document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1171816>Question 1171816</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In the diagram, Line AB=Line AC, angle A= 60 degrees. If Line AD = 1/4 Line AC, and AE:EB=K:5, find the value of k, with all values in cm.\r<BR>\n" );
document.write( "\n" );
document.write( "A) 5<BR>\n" );
document.write( "B) 2<BR>\n" );
document.write( "C) 10<BR>\n" );
document.write( "D) 2 1/2<BR>\n" );
document.write( "E) 3 1/2\r<BR>\n" );
document.write( "\n" );
document.write( "https://ibb.co/K55rshc<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #796724 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=796724\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "We're given that AB = AC, so triangle ABC is isosceles. This means the base angles B and C are congruent, as they are the base angles. Let y = angle B = angle C.<BR>\n" );
document.write( "We're also given angle A is 60 degrees.<BR>\n" );
document.write( "So,<BR>\n" );
document.write( "A+B+C = 180<BR>\n" );
document.write( "60+y+y = 180<BR>\n" );
document.write( "2y+60 = 180<BR>\n" );
document.write( "2y = 180-60<BR>\n" );
document.write( "2y = 120<BR>\n" );
document.write( "y = 120/2<BR>\n" );
document.write( "y = 60<BR>\n" );
document.write( "Each angle of triangle ABC is 60 degrees, so triangle ABC is equilateral.<BR>\n" );
document.write( "This tells us that AB = BC = AC.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Draw a line from E to C<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" width=\"50%\" src = \"https://i.imgur.com/yyhKTzN.png\"><BR>\n" );
document.write( "We have similar triangles ADE and EDC<BR>\n" );
document.write( "This means we can form the proportion<BR>\n" );
document.write( "AD/DE = DE/DC<BR>\n" );
document.write( "and that can be rearranged into<BR>\n" );
document.write( "DE = sqrt(AD*DC)<BR>\n" );
document.write( "This shows DE is the geometric mean of AD and DC\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Let x = AD<BR>\n" );
document.write( "Since AD is 1/4 the length of AC, we know that AC is 4 times longer compared to AD, so AC = 4*AD = 4x<BR>\n" );
document.write( "This leads to <BR>\n" );
document.write( "AD+DC = AC<BR>\n" );
document.write( "x+DC = 4x<BR>\n" );
document.write( "DC = 4x-x<BR>\n" );
document.write( "DC = 3x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Applying the geometric mean formula gets us<BR>\n" );
document.write( "DE = sqrt(AD*DC)<BR>\n" );
document.write( "DE = sqrt(x*3x)<BR>\n" );
document.write( "DE = sqrt(3x^2)<BR>\n" );
document.write( "This simplifies to x*sqrt(3), but we won't need to do this\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "With AD = x and DE = sqrt(3x^2), we can find AE through the pythagorean theorem<BR>\n" );
document.write( "(AD)^2 + (DE)^2 = (AE)^2<BR>\n" );
document.write( "(x)^2 + (sqrt(3x^2))^2 = (AE)^2<BR>\n" );
document.write( "x^2+3x^2 = (AE)^2<BR>\n" );
document.write( "4x^2 = (AE)^2<BR>\n" );
document.write( "(AE)^2 = 4x^2<BR>\n" );
document.write( "AE = sqrt(4x^2)<BR>\n" );
document.write( "AE = sqrt((2x)^2)<BR>\n" );
document.write( "AE = 2x\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Now we can say<BR>\n" );
document.write( "AE+EB = AB<BR>\n" );
document.write( "2x+EB = 4x<BR>\n" );
document.write( "EB = 4x-2x<BR>\n" );
document.write( "EB = 2x<BR>\n" );
document.write( "We know that AB = 4x because triangle ABC is equilateral.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The last step is to solve the proportion below<BR>\n" );
document.write( "AE:EB = k:5<BR>\n" );
document.write( "AE/EB = k/5<BR>\n" );
document.write( "(2x)/(2x) = k/5<BR>\n" );
document.write( "1 = k/5<BR>\n" );
document.write( "k/5 = 1<BR>\n" );
document.write( "k = 1*5<BR>\n" );
document.write( "k = 5\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "So AE:EB = k:5 turns into AE:EB = 5:5 and we could reduce that ratio to 1:1<BR>\n" );
document.write( "AE:EB = 1:1 indicates AE and EB are the same length.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Answer: A) 5<BR>\n" );
document.write( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );