document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=916978>Question 916978</A>: <I><SPAN STYLE=\"word-break: break-all;\"> David's garden plot is in the shape of a quadrilateral. If the corner points are A, B, C and D then the angles at A and C are 120 degrees and 60 degrees respectively. AD = 16m, BC = 25m, and (DC) is 5m longer than (AB). A fence runs around the entire boundary of the plot. How long is the fence? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #645858 by </B><A HREF=/tutors/aboutme.mpl?userid=Bakr.R><B>Bakr.R(8)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Bakr.R><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=Bakr.R><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=645858\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To start at first, we may lay out a diagram by the given information:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=Bakr.R&illustration_name=Diagram_of_David_s_garden_plot\">.<BR>\n" );
document.write( "Diagram source: Haesemathematics\r<BR>\n" );
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document.write( "The question requires us to find the total length of the fence, they provide us with two angles at opposing sides, and a definite side along with each angle, and that allows us to cut the quadrilateral into two triangles without affecting any values:<BR>\n" );
document.write( " <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=Bakr.R&illustration_name=Triangular_cut\">.\r<BR>\n" );
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document.write( "Now that we have cut the shape into two triangles, trigonometric calculations (methods) may be used.\r<BR>\n" );
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document.write( "Two sides are unknown, but in the question it is mentioned that the length of DC is 5 m longer than AB, and through that we may deduce an expression for each side, let x be the unknown value:<BR>\n" );
document.write( " <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/display-illustration.mpl?tutor=Bakr.R&illustration_name=Unknown_values\">.\r<BR>\n" );
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document.write( "In each triangle, there is only one angle and a definite side, with two unknowns (two sides), and so no rule may be used to solve for the unknown, but, the two triangles have something in common, being that two triangles share one side, being equal for both, which is the length of DB.<BR>\n" );
document.write( "From having DB the common side, we can deduce for each triangle an equation, using the cosine rule <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%5E2=b%5E2%2Bc%5E2-2bccosA ALIGN=MIDDLE  ALT=\"a%5E2=b%5E2%2Bc%5E2-2bccosA\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "For Triangle ABD we may deduce the equation: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B16%5E2-2%2Ax%2A16%2Acos%28120%29 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B16%5E2-2%2Ax%2A16%2Acos%28120%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "For Triangle BCD we may deduce the equation: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=%28x%2B5%29%5E2%2B25%5E2-2%2A%28x%2B5%29%2A25%2Acos%2860%29 ALIGN=MIDDLE  ALT=\"DB%5E2=%28x%2B5%29%5E2%2B25%5E2-2%2A%28x%2B5%29%2A25%2Acos%2860%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "It is apparent now, that we can equal these two equations and find the unknown, but to make the method more simple, let us simplify each equation first.<BR>\n" );
document.write( "The equation for triangle ABD;<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B16%5E2-2%2Ax%2A16%2Acos%28120%29 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B16%5E2-2%2Ax%2A16%2Acos%28120%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B265-2%2Ax%2A-8 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B265-2%2Ax%2A-8\"   DISABLED_style= \"max-width:90%; height:auto;\" > cos(120)=-1/2 or -0.5<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B16x%2B256 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B16x%2B256\"   DISABLED_style= \"max-width:90%; height:auto;\" > Now we have a simplified equation.<BR>\n" );
document.write( "The equation of triangle BCD:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=%28x%2B5%29%5E2%2B25%5E2-2%2A%28x%2B5%29%2A25%2Acos%2860%29 ALIGN=MIDDLE  ALT=\"DB%5E2=%28x%2B5%29%5E2%2B25%5E2-2%2A%28x%2B5%29%2A25%2Acos%2860%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B10x%2B25%2B625%2B%28-2x-10%29%2A12.5 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B10x%2B25%2B625%2B%28-2x-10%29%2A12.5\"   DISABLED_style= \"max-width:90%; height:auto;\" > cos(60)=1/2 or 0.5<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2%2B10x%2B650-25x-125 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2%2B10x%2B650-25x-125\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=x%5E2-15x%2B525 ALIGN=MIDDLE  ALT=\"DB%5E2=x%5E2-15x%2B525\"   DISABLED_style= \"max-width:90%; height:auto;\" > Now we have a simplified equation for both triangles.<BR>\n" );
document.write( "From here, we can also (obviously) deduce that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=DB%5E2=DB%5E2 ALIGN=MIDDLE  ALT=\"DB%5E2=DB%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and so through that, using both of the simplified equations:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2B16x%2B256=x%5E2-15x%2B525 ALIGN=MIDDLE  ALT=\"x%5E2%2B16x%2B256=x%5E2-15x%2B525\"   DISABLED_style= \"max-width:90%; height:auto;\" > And now we further simplify and solve to find the value of the unknown.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=31x-269=0 ALIGN=MIDDLE  ALT=\"31x-269=0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=31x=269 ALIGN=MIDDLE  ALT=\"31x=269\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=269%2F3 ALIGN=MIDDLE  ALT=\"x=269%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Now that we've found a value for x, we may solve the problem by substituting it into the equation for finding the total length of the fence, by finding the perimeter of the quadrilateral:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=Fence+Length=16%2B25%2B2x%2B5 ALIGN=MIDDLE  ALT=\"Fence+Length=16%2B25%2B2x%2B5\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16%2Bx%2B%28x%2B5%29%2B25 ALIGN=MIDDLE  ALT=\"16%2Bx%2B%28x%2B5%29%2B25\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=63.35 ALIGN=MIDDLE  ALT=\"x=63.35\"   DISABLED_style= \"max-width:90%; height:auto;\" > rounded to hundredth d.p. </SPAN>\n" );
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