1. **State the problem:** We need to find the area of a blue irregular quadrilateral with given side lengths and heights, where the base is 3 cm, a vertical height from the base to the upper vertex is 5 cm, a small vertical segment on the left side is 1 cm, and an upper slant height is 4 cm. The shape is not a trapezoid. 2. **Analyze the shape:** The quadrilateral is divided by a diagonal dashed line into two right triangles because the diagonal forms two right angles with the base. 3. **Identify the triangles:** - The right triangle on the right has a base of 3 cm and height 5 cm. - The right triangle on the left has a vertical height of 1 cm and a slant height of 4 cm (hypotenuse). 4. **Calculate the base of the left triangle:** Use the Pythagorean theorem for the left triangle: $$4^2 = 1^2 + b^2 \implies 16 = 1 + b^2 \implies b^2 = 15 \implies b = \sqrt{15}$$ 5. **Calculate the area of each triangle:** - Right triangle on the right: $$\text{Area}_1 = \frac{1}{2} \times 3 \times 5 = \frac{15}{2} = 7.5$$ - Right triangle on the left: $$\text{Area}_2 = \frac{1}{2} \times \sqrt{15} \times 1 = \frac{\sqrt{15}}{2}$$ 6. **Calculate total area:** $$\text{Area}_{total} = 7.5 + \frac{\sqrt{15}}{2} \approx 7.5 + 1.9365 = 9.4365$$ 7. **Final answer:** The area of the blue irregular quadrilateral is approximately $9.44$ square centimeters.