Calculate the y-coordinate of the centroid of the composite lamina with respect to the line AB.

The moment of each component is its weight multiplied by the distance of its centroid from AB.

• For triangle ADE: Moment$_1 = \left(\frac{\sqrt{3}}{2} W\right) \times \left(\frac{\sqrt{3}}{3}\right) = \frac{3}{6} W = \frac{1}{2} W$

• For rectangle EFCD: Moment$_2 = \left(2\sqrt{3} W\right) \times \left(\frac{\sqrt{3}}{2}\right) = \frac{2 \times 3}{2} W = 3 W$

• For triangle F1CB: Moment$_3 = \left(\frac{\sqrt{3}}{2} W\right) \times \left(\frac{\sqrt{3}}{3}\right) = \frac{3}{6} W = \frac{1}{2} W$

Total Moment = Moment$_1$ + Moment$_2$ + Moment$_3$

$$\text{Total Moment} = \frac{1}{2} W + 3 W + \frac{1}{2} W = 4 W$$

The y-coordinate of the centroid is the total moment divided by the total weight of the lamina.
The total weight of the lamina is given in your table as $3\sqrt{3} W$.

$$\bar{y} = \frac{\text{Total Moment}}{\text{Total Weight}} = \frac{4 W}{3\sqrt{3} W}$$

Cancel out $W$:

$$\bar{y} = \frac{4}{3\sqrt{3}}$$

Rationalize the denominator:

$$\bar{y} = \frac{4}{3\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3 \times 3} = \frac{4\sqrt{3}}{9}$$

The y-coordinate of the centroid from AB is $\frac{4\sqrt{3}}{9}$ m.

The final answer is $\frac{4\sqrt{3}}{9} \text{ m}$.