In a coordinate system, the vertices of triangle $ABC$ are: $A(0 ; 4), B(3 ; 0), C(c ; 6)$. The area of the triangle is 7. What is $c$, given that $0<c<3$?
Solution. The area of the rectangle is equal to the sum of the areas of the four triangles visible in the diagram:

$$
\begin{aligned}
& 18=\frac{3 \cdot 4}{2}+\frac{6(3-c)}{2}+\frac{2 c}{2}+7 \\
& 11=3 \cdot 2+3(3-c)+c
\end{aligned}
$$

from which \( c=2 \).

![](https://cdn.mathpix.com/cropped/2024_05_02_a74f21f127857d8681dbg-1.jpg?height=624&width=415&top_left_y=482&top_left_x=844)

For verification, the areas of the triangles are:

$$
\begin{array}{ll}
T_{1}=2, & T_{2}=6, \\
T_{3}=3, & T_{4}=18-(2+6+3)=7 .
\end{array}
$$