1. The problem states that the area of triangle ABC is twice the area of rectangle EFCD. 2. The area of rectangle EFCD is given by the formula $$\text{Area}_{\text{rectangle}} = \text{width} \times \text{height} = h \times 5 = 5h.$$ 3. The area of triangle ABC is given by $$\text{Area}_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 18 \times 6 = 54.$$ 4. According to the problem, $$\text{Area}_{\text{triangle}} = 2 \times \text{Area}_{\text{rectangle}}$$ so $$54 = 2 \times 5h$$ 5. Simplify and solve for $h$: $$54 = 10h$$ $$\cancel{10}h = \frac{54}{\cancel{10}}$$ $$h = 5.4$$ 6. Therefore, the value of $h$ is 5.4 cm.