2022 AMC 10A Problem 15

Notice that $7^2 + 24^2$ and $15^2 + 20^2$ are both the same. This forces $AC = 25$ since otherwise $\angle B$ and $\angle D$ would both be acute or obtuse, violating the fact that their sum is $180^{\circ}.$ Also since $\angle B$ is right, we know that $AC$ is the diameter of the circle. The area of the circle is then $\dfrac{625}{4} \pi.$ To find the area of the quadrilateral, we can find the area of each of the triangles, which is $$\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$$ $$84 + 150 = 234.$$ To find the area outside the quadrilateral, we subtract to get $$\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$$ Therefore, $$a + b + c = 625 + 936 + 4$$ $$= 1565.$$ Thus, D is the correct answer.