document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1207717>Question 1207717</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Activity: The sum of the lengths of any two sides of a triangle is greater than the length of the third<BR>\n" );
document.write( "side. What can you conclude about AC in ABC? (Hint: Use the above example as a guideline.)<BR>\n" );
document.write( "2. BC = 7; AC = 2 + AB<BR>\n" );
document.write( "3. AB + AC = 5; AC + BC = 4<BR>\n" );
document.write( "4. BC + AC = 22; AB + BC = 12\r<BR>\n" );
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document.write( "These are the only questions I have. I don't want to make three different emails of practically the same thing. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #845745 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=845745\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Triangle Inequality Theorem<BR>\n" );
document.write( "The sum of the lengths of any two sides of a triangle is greater than the length of the third side.\r<BR>\n" );
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document.write( "AB, BC, and AC are the three sides of triangle ABC.\r<BR>\n" );
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document.write( "The Triangle Inequality Theorem generates these 3 inequalities<BR>\n" );
document.write( "AB + BC > AC<BR>\n" );
document.write( "AB + AC > BC<BR>\n" );
document.write( "AC + BC > AB<BR>\n" );
document.write( "I'll refer to them as inequality (1) through (3).\r<BR>\n" );
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document.write( "Problem 2\r<BR>\n" );
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document.write( "We're given that<BR>\n" );
document.write( "AC = 2 + AB\r<BR>\n" );
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document.write( "Let's solve for AB<BR>\n" );
document.write( "AC = 2 + AB<BR>\n" );
document.write( "AB = AC - 2\r<BR>\n" );
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document.write( "Then apply this to inequality (1)<BR>\n" );
document.write( "AB + BC > AC<BR>\n" );
document.write( "AC - 2 + 7 > AC<BR>\n" );
document.write( "AC + 5 > AC<BR>\n" );
document.write( "5 > 0<BR>\n" );
document.write( "The last inequality is always true, so the first inequality is always true.<BR>\n" );
document.write( "We don't really generate anything interesting here, so let's move on.\r<BR>\n" );
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document.write( "Now move to inequality (2)<BR>\n" );
document.write( "AB + AC > BC<BR>\n" );
document.write( "AC-2 + AC > 7<BR>\n" );
document.write( "2AC-2 > 7<BR>\n" );
document.write( "2AC > 7+2<BR>\n" );
document.write( "2AC > 9<BR>\n" );
document.write( "AC > 9/2<BR>\n" );
document.write( "AC > 4.5<BR>\n" );
document.write( "Inequality (2) is only true when AC is larger than 4.5\r<BR>\n" );
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document.write( "Lastly inequality (3)<BR>\n" );
document.write( "AC + BC > AB<BR>\n" );
document.write( "AC + 7 > AC-2<BR>\n" );
document.write( "7 > -2<BR>\n" );
document.write( "This is always true.<BR>\n" );
document.write( "Like with the first case shown above, nothing interesting is here.\r<BR>\n" );
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document.write( "To wrap up problem 2, if triangle ABC has sides BC = 7 and AC = 2+AB, then we conclude that <font color=red>AC > 4.5</font>\r<BR>\n" );
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document.write( "Problem 3\r<BR>\n" );
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document.write( "AB + AC = 5 solves to AB = 5 - AC<BR>\n" );
document.write( "AC + BC = 4 solves to BC = 4 - AC\r<BR>\n" );
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document.write( "Plug them into inequality (1)<BR>\n" );
document.write( "AB + BC > AC<BR>\n" );
document.write( "5-AC + 4-AC > AC<BR>\n" );
document.write( "9-2AC > AC<BR>\n" );
document.write( "9 > 3AC<BR>\n" );
