document.write( "Question 919088: what is the area of the circle inscribed inside an equilateral triangle whose area is 4 square root of 3 square centimeters? \n" ); document.write( "
Algebra.Com's Answer #557481 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "
\r\n" );
document.write( "Draw the altitude AD, which divides the base BC into two equal parts,\r\n" );
document.write( "BD = DC.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Since TABC is equilateral, AB = BC = 2·BD. \r\n" );
document.write( "\r\n" );
document.write( "By the Pythagorean theorem on right triangle BAD,\r\n" );
document.write( "\r\n" );
document.write( "BDČ+ADČ=ABČ\r\n" );
document.write( "BDČ+ADČ=(2·BD)Č\r\n" );
document.write( "BDČ+ADČ=4·BDČ\r\n" );
document.write( "    ADČ=4·BDČ-BDČ\r\n" );
document.write( "    ADČ=3·BDČ\r\n" );
document.write( "     AD=√3·BD\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "The Area of TABC = \"1%2F2\"(BC)(AD)\r\n" );
document.write( "                 = \"1%2F2\"(2·BD)(√3·BD)\r\n" );
document.write( "                 = 4√3·BDČ\r\n" );
document.write( "\r\n" );
document.write( "We are told that the area of TABC is 4·√3\r\n" );
document.write( "So we have the equation \r\n" );
document.write( "\r\n" );
document.write( "      4√3 = √3·BDČ\r\n" );
document.write( "\r\n" );
document.write( "Divide both sides by √3\r\n" );
document.write( "\r\n" );
document.write( "        4 = BDČ\r\n" );
document.write( "        2 = BD\r\n" );
document.write( "\r\n" );
document.write( "Draw AO, where O is the center of the circle.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "OB bisects < ABC which is 60°, so < OBD is 30° and  BOD = 60°,\r\n" );
document.write( "so TBOD is a 30°-60°-90° triangle and so OB = 2·OD\r\n" );
document.write( "\r\n" );
document.write( "By the Pythagorean theorem applied to TBOD,\r\n" );
document.write( "\r\n" );
document.write( "BDČ+ODČ=OBČ\r\n" );
document.write( " 2Č+ODČ=(2·OD)Č\r\n" );
document.write( "  4+ODČ=4·ODČ\r\n" );
document.write( "  4+ODČ=4·ODČ\r\n" );
document.write( "      4=3·ODČ\r\n" );
document.write( "      \"4%2F3\"=ODČ\r\n" );
document.write( "      \r\n" );
document.write( "The area of a circle is given by\r\n" );
document.write( "\r\n" );
document.write( "    Area = p·(radius)Č\r\n" );
document.write( "         = p·ODČ               \r\n" );
document.write( "         = p·\"4%2F3\"\r\n" );
document.write( "         = \"4pi%2F3\"\r\n" );
document.write( "\r\n" );
document.write( "Edwin
\n" ); document.write( "
\n" );