2021 Fall AMC 10A Problem 19

The side length of the inner square traced out by the inner circle is $s - 4.$ There are also the small pieces remaining in the corner. These form a total area of $$(1 + 1)^2 - \pi 1^2 = 4 - \pi.$$ Therefore, $$A = s^2 - (s - 4)^2 - (4 - \pi)$$$$= 8s - 20 + \pi.$$ The outer disk traces out an area that is comprised of $4$ rectangles and $4$ quarter-circles. The rectangles have area $s \cdot 2 = 2s$ and the quarter-circles form a circle with radius $2$ and area $4 \pi.$ This gives us $$2A = 8s + 4 \pi.$$ Equating the two equations we get $$8s + 4 \pi = 2(8s - 20 + \pi).$$ Solving yields $$8s = 40 + 2 \pi$$$$s = 5 + \dfrac{\pi}{4}.$$ Thus, A is the correct answer.