Answer
$P=\frac{\pi}{4}$
Work Step by Step
In this problem, we are dealing with how much of the square is covered by the coin so the probability is the ratio of the target area of the coin to the area of the square:
$P=\frac{A_{c}}{A_{s}}$
In order to cover a vertex, the coin's center must be d/2 units away from a vertex. At most, the center would lie in the intersection of the lines so that it would cover 4 quarters of a circle which have an area of:
$A_{c}=4\left(\frac{1}{4}\right)\left[\pi\left(\frac{d}{2}\right)^{2}\right]=\frac{\pi d^{2}}{4}$
The area of a square at any point the where the coin lands is:
$A_{s}=d^{2}$
The probability is:
$P=\frac{\pi d^{2} / 4}{d^{2}}$
$P=\frac{\pi}{4}$
