1. **State the problem:** Find the total area of a figure composed of a parallelogram and a semicircle attached to its left side. 2. **Identify given dimensions:** - Parallelogram base $b = 12$ units - Parallelogram height $h = 6$ units - Semicircle diameter $d = 6$ units (same as parallelogram height) 3. **Formulas to use:** - Area of parallelogram: $A_{parallelogram} = b \times h$ - Area of semicircle: $A_{semicircle} = \frac{1}{2} \pi r^2$ where $r = \frac{d}{2}$ 4. **Calculate the area of the parallelogram:** $$ A_{parallelogram} = 12 \times 6 = 72 $$ 5. **Calculate the radius of the semicircle:** $$ r = \frac{6}{2} = 3 $$ 6. **Calculate the area of the semicircle:** $$ A_{semicircle} = \frac{1}{2} \pi (3)^2 = \frac{1}{2} \pi \times 9 = \frac{9\pi}{2} $$ 7. **Calculate the total area:** $$ A_{total} = A_{parallelogram} + A_{semicircle} = 72 + \frac{9\pi}{2} $$ 8. **Approximate the total area using $\pi \approx 3.1416$:** $$ A_{total} \approx 72 + \frac{9 \times 3.1416}{2} = 72 + \frac{28.2744}{2} = 72 + 14.1372 = 86.1372 $$ 9. **Round to the nearest tenth:** $$ A_{total} \approx 86.1 $$ **Final answer:** The total area of the figure is approximately **86.1** square units.