Circles tell the area

The key to this problem is knowing that $B_1, B_2, B_3, B_4$ is cyclic, which can be easily proved using inversion. Denote $A'_k$ to be the point which $A_k$ maps to after inversion.

Inverting through $A_1$ maps the two circles which include $A_1$ into two sides of a triangle with vertices $B'_1, A'_2, A'_4$ . Since this inversion must map the circumcircle of $A_1A_2A_3A_4$ into a line, therefore, $A'_3$ must lie on line $A'_2A'_4$ .

The two circles which include $A_3$ must then map into two circles which pass through $(A'_2, A'_3),(A'_3, A'_4)$ , respectively. The circles intersect lines $A_1A'_2$ , $A_1A'_4$ at $B'_2$ and $B'_4$ respectively, and intersect at $B'_3$ . By simple angle chasing, we can see that $B'_1B'_2B'_3B'_4$ is a cyclic quadrilateral, therefore reinverting gives us the desired result.

This implies that the circumradius of the triangle $B_1B_3B_4$ is also 5.

And, the formula for the area of a triangle is given by

$Area = \frac{ABC}{4R}$ , where $ABC$ denotes the product of the sides, and $R$ denotes the circumradius.

Our answer is then $\frac{40}{(4)(5)} = \boxed{2}$ .