Calculation of Area of right angled triangle - Apostol exercise 1.7 problem 2 "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."
We have to make use of the axioms for  area as a set function . 
Well , I was able to show that a right angled triangle is measurable.
As a right angled triangle can be shown to be the intersection of two rectangles.
One which shares its two sides with the two non-hypotenuse sides of the right angled triangle. And the other rectangle has the hypotenuse of the triangle as its side.
Now my query is do we use these two rectangles to calculate the area of the triangle ?
I couldn't do it that way. 
Or do we have to show that a rectangle is composed of two right angled triangles which are congruent. 
(This one seems easier. But in Apostol's Calculus , the notion of 'congruence' is given in terms of sets of points. I.e two sets of points are congruent if their points can be put in one-to-one correspondence in such a way that distances are preserved.  Proving that the two triangles are congruent in this way seems to me to be too hard. I think , Apostol doesn't want us to use the congruence proofs of Euclidean Geometry , which is justified I guess.)
In the end,  I couldn't use either of the above two methods , hence I used the method of exhaustion . 
(Apostol used this method to calculate the area under a parabola. I applied the same method by calculating the area under a straight line , i.e the hypotenuse.)
I wanted to know how to find the area using congruence. Do we have to use the notion of 'congruence' as Apostol has given it ?
 A: Since you clearly state "We have to make use of the axioms for area as a set function," lets list them: 
DEFINITION/AXIOMS 
We assume that there exists a class $M$ 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 $T − S\subseteq M$ 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. A step region is formed from a finite union of adjacent rectangles resting on a common base, i.e. $S\subseteq Q \subseteq T$. If there is a unique number $c$ such that $A(S) \le c \le A(T)$ for all such step regions $S$ and $T$, then $A(Q) = c$.

So we know by 2) that the intersection of two rectangles $R_1 \subseteq M$, $R_2 \subseteq M$ is in $M$. Use the axioms/properties listed above to determine what this means for the triangle formed by the intersection of two such rectangles.
Note also that Apostol defines a rectangle as follows: $R =  \{ (x,y)| 0\le x\leq h, \text{ and}\; 0\le y\le k \}$. 
Note that the union of two congruent right triangles can be positioned to form a rectangle, and alternatively, a rectangle's diagonal forms two congruent right triangles. 
Refer to how Apostol defines congruence: to show two triangles (or two rectangles, or two sets of points, in general) are congruent, show that there exists a bijection between the two sets of points such that distances are preserved. 

Note that this problem can be generalized to all triangles, not just right triangles. To see this, you need only note that every triangle is the union of two right triangles.
A: The notion of 'congruence' as Apostol has given is the well-known "Side-Side-Side" formula as mentioned here.
If we split a rectangle by anyone of the diagonal, we can prove the two right triangles produced, are congruent and hence of the same area.
So, the area of a right triangle= $\frac12$ the area of the rectangle.
