Question 823461
The side length of the hexagon and two sides from the central angle form an isosceles triangle, and you want the area of SIX of these triangles.  The circle is cut into sectors for each central angle to be 360/6=60 degrees.  


Now consider each isosceles triangle is composed of TWO right triangles of the 30-60-90 type.  You have TWELVE of these triangles making the hexagon.  The hypotenuse is 15 units.  The long leg is 2 times the short leg.  If the short leg is x, and the long leg is 2x, then:


{{{x^2+(2x)^2=15^2}}}
{{{4x^2+x^2=15^2}}}
{{{5x^2=15*15}}}
{{{x^2=45}}}
{{{x=3*sqrt(5)}}}


The area of ONE isosceles triangle is {{{(1/2)(base)(height)}}}; which is
{{{(1/2)(2*3*sqrt(5))(2*3*sqrt(5))}}}
{{{(1/2)36*5}}}
{{{18*5}}}
90 square units.
'
Six of these is {{{6*90=270}}} square units.