2018 AMC 8 Problem 15

Compare the radius of a smaller circle to the radius of the larger circle.

Area scales by the square of the scale factor.

Let $A$ be the area of the large circle.

Since the diameter of each of the two smaller circles is itself the radius of the larger circle, the radius of each smaller circle is half that of the larger circle.

Symbolically, if we allow $r_l$ to be the radius of the large circle and $r_s$ to be the radius of each of the smaller circles: $$r_s = \dfrac12 r_l$$ As the area of the larger circle is equal to $A=\pi r_l^2,$ the area of the smaller circles are equal to $$\begin{align*}\pi r_s^2 &= \pi \left(\dfrac12 r_l\right)^2 \\&= \dfrac14 (\pi r_l^2)\\&=\dfrac14 A.\end{align*}$$ As the area of two of these smaller circles combined is equal to $1$ square unit, then it follows that $2\cdot \dfrac14 A=1$ square unit, implying that $A=2$ square units.

As the area of the shaded region is equal to the area of the larger circle $(A)$ minus the combined area of the two smaller circles $(1),$ the area of the shaded region is $A - 1=2-1=1$ square unit.

Thus, the correct answer is D