document.write( "Question 134189: Find the area of the shaded region!
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\n" ); document.write( "PLEASE HELP!\r
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Algebra.Com's Answer #98126 by checkley71(8403)\"\" \"About 
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THE AREA OF THE CIRCLE IS:
\n" ); document.write( "A=PIR^2
\n" ); document.write( "A=3.14*12^2
\n" ); document.write( "A=3.14*144
\n" ); document.write( "A=452.16
\n" ); document.write( "452.16(60/360)
\n" ); document.write( "452.16*(1/6)=75.36 IS THE AREA OF THE 60 DEGREE ARC.
\n" ); document.write( "THE HEIGHT IF THE EQUILATERAL TRIANGLE IS:
\n" ); document.write( "6^2+X^2=12^2 WHERE X=HEIGHT.
\n" ); document.write( "36+X^2=144
\n" ); document.write( "X^2=144-36
\n" ); document.write( "X^2=108
\n" ); document.write( "X=SQRT108
\n" ); document.write( "X=10.39 FOR THE HEIGHT.
\n" ); document.write( "THE EQUILATERAL TRIANGLE AREA IS:
\n" ); document.write( "A=BH/2
\n" ); document.write( "A=12*10.39/2
\n" ); document.write( "A=124.68/2
\n" ); document.write( "A=62.34
\n" ); document.write( "NOW SUBTRACT THE TRIANGLE AREA FROM THE 60 DEGREE ARC AREA:
\n" ); document.write( "75.36-62.34=13.02 ANSWER FOR THE SHADED AREA.\r
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