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Square Barn - Final
 

Okay. Here's the whole picture. Most regular polygons follow the same procedure. And it doesn't matter where the attachment point is, they all are done by using partial circles and result in a final triangle, from which you may have to subtract a portion of the barn. The base of the triangle may take a little more figuring and there may be a different angle from each side, but it is all doable. I believe this to be a universal solution. If you have any suggestions or questions let me hear from you. I have also included links to other published problems.
Tom Laidlaw

Angle A
= acos((b2+c2-a2)/(2bc))
= 79.82o

Area K = b*c*sinA/2
K = 40*14.14*.984/2 =278.385

or sqrt(s(s-a)(s-b)(s-c))
where s= (a+b+c)/2 = 278.385

blue quadrilateral
= triangle -1/2 barn
 278.385 - 50 =228.385

angle of wedge
= 135 - 79.82 = 55.18

So: total grazing area
 = 3/4 of a 50 ft. circle
plus 2 x 55.18/360 of a 40 foot circle plus quadrilateral

5890.486
  770.458
  770.458
  228.385
7659.787 sq. ft.