Within a farmer's pastures, there is a one-acre tract of land shaped like a right triangle. At the midpoint of the hypotenuse is a post, to which a dog is tethered with just enough rope to reach the endpoints of the hypotenuse. There are also posts at the midpoints of the legs of the triangle; to each of which is tethered a sheep with just enough rope to reach the endpoints of its respective leg of the triangle.
How much space do the sheep have to graze in (collectively) without having to worry about the dog reaching them?
The situation can be re-described thusly:
Take a circle and draw the horizontal diameter.
Then stick at the center of the circle... stick a post and tether the dog to it.
From the end points of the diameter, draw line segments to ANY point in the top half of the circle. These two line segments and the diameter will form a right triangle, and if we scale it up to the size of one acre, we describe the farmer's pasture. The circle represents the range of the dog.
Now, you will note the the COMPLETE triangle (tract of land) is within range of the dog. So, the sheep must content themselves to their respective circles OUTSIDE the larger (dog's circle).
Since I will assume that this is the same answer regardless of whether or not the respective legs are equal to each other... I will assume (for simplicity) that the two legs are √2 in length (and the diameter is 2 units in length).... then the area of the triangle tract is 1/2 * √2 * √2 (1/2 x base x height) = 1 acre.
One sheep's respective semicircle is 1/2 x π x (√2/2)² = π/4.
But we must subtract the area OF THE SHEEP'S SEMICIRCLE that the dog can reach... which is a quarter of the dog's circle less half the triangular tract.
One quarter of the dog's region is: π/4. And half the triangular area is 1/2.
so... the region of interest is π/4 - 1/2.
Subtracting the region of interest from the sheep's semicircle is:
π/4 - (π/4 - 1/2)
= 1/2 acre
Since both sheep can get to its own 1/2 acre... the total is twice that...
= 1 acre