document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=192077>Question 192077</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Given (triangle)ABC~(tringle)XYZ, AB=9, BC=18 and AC-21.  If the perimeter of (triangle)XYZ is 112, find the measures of the three sides of (triangle)XYZ. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #144216 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=144216\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw  a  rough  sketch  of  two  similar  triangles<BR>\n" );
document.write( "label  sides  carefully\r<BR>\n" );
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document.write( "as  triangles  are  similar,  sides  are  proportional\r<BR>\n" );
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document.write( "per  of  given  triangle  abc  =  9 + 18 + 21 = 48.\r<BR>\n" );
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document.write( "per  of  new  triangle  is  112\r<BR>\n" );
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document.write( "as  perimeters  vary  as  scale  factor  to  1st  power\r<BR>\n" );
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document.write( "sf  =  112/48  =2 1/3  =7/3\r<BR>\n" );
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document.write( "xy  =  (7/3)ab =7/3 * 9 = 21<BR>\n" );
document.write( "yz  = (7/3) bc  = 7/3*18 = 42<BR>\n" );
document.write( "xz = (7/3) ac = 7/3*21 = 49\r<BR>\n" );
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document.write( "checking  per  2nd  triangle  = 21+42+49= 112  ok </SPAN>\n" );
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