SOLUTION: 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 tr

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.
Thanks for your help.

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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.
***
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
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