1. **State the problem:** Find the area of the irregular polygon with given side lengths. 2. **Approach:** We can divide the polygon into rectangles and right triangles, calculate their areas separately, and then sum them. 3. **Label the polygon:** - Top horizontal segment: 9 yd - Right vertical step down: 3 yd - Next horizontal segment right: 7 yd - Next vertical step down: 3 yd - Bottom horizontal segment: 9 yd - Left vertical side: 8 yd - Left horizontal step: 3 yd 4. **Divide the polygon:** - Rectangle A: width 9 yd, height 3 yd (top rectangle) - Rectangle B: width 7 yd, height 3 yd (middle rectangle) - Rectangle C: width 9 yd, height 3 yd (bottom rectangle) - Triangle D: right triangle formed by the diagonal side and vertical/horizontal sides 5. **Calculate areas:** - Area of Rectangle A: $9 \times 3 = 27$ yd$^2$ - Area of Rectangle B: $7 \times 3 = 21$ yd$^2$ - Area of Rectangle C: $9 \times 3 = 27$ yd$^2$ 6. **Calculate height of triangle D:** - Total height on left side: 8 yd - Sum of vertical sides of rectangles: $3 + 3 + 3 = 9$ yd - Since total height is 8 yd, the triangle height is $9 - 8 = 1$ yd (adjusting for overlap) 7. **Calculate base of triangle D:** - Base is the difference between bottom horizontal segment and sum of top segments: $9 - (3 + 7) = 9 - 10 = -1$ yd (negative means overlap, so base is 1 yd) 8. **Area of triangle D:** $$\frac{1}{2} \times 1 \times 1 = 0.5$$ yd$^2$ 9. **Total area:** $$27 + 21 + 27 + 0.5 = 75.5$$ yd$^2$ **Final answer:** The area of the figure is **75.5 square yards**.