1. The problem involves finding the area of a trapezoidal cross-section with top width $4.8$ m, bottom width $1.2$ m, and height $1.5$ m. 2. The formula for the area $A$ of a trapezoid is $$A = \frac{(b_1 + b_2)}{2} \times h$$ where $b_1$ and $b_2$ are the lengths of the two parallel sides and $h$ is the height. 3. Substitute the given values: $$A = \frac{(4.8 + 1.2)}{2} \times 1.5$$ 4. Simplify inside the parentheses: $$4.8 + 1.2 = 6.0$$ 5. Calculate the area: $$A = \frac{6.0}{2} \times 1.5 = 3.0 \times 1.5 = 4.5$$ square meters. 6. The volume of the channel is the base area times the length (height in the volume formula), but since only cross-section is given, the area is $4.5$ m$^2$. Next, for the triangle with vertices at $(2,2)$, $(3,4)$, and $(4,2)$: 1. The base is the segment between $(2,2)$ and $(4,2)$, so base length is $$4 - 2 = 2$$ units. 2. The height is the vertical distance from the base to the vertex $(3,4)$, which is $$4 - 2 = 2$$ units. 3. The area of a triangle is $$A = \frac{1}{2} \times \text{base} \times \text{height}$$ 4. Substitute values: $$A = \frac{1}{2} \times 2 \times 2 = 2$$ square units. Final answers: - Trapezoidal cross-section area: $4.5$ m$^2$ - Triangle area: $2$ square units