ABDG is a square. Given that $BC = DE = EF = FG = AH$, what fraction of the figure is shaded? Ans: ___

1. Determine the side length of the square in terms of units.

Let $BC = DE = EF = FG = AH = 1 \text{ unit}$.
Since $GD = GF + FE + ED$, the side length of the square $ABDG$ is $1 + 1 + 1 = 3 \text{ units}$.

2. Calculate the area of the top shaded triangle.

The base of the top shaded triangle is $AB = 3 \text{ units}$.
Its height is the distance from $AB$ to $HC$, which is $AH = 1 \text{ unit}$.
Area of top shaded triangle $= \frac{1}{2} \times 3 \times 1 = 1.5 \text{ square units}$.

3. Calculate the area of the two bottom shaded triangles.

For $\triangle HGF$, the base $GF = 1 \text{ unit}$ and the height $HG = AG - AH = 3 - 1 = 2 \text{ units}$.
Area of $\triangle HGF = \frac{1}{2} \times 1 \times 2 = 1 \text{ square unit}$.
For $\triangle CDE$, the base $ED = 1 \text{ unit}$ and the height $CD = BD - BC = 3 - 1 = 2 \text{ units}$.
Area of $\triangle CDE = \frac{1}{2} \times 1 \times 2 = 1 \text{ square unit}$.

4. Find the total shaded area and the fraction of the figure that is shaded.

Total shaded area $= 1.5 + 1 + 1 = 3.5 \text{ square units}$.
Total area of the square $= 3 \times 3 = 9 \text{ square units}$.
Fraction shaded $= \frac{3.5}{9} = \frac{7}{18}$.