1. **Stating the problem:** We are given an irregular polygon with the following side lengths: top and bottom horizontal sides are 9 ft each, left vertical side is 6 ft, right vertical side is 7 ft, and there are stepped-in segments at the top-right corner measuring 3 ft and 2 ft horizontally and 2 ft vertically. 2. **Goal:** Calculate the area of this irregular polygon. 3. **Approach:** We can divide the polygon into simpler shapes (rectangles and squares), calculate their areas, and then sum or subtract as needed. 4. **Step 1: Break down the polygon:** - The main rectangle has width 9 ft and height 6 ft. - The stepped-in part on the top-right can be seen as a smaller rectangle of width 3 ft and height 2 ft, and a smaller rectangle of width 2 ft and height 2 ft. 5. **Step 2: Calculate areas:** - Area of main rectangle: $$9 \times 6 = 54$$ sq ft. - Area of the stepped-in rectangle (3 ft by 2 ft): $$3 \times 2 = 6$$ sq ft. - Area of the smaller stepped-in rectangle (2 ft by 2 ft): $$2 \times 2 = 4$$ sq ft. 6. **Step 3: Calculate total area:** - Since the stepped-in parts are cut out from the main rectangle, subtract their areas: $$54 - 6 - 4 = 44$$ sq ft. 7. **Step 4: Adjust for the right vertical side:** - The right vertical side is 7 ft, which is 1 ft taller than the main rectangle height (6 ft). - This extra 1 ft height with width 2 ft (bottom step) adds an area: $$2 \times 1 = 2$$ sq ft. 8. **Step 5: Final area:** $$44 + 2 = 46$$ sq ft. **Answer:** The area of the irregular polygon is **46 sq ft**.