document.write( "Question 165697: Triangle ABC is isoceles with ABis equal to AC 7.5cm and BC is 9cm. The height AD from A to BC , is 6cm. Find the area of triangle ABC. What will the height be from C to AB i.e CE \n" ); document.write( "
Algebra.Com's Answer #122151 by jojo14344(1513)\"\" \"About 
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\n" ); document.write( "I'll leave the sketch with you okay, & hopefully we can follow together:
\n" ); document.write( "We find first the \"Area=%281%2F2%29bh\"
\n" ); document.write( "where, \"BC=base=9cm\", & \"AD=height=6cm\"
\n" ); document.write( "So, \"A=%281%2F2%29%289%29%286%29\"
\n" ); document.write( "\"A=27cm%5E2\"
\n" ); document.write( "Now , next is interesting in finding CE?
\n" ); document.write( "We first get \"angle%28B%29\" by trigo function of right triangle. How?
\n" ); document.write( "Since AD split the base(BC) in to half, \"side%28BD%29=9%2F2=4.5cm\" right? Same as \"side%28DC%29\".
\n" ); document.write( "That forms a right triangle (ADB).
\n" ); document.write( "\"sin%28beta%29=opp%2Fhyp=AD%2FAB=6%2F7.5=0.80\"
\n" ); document.write( "\"%28beta%29=sin%5E-1%280.80%29\"
\n" ); document.write( "\"highlight%28%28beta%29=53.13%5Eo%29\"
\n" ); document.write( ".
\n" ); document.write( "Now we know line CE(?) cuts line AB into half:\"7.5%2F2=3.75cm=BE\", right?
\n" ); document.write( "That forms a triangle \"BCE\" and we'll use Cosine Law in getting line CE.
\n" ); document.write( "\"c%5E2=a%5E2%2Bb%5E2-2abcos%28beta%29\"
\n" ); document.write( "where ---\"system%28a=BC=9cm%2Cb=BE=3.75cm%2Cc=CE%29\"
\n" ); document.write( "Continuing,
\n" ); document.write( "\"c%5E2=9%5E2%2B3.75%5E2-2%289%29%283.75%29cos%2853.13%29\"
\n" ); document.write( "\"c%5E2=81%2B14.0625-40.50=95.0625-40.50\"
\n" ); document.write( "\"c=sqrt%2854.5625%29\"
\n" ); document.write( "\"highlight%28c=7.38cm=CE%29\"
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo\r
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