1. **State the problem:** Find the area of the compound figure composed of rectangles and triangles with given side lengths. 2. **Analyze the figure:** The figure can be divided into simpler shapes: rectangles and triangles. 3. **Break down the figure:** - The left vertical side is 7 m. - From the top left corner, going down 4 m, then right 5 m, then down 3 m, then a diagonal 6 m to the right, and finally a vertical 9 m segment on the right side with a 9 m horizontal base. 4. **Calculate areas of simpler shapes:** - Rectangle 1 (top left): width 7 m, height 4 m $$\text{Area}_1 = 7 \times 4 = 28$$ - Rectangle 2 (middle right): width 5 m, height 3 m $$\text{Area}_2 = 5 \times 3 = 15$$ - Triangle formed by the diagonal 6 m segment and vertical 3 m segment: The base of this triangle is 6 m (horizontal), height is 3 m (vertical drop). $$\text{Area}_3 = \frac{1}{2} \times 6 \times 3 = 9$$ - Rectangle 3 (bottom right): width 9 m, height 9 m $$\text{Area}_4 = 9 \times 9 = 81$$ 5. **Sum all areas:** $$\text{Total Area} = 28 + 15 + 9 + 81 = 133$$ 6. **Final answer:** The area of the compound figure is **133 square meters**.