Area under a Chord

The chord line and the the two  radii of 8  form a triangle of which one angle is 125 degrees and the other two are

27.5 degrees   (must add to 180 degrees)

  Draw a line of length 'h' perpindicular to the chord which bisects the 125 degree angle .

Now you have two equal triangles with angles of 90  125/2   and 27.5 degree angles

Now use law of sines to find 'h'  

sin 90 /8  = sin 27.5 /h = sin 62.5/(1/2chord)         

h = 3.694    and chord = 14.192

Then area of the triangle = 1/2 b h      where b is the chord

Area = 1/2 (14.192)(3.694) = 26.22 units^2

Then just find the area of the sector and subtract the area of the triangle

sector area     =   125/360 x  pi x 8^2 = 69.813 sq units

Now subtract the triangle for the area of the segment.....

Area of sector - Area of triangle  = Desired Area

The "formula" for the area is :

 r^2  [ (pi * 125/360)  -  (1/2) sin (125°)  ]  =

8^2 [  (pi * 125/360)  - (1/2)sin (125° ) ]  ≈  43.6  units^2