A triangle has vertex $A=(3,0), B=(0,3)$ and $C$, where $C$ is on the line $x+y=7$. What is the area of the triangle?
Observe that the height $h$, relative to the side $A B$, of all triangles $A B C$ that have the vertex $C$ on the line $x+y=7$, is the same, since the latter line is parallel to the line passing through $A$ and $B$. Therefore, these triangles all have the same area, namely:

$$
\frac{A B \times h}{2}
$$

We need to determine $A B$ and $h$. Since $A B$ is the hypotenuse of a right triangle that has both legs equal to $7-4=3$, it follows from the Pythagorean Theorem that:

$$
A B=\sqrt{3^{2}+3^{2}}=\sqrt{18}=3 \sqrt{2}
$$

It remains to calculate $h$, which is the distance between the parallel lines. The line $x+y=7$ is determined by the points $C=(7,0)$ and $D=(0,7)$. The line $x=y$ is perpendicular to the parallel lines above and forms a $45^{\circ}$ angle with the $O Y$ axis. Let $M$ be the foot of the perpendicular from $B$ to the line $x+y=7$. Therefore, the triangle $B M C$ is a right isosceles triangle with legs equal to $h$ and hypotenuse $7-3=4 \text{ cm}$.

From the Pythagorean Theorem it follows that:

$$
4^{2}=h^{2}+h^{2} \Rightarrow h=2 \sqrt{2}
$$

Finally, the area sought is:

$$
\frac{3 \sqrt{2} \times 2 \sqrt{2}}{2}=6
$$