BC is the diameter of circle C2 so we can say that $$\angle BAC=90^{\circ\ }$$ as angle in the semi circle is $$90^{\circ\ }$$
Therefore overlapping area = $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
AB= AC = 8 cm and as $$\angle BAC=90^{\circ\ }$$, so we can conclude that BC= $$8\sqrt{2}$$ cm
Radius of C2 = Half of length of BC = $$4\sqrt{2}$$ cm
Area of C2 = $$\pi\left(4\sqrt{2}\right)^2=32\pi$$ $$cm^2$$
A is the centre of C1 and C1 passes through B, so AB is the radius of C1 and is equal to 8 cm
Area of the minor sector made be BC in C1 = $$\frac{1}{4}$$(Area of circle C1) - Area of triangle ABC = $$\frac{1}{4}\pi\left(8\right)^2-\left(\frac{1}{2}\times8\times8\right)=16\pi-32$$ $$cm^2$$
Therefore,
Overlapping area between the two circles= $$\frac{1}{2}$$(Area of circle C2) + Area of the minor sector made be BC in C1
= $$\frac{1}{2}\left(32\pi\right)\ +\left(16\pi-32\right)=32\left(\pi-1\right)$$ $$cm^2$$