Let the two sides of the triangle have length x and y, and the hypotenuse z. The area outside the field the sheep can cover is pi*(x/2)2/2 + pi*(y/2)2/2.
The area inside the field, which can be covered by at least one sheep, is xy/2.
The part of the sheep's area which the dog can cover is everything on the side of the hypotenuse containing the field, or pi*(z/2)2/2.
So the area the sheep have to themselves is pi*(x/2)2/2 + pi*(y/2)2/2 + xy/2 - pi*(z/2)2/2 =
(pi/2)*((x/2)2 + (y/2)2 - (z/2)2) + xy/2 =
(pi/8)*(x2 + y2 - z2) + xy/2 =
(pi/8)*0 + xy/2 (because in a right triange x2 + y2 = z2) =
xy/2
The area of the field is also xy/2, which is one acre, so the answer is one acre.
Here is a simpler solution provided by David Nixon:
Pythagoras really states that the areas of similar shapes on two sides of a right triangle add to a third similar shape on the hypoteneuse. Here, we have the semicircle described by the dog on the hypoteneuse, and the two sheep on the other two sides. The area impinging on the sheep is the dog's semi-circle less the triangle. So the solution is Sheep - shaded areas = Sheep - (Dog -1) = 1.
Michael Shackleford, A.S.A.