Random Quadrilaterals

Inspired by Delta Airlines, you and a graphic designer colleague are crafting a new logo for Fiddler Airlines. Your colleague starts with an equilateral triangle with a side length of $1$ unit. Next, you pick a random point inside the triangle, giving you four points: the three corners of the triangle and this fourth point inside. There are several (simple) quadrilaterals that can be drawn whose four vertices are these four points. Your colleague picks one of these quadrilaterals at random, with each having an equal probability. This quadrilateral will be the new logo.

On average, what can you expect the area of the quadrilateral to be?

Since you start with an equilateral triangle with unit side lengths, the total aread of the triangle is $A = \frac{\sqrt{3}}{4}$ square units. For any random chosen point, say $x,$ within this equilateral triangle, there are $\binom{3}{2} = 3$ possible choices of triangles to remove from the larger equilateral triangle to form our simple quadrilateral logo. Let's assume that we enumerate the areas of these three resulting, nonoverlapping triangles as $A_1(x),$ $A_2(x)$ and $A_3(x),$ then by design we have $$A_1(x) + A_2(x) + A_3(x) = A.$$ In order to obtain a quadrilateral, your colleague will uniformly randomly select one of these three triangles to remove from the initial unit equilateral triangle. Let's denote the area of the resulting random quadrilateral, conditioned on your initial choice of the point $x,$ as $Q \mid x,$ then we have \begin{align*} \mathbb{E} \left[ Q \mid x \right] &= \frac{1}{3} ( A - A_1(x) ) + \frac{1}{3} (A - A_2(x)) + \frac{1}{3} (A - A_3(x)) \\ &= A - \frac{A_1(x)+A_2(x)+A_3(x)}{3} \\ &= \frac{2}{3} A = \frac{\sqrt{3}}{6}.\end{align*} Since this equation holds regardless of your choice of $x,$ the expected area of the quadrilateral obtained is $$\mathbb{E}[Q] = \frac{\sqrt{3}}{6}.$$