document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=126833>Question 126833</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A garden in the shape of a right triangle has sides measuring 60, 80 and 100 ft.  The owner of the garden adds a new section of fence that runs from the right angle to the hypoteneuse and separates the garden into two parts of equal perimeter.  Find the length of the new section of the fence.  Problem is not out of a textbook, rather on a worksheet. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #93149 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=93149\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A garden in the shape of a right triangle has sides measuring 60, 80 and 100 ft. The owner of the garden adds a new section of fence that runs from the right angle to the hypotenuse and separates the garden into two parts of equal perimeter. Find the length of the new section of the fence. <BR>\n" );
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document.write( "Draw the triangle ABC, (B=90) with a=60, b=100; c=80<BR>\n" );
document.write( "Let D = point on side b where the new section of fence BD will be joined<BR>\n" );
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document.write( "The perimeters are give as equal so we can say:<BR>\n" );
document.write( "CD + 60 + BD = AD + 80 + BD<BR>\n" );
document.write( "Subtract BD from both sides<BR>\n" );
document.write( "CD + 60 = AD + 80<BR>\n" );
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document.write( "We know that CD + AD = 100<BR>\n" );
document.write( "CD = (100-AD)<BR>\n" );
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document.write( "Substitute (100-AD for CD; find AD<BR>\n" );
document.write( "100 - AD + 60 = AD + 80<BR>\n" );
document.write( "160 = AD + AD + 80<BR>\n" );
document.write( "160 - 80 = 2AD<BR>\n" );
document.write( "AD = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=80%2F2 ALIGN=MIDDLE  ALT=\"80%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "AD = 40, <BR>\n" );
document.write( "then <BR>\n" );
document.write( "100-40 = 60 is CD<BR>\n" );
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document.write( "We have to find the length of BD which is the new fence<BR>\n" );
document.write( "Cos C = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60%2F100 ALIGN=MIDDLE  ALT=\"60%2F100\"   DISABLED_style= \"max-width:90%; height:auto;\" > .6<BR>\n" );
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document.write( "Using triangle BCD, find BD, (c in the formula)<BR>\n" );
document.write( "Using the law of cosines:<BR>\n" );
document.write( "c^2 = a^2 + b^2 - 2(ab)*Cos C<BR>\n" );
document.write( "BD^2 = 60^2 + 60^2 - 2(60*60) * .6<BR>\n" );
document.write( "BD^2 = 3600 + 3600 - 2(3600) *.6<BR>\n" );
document.write( "BD^2 = 7200 .6(7200)<BR>\n" );
document.write( "BD^2 = 7200 - 4320<BR>\n" );
document.write( "BD = SqRT(2880)<BR>\n" );
document.write( "BD = 53.6656, say 54 ft of additional fencing required\r<BR>\n" );
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