1. **State the problem:** Find the area of the irregular polygon with given side lengths. 2. **Analyze the figure:** The polygon has vertical and horizontal segments with one diagonal forming a triangular point. 3. **Break the figure into rectangles and triangles:** - The total height is 11 mm. - The left vertical side is 11 mm. - The polygon can be divided into a large rectangle and a small right triangle formed by the diagonal. 4. **Calculate the area of the large rectangle:** - Width = 7 mm (4 mm + 3 mm horizontal segments) - Height = 11 mm - Area = width \times height = $7 \times 11 = 77$ mm$^2$ 5. **Calculate the area of the small triangle:** - The triangle is right-angled with legs 3 mm (horizontal) and 4 mm (vertical) - Area = $\frac{1}{2} \times 3 \times 4 = 6$ mm$^2$ 6. **Subtract the triangle area from the rectangle area:** $$\text{Area} = 77 - 6 = 71 \text{ mm}^2$$ 7. **Final answer:** The area of the figure is **71 square millimeters**.