1. **State the problem:** We are given an irregular polygon with labeled sides and need to find its area in mm². 2. **Analyze the shape:** The figure has two vertical segments labeled 9.1 mm and 2 mm, both marked with arrows showing they are equal lengths (so the side above the 4.3 mm segment is 9.1 mm). The top horizontal segment is 12 mm, the bottom horizontal segment is 4.3 mm, and there is an angled segment labeled 8 mm. 3. **Break down the polygon:** To find the area, split the shape into simpler shapes (rectangles and triangles). 4. **Calculate the rectangle area:** The rectangle formed by the vertical sides 9.1 mm and the bottom horizontal 4.3 mm has an area: $$\text{Area}_{rectangle} = 9.1 \times 4.3 = 39.13 \ \text{mm}^2$$ 5. **Calculate the remaining area:** Consider the top right triangle formed by the difference in bases and height. The top horizontal segment is 12 mm and the bottom horizontal segment under this triangle is 4.3 mm, so the base length of the triangle is: $$12 - 4.3 = 7.7 \ \text{mm}$$ The vertical height of this triangle is 2 mm (the given side labeled 2 mm). 6. **Triangle area:** $$\text{Area}_{triangle} = \frac{1}{2} \times 7.7 \times 2 = 7.7 \ \text{mm}^2$$ 7. **Total area:** $$\text{Area}_{total} = 39.13 + 7.7 = 46.83 \ \text{mm}^2$$ **Final answer:** The area of the shape is **46.83 mm²**.