The third type of lune is constructed with a circle circumscribing a square.  We begin with the square ABCD and it's diagonals AC and BD with intersection O₁.

Now, we construct a circle whose center is O₁ and radius AO₁.


Next we find the midpoint, O₂, of AB and construct a second circle with center O₂ and radius AO₂.


Now the lune we are seeking is the area of circle O₂ that is not contained within circle O₁.


Now we will prove that this yellow lune is in fact quadrable by showing that it is equal in area to triangle ABO₁, which we know is quadrable since all triangles are squareable.  Begin by looking at triangle ABC.  We know that angle ABC is a right angle and that AB=BC so we can use the pythagorean theorem to say that (AC)²=(AB)²+(BC)²=2(AB)².  Now let's look at the ratio of the area of the sector AFB with the area of ABC.  Area(AFB)=(pi)(AB)² and Area(ABC)=(pi)(2)(AB)², so the ratio of Area(AFB):Area(ABC)=1:2.  This implies that the area of the quadrant AEBO₁ is half the area of the semicircle ABC, therefore the area of the semicircle AFB is equal to the area of quadrant AEBO₁.  So we have that Area(AFB)-Area(AEBO₁)=Area(AEBO₁)-Area(AEBO₂).  Therefore Area(LuneAFBE)=Area(TriangleABO₁) giving us that the lune is in face quadrable.

 
To view an interactive GSP sketch of this lune click here.