document.write( "Question 1154274: The lengths of the sides of triangle ABC are often abbreviated by writing a = BC, b = CA, and c = AB. Notice that lower-case sides oppose upper-case vertices. Suppose now that angle BCA is right, so that a2 + b2 = c2. Let F be the foot of the perpendicular drawn fromCtothehypotenuseAB. Ifa=5,b=12andc=13,whatarethelengthsofFA, FB,andFC? Doesc=FA+FB? \n" ); document.write( "
Algebra.Com's Answer #776845 by mananth(16946)![]() ![]() You can put this solution on YOUR website! ![]() \n" ); document.write( "\n" ); document.write( "Triangle ACF ~ triangle ABC\r \n" ); document.write( "\n" ); document.write( "AC/AB =AF/AC\r \n" ); document.write( "\n" ); document.write( "12/13 = AF/12 \n" ); document.write( "AF =11.08\r \n" ); document.write( "\n" ); document.write( "In BCF and BAC\r \n" ); document.write( "\n" ); document.write( "BC/AB = BF/BC\r \n" ); document.write( "\n" ); document.write( "Plug values to get BF = 1.92\r \n" ); document.write( "\n" ); document.write( "CF^2 = BF * AF ( geometric mean property)\r \n" ); document.write( "\n" ); document.write( "Plug values to get CF = 4.61\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |