document.write( "Question 1086357: rod of 20 m length has an isosceles triangular base, where the equal sides are 12 cm each.of the included angle in the base is 40°,calculate:
\n" ); document.write( "a)thearea of the cross-section.
\n" ); document.write( "b)The volume of the mental rod.
\n" ); document.write( "c)the mass of the mental rod,if it is made of copper ,whose density is 9000kg/m3
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Algebra.Com's Answer #700525 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
.
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\n" ); document.write( "\"A=%281%2F2%29Bh\"
\n" ); document.write( "and
\n" ); document.write( "\"B%2F2=12cos%2840%29\"
\n" ); document.write( "\"B%2F2=9.19\"
\n" ); document.write( "\"B=18.385\"\"cm\"
\n" ); document.write( "and
\n" ); document.write( "\"h%5E2%2B%289.1925%29%5E2=12%5E2\"
\n" ); document.write( "\"h%5E2=59.4973\"
\n" ); document.write( "\"h=7.71\"
\n" ); document.write( "So then,
\n" ); document.write( "\"A=%281%2F2%29%287.71%29%2818.385%29\"
\n" ); document.write( "\"A=70.9\"\"cm%5E2\"
\n" ); document.write( "Convert the area to \"m%5E2\",
\n" ); document.write( "\"A=70.9%2F%28100%5E2%29\"
\n" ); document.write( "\"A=.00709\"\"m%5E2\"
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\n" ); document.write( "\"V=A%2AL\"
\n" ); document.write( "\"V=0.00709%2A20\"
\n" ); document.write( "\"V=0.1418\"\"m%5E3\"
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\n" ); document.write( ".
\n" ); document.write( "\"m=rho%2AV\"
\n" ); document.write( "\"m=9000%2A0.1418\"
\n" ); document.write( "\"m=1276\"\"kg\"
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