document.write( "Question 1039827: Thank you all beforehand for helping me out. I really appreciate your time and effort.\r
\n" ); document.write( "\n" ); document.write( "Find the area of triangle ABC if AB = BC = 12 and angle ABC = 120.
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Algebra.Com's Answer #654568 by josgarithmetic(39799)\"\" \"About 
You can put this solution on YOUR website!
Cut into two right triangles from point B to midpoint of AC.\r
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\n" ); document.write( "\n" ); document.write( "Look at either of these right triangles. BC is opposite a 90 degree angle, a base leg is opposite a 60 degree angle, and altitude of the isosceles is opposite the 30 degree angle. SPECIAL RIGHT TRIANGLE.\r
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\n" ); document.write( "\n" ); document.write( "Let 6 be the short leg of the special 30-60-90 right triangle.
\n" ); document.write( "12 is the length of the hypotenuse.
\n" ); document.write( "Let y be altitude or the other leg.\r
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\n" ); document.write( "\n" ); document.write( "\"y%5E2%2B6%5E2=12%5E2\"
\n" ); document.write( "\"y%5E2=144-36\"
\n" ); document.write( "\"y%5E2=108\"
\n" ); document.write( "\"y%5E2=2%2A54=2%2A2%2A27=2%2A2%2A3%2A3%2A3\"
\n" ); document.write( "\"highlight_green%28y=6sqrt%283%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Put attention onto the original isosceles triangle.\r
\n" ); document.write( "\n" ); document.write( "You want the area.
\n" ); document.write( "\"%281%2F2%2912%2A6sqrt%283%29\", one-half of base times height;
\n" ); document.write( "\"highlight%2836%2Asqrt%283%29%29\"
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