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 $bh \over 2$.

  • $\begingroup$ Hint: the rectangles will not have parallel sides. Once you show that half of a rectangle is measurable, the other half will be too, and as both regions are congruent, the formula for the area will follow from 4 and 5 $\endgroup$ – Barry Smith Apr 18 '12 at 21:39

Let $ABC$ be a right triangle, with right angle at $C$. Form two rectangles $R$ and $S$ as follows.

Rectangle $R$ is the natural one with $AB$ as a diagonal that splits $R$ into two halves, with $\triangle ABC$ as one of the halves.

Rectangle $S$ has $AB$ as one side, and the side parallel to $AB$ passes through $C$.

The intersection of $R$ and $S$ is $\triangle ABC$.

To decompose any triangle into two right triangles with uninteresting intersection, there is a natural construction. Be a little careful in describing the construction.

  • $\begingroup$ Well, that's the first piece of the problem nicely done. $\endgroup$ – Gerry Myerson Apr 19 '12 at 0:13

For the second piece of the problem, can you see that every triangle is the union of two right triangles?

EDIT: For the last part of the problem, a right triangle is half a rectangle, so properties 2) and 5) should get you the formula for right triangles. Then as noted every triangle is a union of two right triangles, so property 2 and the result for right triangles should get you the result for all triangles.

  • $\begingroup$ If a base angle is obtuse, the triangle is also the difference of two right triangle. $\endgroup$ – marty cohen Apr 19 '12 at 2:27

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