Bases $AB$ and $CD$ of trapezoid $ABCD$ are each perpendicular to leg $AD$. If $CD = 6$, $AB = 8$, and $AD = 10$, find the average of the areas of all noncongruent triangles two of whose vertices are $A$ and $D$ and whose third vertex is a point on $BC$.
What I tried: My first thought was to use an integral as a vertex $P$ travels from $C$ to $B$, and find some formula for the area of $\triangle DPA$ in terms of the distance $P$ has traveled. The "noncongruent" part is making it tricky, though; I don't know at what point we're going to start running into triangles we've already covered.
Note that you will never run into triangle you have already covered, as areas of all triangles will be different, as the height changes but base does not!
Also, the area varies linearly from $C $ to $ B$. So you can say the required average is half of the sum of areas of $\triangle ADC$ and $\triangle ADB$:
$$A = \dfrac{6\cdot10+8\cdot 10}{2\times2} = 35 \text{ sq. units}$$