Prove that every right triangular region is measurable because it can be obtained as the intersection of two rectangles. Prove that every triangular region is measurable and its area is one half the product of its base and altitude.
I'm supposed to prove the above statement from the following area axioms:
We assume that there exists a class of measurable sets in the plane and a set function $a$ whose domain is $M$ with the following properties:
1) $a(S) \geq 0$ for each set $S$ in $M$.
2) If $S$ and $T$ are two sets in $M$ their intersection and union is also in $M$ and we have: $a(S \cup T) = a(S) + a(T) - a(S \cap T)$
3)If $S$ and $T$ are in $M$ with $S \subseteq T$ then $T − S$ is in $M$ and $a(T − S) = a(T) − a(S)$.
4) If a set $S$ is in $M$ and $S$ is congruent to $T$ then $T$ is also in $M$ and $a(S) = a(T)$.
5) Every rectangle $R$ is in $M$. If the rectangle has length $h$ and breadth $k$ then $a(R) = hk$.
6) Let $Q$ be a set enclosed between two step regions $S$ and $T$ so that $S \subseteq Q \subseteq T$. If there is a unique number $c$ such that $a(S) \leq c \leq a(T)$ for all such step regions $S$ and $T$, then $a(Q) = c$.
I can see from axiom 2 that the intersection of 2 rectangles is measurable, but I can't think of how to use that to get that the area of the intersection is $bh \over 2$.