Question 873359
If the vertex angle of a square with sides measuring 8 units is trisected.
Find the area of the middle triangle (has essentially four sides) and find the areas of the two right triangles. 
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Working with (30-60) right triangles with leg of 8 and leg x opposite 30˚
tan 30˚=x/8
x=8*tan30˚=8*1/√3=8/√3
Area for one of the right triangles=1/2*base*height=1/2*(8/√3)*8=32√3 sq units
Area for two of the right triangles=64√3
Total area of square=8^2=64 sq units
Area of middle triangle=64-64/√3=(√3*64-64)/√3=27.04 sq units