document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=672313>Question 672313</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the areas of similar triangles are 25 and 16. The perimeter oh the first is 15, find the second perimeter.\r<BR>\n" );
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document.write( "I am confused on how to solve this. Every time I get some weird number that isn't the answer. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #417915 by </B><A HREF=/tutors/aboutme.mpl?userid=sachi><B>sachi(548)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=sachi><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=sachi><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=417915\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> In two similar triangles, the ratio of their areas is the square of the ratio of their sides =25/16\r<BR>\n" );
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document.write( "in Similar Triangles - ratios of parts, the perimeter, sides, altitudes and medians are all in the same ratio. Therefore, the area ratio will be the square of any of these ratios too.\r<BR>\n" );
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document.write( "so the ratio of the perimeters=sqrt(25/16)=5/4<BR>\n" );
document.write( "one perimeter=15<BR>\n" );
document.write( "so the 2nd perimeter=15*4/5=12<BR>\n" );
document.write( "ans </SPAN>\n" );
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