How to establish the formula for area of a triangle using the axioms of area? We have the following definition: 
AXIOMATIC DEFINITION OF AREA
We assume there exists a class of $M$ of measurable sets in the plane (i.e. subsets of the plane whose area can be defined) and a set function $a$, whose domain is $M$, with the following properties: 


*

*Non-Negative Property. For each set $S$ in $M$, we have $a(S) \geq 0$. 

*Additive Property. If $S$ and $T$ are in $M$, then $S \cup T$ and $ S \cap T$ are in $M$, and we have $$ a(S\cup T) = a(S) + a(T) - a(S \cap T).$$ 

*Difference Property. If $S$ and $T$ are in $M$ with $S \subseteq T$, then $T \setminus S$ is in $M$, and we have $$a(T \setminus S) = a(T) - a(S).$$ 

*Invariance Under Congruence. If a set $S$ is in $M$ and if $T$ is congruent to $S$, then $T$ is also in $M$, and we have $$a(T) = a(S). $$

*Choice Of Scale. Every rectangle $R$ is in $M$. If the edges of $R$ have lengths $h$ and $k$, then $a(R) = hk. $

*Exhaustion Property. Let $Q$ be a set that can be enclosed between two step regions $S$ and $T$, so that $$S \subseteq Q \subseteq T.$$ If there is one and only one number $c$ which satisfies the inequalities $$a(S) \leq c \leq a(T) $$ for all step regions $S$ and $T$ satisfying $$S \subseteq Q \subseteq T, $$ then $Q$ is measurable and $a(Q) = c$. 
Now using these axioms, how can we establish the formula for the area of a rectangle, i.e., Area $= bh/2$, where $b$ is the base and $h$ is the altitude? 
 A: Given a generic triangle $T$, draw the altitude $h_c$ orthogonal to the longest side $c$. It splits $T$ into two right triangles $T_A$, $T_B$ with legs $c_A$ and $h_c$, resp., $c_B$ and $h_c$, whereby $c_A+c_B=c$. The triangle $T_A$ is half of a rectangle with side lengths $c_A$ and $h_c$, and similarly $T_B$ is half of a rectangle with side lengths $c_B$ and $h_c$. From your axioms 2., 4., and 5. it then follows that $a(T)=c\cdot h_c$.
What we have glossed over here is that one would need a proof that the edges  of the triangles occurring in this construction have area $0$, so that we don't have to bother about some of these edges being counted twice.
A: Alternative approach:
It looks like you are using Apostol's "Calculus" 2nd edition (1966).  Assuming so, your question is answered by his theorem 1.15:
$\int_0^b x^p dx = \frac{b^{p+1}}{p+1}.$  Thus, $\int_0^b kx dx = \frac{kb^{2}}{2}.$ 
Simply construe each triangle to be the union of two right triangles.  Noting Apostol's definition of congruence (page 58, footnote), the right triangle whose hypotenuse has a downward slope will need to be rotated 180 degrees about the y axis, so that its slope is now upward.  Theorem 1.15 may then be applied to the rotated right triangle.
