1. **State the problem:** Find the total area of the irregular quadrilateral composed of a rectangle and a right triangle. 2. **Identify shapes and dimensions:** - Rectangle: height = 9 cm, width = 4 cm - Triangle: height = 12 cm, base = 4 cm (same as rectangle's width) 3. **Formulas used:** - Area of rectangle = $\text{width} \times \text{height}$ - Area of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$ 4. **Calculate area of rectangle:** $$\text{Area}_{rectangle} = 4 \times 9 = 36 \text{ cm}^2$$ 5. **Calculate area of triangle:** $$\text{Area}_{triangle} = \frac{1}{2} \times 4 \times 12 = 24 \text{ cm}^2$$ 6. **Calculate total area:** $$\text{Area}_{total} = 36 + 24 = 60 \text{ cm}^2$$ 7. **Final answer:** The area of the shape is **60 cm²**.