2016 AMC 10A Problem 15

The scrap is the dough minus the seven cookies, so first find the size of the dough.

1. Walk from the center of the dough straight out to its edge, through one outer cookie. You cross the center cookie's radius ($1$ inch), then the whole width of an outer cookie ($2$ inches), and then you are at the edge of the dough.
2. So the dough has radius $1 + 2 = 3$ inches and area $\pi \cdot 3^2 = 9\pi$.
3. Each cookie has area $\pi \cdot 1^2 = \pi$, so the seven cookies use $7\pi$.
4. The scrap has area $9\pi - 7\pi = 2\pi$.
5. The scrap cookie has the same thickness, so its top has the same area, $2\pi$. If its radius is $r$, then $\pi r^2 = 2\pi$, so $r^2 = 2$ and $r = \sqrt{2}$.

The answer is $\boxed{\textbf{(A)}\ \sqrt{2}}$.

"Same thickness" means volume is area times a fixed height, so you can compare areas only. When circles touch, their lengths add up along the straight line through their centers, which is how the dough's radius comes out as radius plus diameter.

Find the radius of the dough first. Going from the center of the dough straight outward: the center cookie has radius $1$, and an outer cookie tangent to it has diameter $2$, ending at the edge of the dough. So the dough has radius $1 + 2 = 3$ and area $9\pi$.

The seven cookies have total area $7\cdot\pi\cdot 1^2 = 7\pi$, leaving scrap of area $9\pi - 7\pi = 2\pi$.

Since the thickness is unchanged, the scrap cookie has the same area $2\pi$, so $\pi r^2 = 2\pi$ and $r = \sqrt 2$.

The answer is $\boxed{\textbf{(A)}\ \sqrt{2}}$.