In the figure, ABC and DEF are right-angled triangles. The base of triangle DEF is twice as long as the base of triangle ABC. The height of triangle DEF is also twice as long as the height of triangle ABC. The difference in the area between the 2 unshaded parts is 18 cm^2.

1. Analyse the given information regarding the positions of points A, F, and B.

The problem states that points A, F, B, and D lie on a straight horizontal baseline. However, there is no information given about the specific distance between A and F, or F and B.

2. Conclude on the validity of Statement 1.

Since the exact position of point F along the line segment AB is not constrained by any of the given areas or lengths, the ratio of AF to AB cannot be determined.

1. Express the area of triangle DEF in terms of the area of triangle ABC

Let the base of triangle ABC be $b$ and its height be $h$.
Area of triangle ABC = $\frac{1}{2}bh$
Base of triangle DEF = $2b$
Height of triangle DEF = $2h$
Area of triangle DEF = $\frac{1}{2}(2b)(2h) = 2bh = 4 \times (\frac{1}{2}bh) = 4 \times \text{Area of triangle ABC}$

2. Evaluate the truth of the statement

Since the area of triangle DEF is 4 times the area of triangle ABC, the statement that it is twice the area is false.

1. Find the ratio of the area of triangle DEF to triangle ABC.

Let the base and height of triangle ABC be $b$ and $h$ respectively.
Area of triangle ABC $= \frac{1}{2}bh$
Base of triangle DEF $= 2b$, Height of triangle DEF $= 2h$
Area of triangle DEF $= \frac{1}{2}(2b)(2h) = 4 \times (\frac{1}{2}bh) = 4 \times \text{Area of triangle ABC}$

2. Express the difference in unshaded areas in terms of the total areas.

Let the area of the shaded part be $S$.
Unshaded area of ABC $= \text{Area of ABC} - S$
Unshaded area of DEF $= \text{Area of DEF} - S$
Difference in unshaded areas $= (\text{Area of DEF} - S) - (\text{Area of ABC} - S) = \text{Area of DEF} - \text{Area of ABC}$

3. Solve for the area of triangle DEF.

We know the difference is $18\text{ cm}^2$.
$4 \times \text{Area of ABC} - \text{Area of ABC} = 18$
$3 \times \text{Area of ABC} = 18$
$\text{Area of ABC} = 6\text{ cm}^2$
$\text{Area of DEF} = 4 \times 6 = 24\text{ cm}^2$
Since the calculated area matches the statement, the statement is True.