Question 134189
THE AREA OF THE CIRCLE IS:
A=PIR^2
A=3.14*12^2
A=3.14*144
A=452.16 
452.16(60/360)
452.16*(1/6)=75.36 IS THE AREA OF THE 60 DEGREE ARC.
THE HEIGHT IF THE EQUILATERAL TRIANGLE IS:
6^2+X^2=12^2 WHERE X=HEIGHT.
36+X^2=144
X^2=144-36
X^2=108
X=SQRT108
X=10.39 FOR THE HEIGHT.
THE EQUILATERAL TRIANGLE AREA IS:
A=BH/2
A=12*10.39/2
A=124.68/2
A=62.34
NOW SUBTRACT THE TRIANGLE AREA FROM THE 60 DEGREE ARC AREA:
75.36-62.34=13.02 ANSWER FOR THE SHADED AREA.