document.write( "AC < 9/3<BR>\n" );
document.write( "AC < 3\r<BR>\n" );
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document.write( "Also, plug them into inequality (2)<BR>\n" );
document.write( "AB + AC > BC<BR>\n" );
document.write( "5-AC + AC > 4-AC<BR>\n" );
document.write( "5 > 4-AC<BR>\n" );
document.write( "5-4 > -AC<BR>\n" );
document.write( "1 > -AC<BR>\n" );
document.write( "AC > -1<BR>\n" );
document.write( "This will always be true since AC is a positive length.<BR>\n" );
document.write( "This subcase isn't useful so we move on.\r<BR>\n" );
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document.write( "Lastly inequality (3)<BR>\n" );
document.write( "AC + BC > AB<BR>\n" );
document.write( "AC + 4-AC > 5-AC<BR>\n" );
document.write( "4 > 5-AC<BR>\n" );
document.write( "4+AC > 5<BR>\n" );
document.write( "AC > 5-4<BR>\n" );
document.write( "AC > 1<BR>\n" );
document.write( "1 < AC\r<BR>\n" );
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document.write( "Combine 1 < AC with AC < 3 to conclude <font color=red>1 < AC < 3</font> is the answer to problem 3.\r<BR>\n" );
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document.write( "Problem 4\r<BR>\n" );
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document.write( "BC + AC = 22 solves to BC = 22 - AC\r<BR>\n" );
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document.write( "Plug this into the other equation so we get something in terms of AC.<BR>\n" );
document.write( "AB + BC = 12<BR>\n" );
document.write( "AB + 22-AC = 12<BR>\n" );
document.write( "AB - AC = 12-22<BR>\n" );
document.write( "AB - AC = -10<BR>\n" );
document.write( "AB = AC-10\r<BR>\n" );
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document.write( "The key equations we'll be using are <BR>\n" );
document.write( "BC = 22-AC and AB = AC-10<BR>\n" );
document.write( "Like with the previous problem, they are useful for substitutions.\r<BR>\n" );
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document.write( "Let's revisit inequality (1)<BR>\n" );
document.write( "AB + BC > AC<BR>\n" );
document.write( "AC-10 + 22-AC > AC<BR>\n" );
document.write( "12 > AC<BR>\n" );
document.write( "AC < 12\r<BR>\n" );
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document.write( "Now focus on inequality (2)<BR>\n" );
document.write( "AB + AC > BC<BR>\n" );
document.write( "AC-10 + AC > 22-AC<BR>\n" );
document.write( "2AC-10 > 22-AC<BR>\n" );
document.write( "2AC+AC > 22+10<BR>\n" );
document.write( "3AC > 32<BR>\n" );
document.write( "AC > 32/3<BR>\n" );
document.write( "AC > 10.6667 approximately\r<BR>\n" );
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document.write( "And now inequality (3)<BR>\n" );
document.write( "AC + BC > AB<BR>\n" );
document.write( "AC + 22-AC > AC-10<BR>\n" );
document.write( "22 > AC-10<BR>\n" );
document.write( "22+10 > AC<BR>\n" );
document.write( "32 > AC<BR>\n" );
document.write( "AC < 32\r<BR>\n" );
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document.write( "Combine or overlap AC < 12 with AC < 32, so we determine overall that AC < 12.<BR>\n" );
document.write( "If some length of AC makes AC < 12 true, then it autometically makes AC < 32 true as well (but not always the other way around).\r<BR>\n" );
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document.write( "Combine 32/3 < AC with AC < 12 to determine <font color=red>32/3 < AC < 12</font> <BR>\n" );
document.write( "In other words, <font color=red>10.6667 < AC < 12</font> when going with the approximate form.\r<BR>\n" );
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document.write( "Answers:<BR>\n" );
document.write( "Problem 2)  <font color=red>AC > 4.5</font><BR>\n" );
document.write( "Problem 3)  <font color=red>1 < AC < 3</font><BR>\n" );
document.write( "Problem 4)  <font color=red>32/3 < AC < 12</font>  (where 32/3 = 10.6667 approximately)<BR>\n" );
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