1. **Problem Statement:** Find the area of the irregular polygon with given side lengths: 15 mm (left vertical), 9 mm (bottom horizontal), 6 mm (right vertical), and smaller segments 2 mm, 2 mm, 2 mm forming a protrusion on the upper right. 2. **Approach:** We can divide the figure into two rectangles: a larger rectangle and a smaller rectangular protrusion on the top right. 3. **Dimensions of the larger rectangle:** - Height: total left side length = 15 mm - Width: bottom side length = 9 mm 4. **Dimensions of the smaller rectangle (protrusion):** - Height: difference between left side and right vertical segment = $15 - 6 = 9$ mm - Width: sum of the two small horizontal segments = $2 + 2 = 4$ mm 5. **Calculate the area of the larger rectangle:** $$\text{Area}_1 = 15 \times 9 = 135 \text{ mm}^2$$ 6. **Calculate the area of the smaller rectangle:** $$\text{Area}_2 = 9 \times 4 = 36 \text{ mm}^2$$ 7. **Total area of the figure:** $$\text{Area} = \text{Area}_1 + \text{Area}_2 = 135 + 36 = 171 \text{ mm}^2$$ **Final answer:** The area of the figure is $171$ square millimeters.