document.write( "Question 1114771: I have a triangle. One of its angles is divided into two by a bisector. One of the two parts of the divided side is x and the total of two parts is 12. One of the other sides of the triangle is 9 and the other is not given. Find x.\r
\n" ); document.write( "\n" ); document.write( "If my explanation of the problem is not clear you can type angle bisector theorem practice into the google and click the first thing that came out. I don’t understand the 10th problem.
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Algebra.Com's Answer #729694 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!
your triangle is right angle triangle\r
\n" ); document.write( "\n" ); document.write( "one leg is \"12\", the other leg is \"9\", so you use Pythagorean theorem to find the length of the hypotenuse \"c\"\r
\n" ); document.write( "\n" ); document.write( "\"c%5E2=12%5E2%2B9%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"c%5E2=144%2B81\"\r
\n" ); document.write( "\n" ); document.write( "\"c%5E2=225\"\r
\n" ); document.write( "\n" ); document.write( "\"c=sqrt%28225%29\"\r
\n" ); document.write( "\n" ); document.write( "\"c=15\"->the hypotenuse\r
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\n" ); document.write( "\n" ); document.write( "since one leg is \"12\", the other leg is \"9\", on the leg (length\"12\") one part \"x\", than the other part is \"12-x\"\r
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\n" ); document.write( "\n" ); document.write( "now use proportion:\r
\n" ); document.write( "\n" ); document.write( "\"15%3A%2812-x%29=9%3Ax\"....solve for \"x\"\r
\n" ); document.write( "\n" ); document.write( "\"15x=%2812-x%299\"\r
\n" ); document.write( "\n" ); document.write( "\"15x=108-9x\"\r
\n" ); document.write( "\n" ); document.write( "\"15x%2B9x=108\"\r
\n" ); document.write( "\n" ); document.write( "\"24x=108\"\r
\n" ); document.write( "\n" ); document.write( "\"x=108%2F24\"\r
\n" ); document.write( "\n" ); document.write( "\"x=4.5\"\r
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