Two $2$-inch squares are placed with their bases on the same line, $1$ inch apart. A third $2$-inch square is centered above the gap and rotated $45^\circ$, then lowered until it touches both squares, as shown. How many inches is the point $B$ from the line?

1. The rotated square has side $2$, so by theorem 25 its diagonal is $2\sqrt2$; let its center be at height $k$. Then $D=(0,k-\sqrt2)$ and $A=(-\sqrt2,k)$, so edge $DA$ has slope $-1$ and equation $y=k-\sqrt2-x$.
2. Edge $DA$ must pass through $G=(-0.5,2)$, the inner top corner of the left square (half of the $1$-inch gap, at height $2$): $2=k-\sqrt2+0.5$, so $k=1.5+\sqrt2$.
3. Point $B$ is $\sqrt2$ above the center, so $BF=k+\sqrt2=1.5+2\sqrt2=\boxed{1.5+2\sqrt2}$ inches.