document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1205456>Question 1205456</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Triangle ABC has sides AB = 7, BC = 8, and AC = 9.  Point C is reflected over line AB to create point C'.  Next, point B is reflected over line AC' to create point B'.  Find the area of triangle B'C'C. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #842275 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=842275\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Diagram<BR>\n" );
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document.write( "Triangle ABC has sides AB = 7, BC = 8, and AC = 9.<BR>\n" );
document.write( "That's equivalent to saying the sides are: c = 7, a = 8, b = 9<BR>\n" );
document.write( "Each lowercase letter represents the side opposite its uppercase letter (eg: side b is opposite angle B)\r<BR>\n" );
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document.write( "I'll use Heron's formula to compute the area of triangle ABC.<BR>\n" );
document.write( "First we need the semiperimeter<BR>\n" );
document.write( "s = (a+b+c)/2<BR>\n" );
document.write( "s = (8+9+7)/2<BR>\n" );
document.write( "s = 12<BR>\n" );
document.write( "Then we can compute the area<BR>\n" );
document.write( "area = sqrt(s*(s-a)*(s-b)*(s-c))<BR>\n" );
document.write( "area = sqrt(12*(12-8)*(12-9)*(12-7))<BR>\n" );
document.write( "area = 26.83281573 square units approximately\r<BR>\n" );
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document.write( "C reflects over line AB to land on C'<BR>\n" );
document.write( "Line segment CC' is perpendicular to AB. <BR>\n" );
document.write( "Let D be the intersection of segments CC' and AB. <BR>\n" );
document.write( "Segment DC is the altitude of triangle ABC when AB is the base. \r<BR>\n" );
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document.write( "We know AB = 7 but we don't know DC. But we can easily find it like so<BR>\n" );
document.write( "area = 0.5*base*height<BR>\n" );
document.write( "26.83281573 = 0.5*7*height<BR>\n" );
document.write( "26.83281573 = 3.5*height<BR>\n" );
document.write( "height = 26.83281573/3.5<BR>\n" );
document.write( "height = 26.83281573/3.5<BR>\n" );
document.write( "height = 7.66651878<BR>\n" );
document.write( "This is the approximate length of segment DC.\r<BR>\n" );
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document.write( "DC' is also this length because of the reflection.<BR>\n" );
document.write( "Then,<BR>\n" );
document.write( "CC' = DC + DC'<BR>\n" );
document.write( "CC' = DC + DC<BR>\n" );
document.write( "CC' = 2*DC<BR>\n" );
document.write( "CC' = 2*7.66651878<BR>\n" );
document.write( "CC' = 15.33303756\r<BR>\n" );
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document.write( "Also because of mirroring, B'C' = 8 since BC = 8. \r<BR>\n" );
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document.write( "Triangle B'C'C has these sides<BR>\n" );
document.write( "BC = 8<BR>\n" );
document.write( "CC' = 15.33303756 (approximate)<BR>\n" );
document.write( "If we could find side B'C, then we'd be able to use Herons formula to get the final answer.\r<BR>\n" );
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document.write( "Technically it is possible to find this missing side length. But it would take too much work I think. Luckily there's another way.\r<BR>\n" );
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document.write( "Focus entirely on triangle ABC. <BR>\n" );
document.write( "Since we know all three sides, we can use the law of cosines to find each of the missing angles. We'll focus on angle A.<BR>\n" );
document.write( "a^2 = b^2 + c^2 - 2*b*c*cos(A)<BR>\n" );
document.write( "cos(A) = (b^2 + c^2 - a^2)/(2bc)<BR>\n" );
document.write( "cos(A) = (9^2 + 7^2 - 8^2)/(2*9*7)<BR>\n" );
document.write( "cos(A) = 0.5238095238095<BR>\n" );
document.write( "A = arccos(0.5238095238095)<BR>\n" );
document.write( "A = 58.4118644948<BR>\n" );
document.write( "This is the approximate measure of angle BAC.<BR>\n" );
document.write( "It's also the approximate measure of angle BAC' because of the reflection.\r<BR>\n" );
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document.write( "Focus on right triangle C'DA.<BR>\n" );
document.write( "We found acute angle C'AD = 58.4118644948 that was computed above.<BR>\n" );
document.write( "The angle AC'D is the complement of this<BR>\n" );
document.write( "90 - 58.4118644948 = 31.5881355052\r<BR>\n" );
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document.write( "I'll skip showing the steps, but using the law of cosines will find that angle B'C'A is roughly 48.18968510421 degrees.\r<BR>\n" );
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document.write( "So,<BR>\n" );
document.write( "angleB'C'C = (angleB'C'A)+(angleAC'D)<BR>\n" );
document.write( "angleB'C'C = (48.18968510421)+(31.5881355052)<BR>\n" );
document.write( "angleB'C'C = 79.77782060941\r<BR>\n" );
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document.write( "-----------------------------------------------\r<BR>\n" );
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document.write( "I apologize if this solution seems a bit cluttered. I'll try to go back and rewrite things when I get the chance. But for now let's just move on.\r<BR>\n" );
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document.write( "To summarize so far we found the following:<ul><li>B'C' = 8</li><li>angle B'C'C = 79.77782060941 degrees (approximate)</li><li>CC' = 15.33303756 (approximate)</li></ul>We have the SAS (side angle side) case going on here. We know two sides of a triangle and the angle between. \r<BR>\n" );
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document.write( "Use the SAS triangle area formula to wrap things up.\r<BR>\n" );
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document.write( "area(triangle) = 0.5*side1*side2*sin(angle between those sides)<BR>\n" );
document.write( "area(B'C'C) = 0.5*(B'C')*(CC')*sin(angle B'C'C)<BR>\n" );
document.write( "area(B'C'C) = 0.5*(8)*(15.33303756)*sin(79.77782060941)<BR>\n" );
document.write( "area(B'C'C) = <font color=red>60.35862404571</font>\r<BR>\n" );
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document.write( "<font color=red>Answer: approximately 60.35862404571 square units</font><BR>\n" );
document.write( "Round that value however needed.\r<BR>\n" );
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document.write( "GeoGebra can be used to verify each claim that I have made above, and can also be used to verify the final answer. <BR>\n" );
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