A carrom board ( 4 ft × 4 ft square) features the queen at its...

Let the centre of the carrom board be the origin $(0,0)$. The board is a square of side $4 \text{ ft}$, so its corners are at $(\pm 2, \pm 2)$.

Let the front edge be the line $y = 2$. The queen is struck from the centre, hits the front edge at some point $(x, 2)$, and rebounds into the hole behind the striking line, which is at $(2, -2)$.

According to the law of reflection, the point of incidence can be found by joining the initial point to the image of the final point across the reflecting edge.

The image of the hole $(2, -2)$ with respect to the front edge $y = 2$ is $(2, 6)$.

The line joining the origin $(0,0)$ to the image $(2, 6)$ has the equation $y = 3x$.

The point of intersection of this line with the front edge $y = 2$ is obtained by substituting $y = 2$, which gives $x = \dfrac{2}{3}$. Thus, the queen strikes the front edge at \left(\dfrac{2}{3}, 2\right)$.<br><br>The displacement of the queen from the front edge to the hole is the distance from \left(\dfrac{2}{3}, 2\right)$ to $(2, -2)$.

The magnitude of this displacement is:

$d = \sqrt{\left(2 - \dfrac{2}{3}\right)^2 + (-2 - 2)^2}$

$d = \sqrt{\left(\dfrac{4}{3}\right)^2 + (-4)^2}$

$d = \sqrt{\dfrac{16}{9} + 16} = \sqrt{\dfrac{160}{9}} = \dfrac{4}{3}\sqrt{10} \text{ ft}$