The figure below is made up of 2 squares, ABCD and DEFG. X is a point on GD and Y is a point on DE. BC = 7 cm, FE = 6 cm and XD = DY = YE.

1. Find the area of the shaded part in the left square ABCD.

The shaded region is triangle ABC, which is exactly half of the square ABCD.
Area of square ABCD $= 7 \times 7 = 49\text{ cm}^2$
Area of shaded triangle ABC $= \frac{1}{2} \times 49 = 24.5\text{ cm}^2$

2. Determine the lengths of the segments XD, DY, and YE in the right square DEFG.

The side length of square DEFG is $6\text{ cm}$, so $DE = 6\text{ cm}$.
Since $Y$ is on $DE$ and $DY = YE$, we have:
$DY = YE = 6 \div 2 = 3\text{ cm}$
We are also given $XD = DY$, so $XD = 3\text{ cm}$.
The length of $GX = GD - XD = 6 - 3 = 3\text{ cm}$.

3. Calculate the area of the shaded triangle FXY by subtracting the unshaded triangles from the square DEFG.

Area of square DEFG $= 6 \times 6 = 36\text{ cm}^2$
The unshaded region in square DEFG consists of three right-angled triangles:

1. Area of $\triangle FGX = \frac{1}{2} \times FG \times GX = \frac{1}{2} \times 6 \times 3 = 9\text{ cm}^2$
2. Area of $\triangle XDY = \frac{1}{2} \times XD \times DY = \frac{1}{2} \times 3 \times 3 = 4.5\text{ cm}^2$
3. Area of $\triangle YEF = \frac{1}{2} \times YE \times EF = \frac{1}{2} \times 3 \times 6 = 9\text{ cm}^2$

Area of shaded $\triangle FXY = 36 - (9 + 4.5 + 9) = 36 - 22.5 = 13.5\text{ cm}^2$

4. Find the total shaded area.

Total shaded area $= \text{Area of } \triangle ABC + \text{Area of } \triangle FXY$
Total shaded area $= 24.5 + 13.5 = 38\text{ cm}^2$

1. Calculate the total area of the figure

Area of square ABCD $= 7 \times 7 = 49 \text{ cm}^2$
Area of square DEFG $= 6 \times 6 = 36 \text{ cm}^2$
Total area $= 49 + 36 = 85 \text{ cm}^2$

2. Calculate the total shaded area

Shaded area in ABCD (Triangle ABC) $= \frac{1}{2} \times 7 \times 7 = 24.5 \text{ cm}^2$
For square DEFG, $DY + YE = 6 \text{ cm}$, so $DY = YE = 3 \text{ cm}$. Thus $XD = 3 \text{ cm}$.
Area of unshaded triangles in DEFG:
Area of $\triangle FGX = \frac{1}{2} \times 6 \times 3 = 9 \text{ cm}^2$
Area of $\triangle XDY = \frac{1}{2} \times 3 \times 3 = 4.5 \text{ cm}^2$
Area of $\triangle FEY = \frac{1}{2} \times 6 \times 3 = 9 \text{ cm}^2$
Shaded area in DEFG (Triangle FXY) $= 36 - 9 - 4.5 - 9 = 13.5 \text{ cm}^2$
Total shaded area $= 24.5 + 13.5 = 38 \text{ cm}^2$

3. Calculate the unshaded area and the required fraction

Unshaded area $= \text{Total area} - \text{Total shaded area} = 85 - 38 = 47 \text{ cm}^2$
Fraction of figure unshaded $= \frac{47}{85}$