Question 1172202
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Three 60 degrees sectors of circles with radius 2 m MAKE one half circle piece of the radius 2 m.


So, compare the area of the equilateral triangle with the side of a= 4 m against the area of the half circle of the radius r = 2 m.


    the difference = {{{a^2*(sqrt(3)/4)}}} - {{{(1/2)*pi*r^2}}} = {{{4^2*(sqrt(3)/4) - (1/2)*3.14*2^2}}} = 0.6648 m^2  (rounded).    <U>ANSWER</U>
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Solved.