Square $A B C D$ has $A$ and $B$ on the $x$-axis and $C$ and $D$ below the $x$-axis on the parabola with equation $y=x^{2}-4$. What is the area of $A B C D$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_d13f9eaf1d614231018ag-3.jpg?height=678&width=550&top_left_y=889&top_left_x=839)
Let $t$ be the $x$-coordinate of $B$ and note that $t>0$. The parabola has reflective symmetry across the $y$-axis, which implies $A B C D$ has reflective symmetry across the $y$-axis. Thus, the $x$-coordinate of $A$ is $-t$.

Therefore, $A B=2 t$, and so the area of $A B C D$ is $(2 t)^{2}=4 t^{2}$.

The point $C$ lies on the parabola with equation $y=x^{2}-4$ and has the same $x$-coordinate as $B$. Therefore, the coordinates of $C$ are $\left(t, t^{2}-4\right)$.

Since $B C$ is vertical, its length is the difference between the $y$-coordinates of $B$ and $C$.

It is given that $C$ is below the $x$-axis, so its $y$-coordinate is negative. The $y$-coordinate of $B$ is 0 , and so we have that $B C=0-\left(t^{2}-4\right)=4-t^{2}$.

Since $A B C D$ is a square, $A B=B C$, and so $2 t=4-t^{2}$.

Rearranging this equation gives $t^{2}+2 t-4=0$, and using the quadratic formula, we get

$$
t=\frac{-2 \pm \sqrt{2^{2}-4(-4)}}{2}=-1 \pm \sqrt{5}
$$

The quantity $-1-\sqrt{5}$ is negative, and we are assuming $t$ is positive, which means $t=-1+\sqrt{5}$. We know that the area of $A B C D$ is $4 t^{2}$, so the area is $4(-1+\sqrt{5})^{2}=4(6-2 \sqrt{5})=24-8 \sqrt{5}$.

ANSWER: $24-8 \sqrt{5}$