In case the description is not clear enough, here is the full question:
"Consider a square with side length one. Choose a uniformly-distributed random point on each side independently. If the random points are connected to form a quadrilateral, what are the mean and variance of the area of the quadrilateral?"
I suppose the point on each side of the square is a uniform $(0,1)$ random variable $X_i$ wtih $i = 1,2,3,4$. So the $4$ vertices would be $p_1 = (0, X_1)$, $p_2 = (X_2, 1)$, $p_3 = (1, X_3)$, and $p_4 = (X_4, 0)$.
Does anyone see a simple, elegant solution to this problem? Thanks!
Notice that the area is $1$ minus the sum of the area of 4 complemenentary triangles:
And if you write down the formula for the area of each triangle, you will find that they have the same distribution. Indeed, let $S_i$ be the area of the triangle with vertices $p_i, a_i, p_{i+1}$ (with the convention that $p_5 = p_1$). Then
\begin{align*} S_1 &= \tfrac{1}{2}(1-X_1)X_2, & S_2 &= \tfrac{1}{2}(1-X_2)(1-X_3), \\ S_3 &= \tfrac{1}{2}X_3(1-X_4), & S_4 &= \tfrac{1}{2}X_1 X_4 \end{align*}
are products of two independent uniform r.v.s multiplied by $\frac{1}{2}$. That is, if $U,V$ are i.i.d. and uniformly distributed on $[0,1]$, then each $S_i$ has the same distribution as $\frac{1}{2}UV$ and hence
$$ \Bbb{E}[\text{area}] = 1 - \Bbb{E}[S_1+S_2+S_3+S_4] = 1 - 4\Bbb{E}\left(\tfrac{1}{2}UV\right) = \tfrac{1}{2}$$
is the answer.