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 O_{1}.

Now, we construct a circle whose center is O_{1} and radius AO_{1}.

Next we find the midpoint, O_{2}, of AB and construct a second
circle with
center O_{2} and radius AO_{2}.

Now the lune we are seeking is the area of circle O_{2} that is
not
contained within circle O_{1}.

Now we will prove that this yellow lune is in fact quadrable by showing
that it is equal in area to triangle ABO_{1}, 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 AEBO_{1} is half the area of the semicircle ABC,
therefore the
area of the semicircle AFB is equal to the area of quadrant
AEBO_{1}. So we have that
Area(AFB)-Area(AEBO_{1})=Area(AEBO_{1})-Area(AEBO_{2}).
Therefore
Area(LuneAFBE)=Area(TriangleABO_{1}) giving us that the lune is
in face
quadrable.

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