A square in the coordinate plane has vertices whose $y$-coordinates are $0$, $3$, $7$, and $4$. What is the area of the square?

1. Opposite vertices of a square have the same sum of $y$-coordinates ($0+7=3+4$), so the vertices with $y=0$ and $y=7$ are opposite. We can assume $A(0,0), B(b, 3), C(c, 7), D(d, 4)$.
2. Then we have $c=b+d$, $AB\bot AD, bd=-12$, $AB=AD$, $b^2+9=d^2+16$, so $b=4, d=-3, c=1$.
3. $AB^2=b^2+9=25$, so the area is $25$.