of QTS.\r<BR>\n" );
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document.write( "Note that the sides of PTQ are in proportion to 5:4:3 because 15/3 = 5, 12/3 = 4, and 9/3 = 3.\r<BR>\n" );
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document.write( "That means that the legs of QTS, namely segments QT and ST, are in proportion <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=9%2F4=x%2F3 ALIGN=MIDDLE  ALT=\"9%2F4=x%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" > where x is the measure of segment ST.\r<BR>\n" );
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document.write( "Solving for x:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=9%2F4=x%2F3 ALIGN=MIDDLE  ALT=\"9%2F4=x%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=4x=27 ALIGN=MIDDLE  ALT=\"4x=27\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=27%2F4 ALIGN=MIDDLE  ALT=\"x=27%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now we know that the sum of the measures of PT and ST equal the measure of the circle diameter, so,\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d=12+%2B+27%2F4=48%2F4%2B27%2F4=75%2F4 ALIGN=MIDDLE  ALT=\"d=12+%2B+27%2F4=48%2F4%2B27%2F4=75%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "But we need the radius which is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=d%2F2 ALIGN=MIDDLE  ALT=\"d%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=r=%281%2F2%29%2875%2F4%29=75%2F8 ALIGN=MIDDLE  ALT=\"r=%281%2F2%29%2875%2F4%29=75%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=9 ALIGN=MIDDLE  ALT=\"9\"   DISABLED_style= \"max-width:90%; height:auto;\" ><IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3%2F8 ALIGN=MIDDLE  ALT=\"3%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" > if you prefer.\r<BR>\n" );
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document.write( "Hope this helps,<BR>\n" );
document.write( "John\r<BR>\n" );
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document.write( " </SPAN>\n" );
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