
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system

Let \( K \) be the midpoint of edge \( BC \), and \( M \) be the center of face \( DCDC_1 \). Since points \( A \) and \( K \) lie in the plane of the bottom face, the line \( AK \) intersects the extension of edge \( DC \) at some point \( O \). Triangles \( ABK \) and \( KCO \) are congruent, hence \( CO = AB = DC \) and \( DO = 2 \cdot DC \). Since points \( M \) and \( O \) lie in the plane of face \( DCDC_1 \), the line \( MO \) intersects edges \( CC_1 \) and \( D_1D \) at some points \( L \) and \( N \). Thus, the section is a quadrilateral \( AKLN \), which is a trapezoid because \( AN \parallel KL \). Given the side length of the cube is 1, it is easy to find by similarity of triangles that \( LC = D_1N = \frac{1}{3} \) and \( DN = \frac{2}{3} \).

Using the Pythagorean theorem, we find the sides of trapezoid \( AN = \sqrt{1 + \frac{4}{9}} = \frac{\sqrt{13}}{3} \), \( AK = \sqrt{1 + \frac{1}{4}} = \frac{\sqrt{5}}{2} \), \( KL = \sqrt{\frac{1}{4} + \frac{1}{9}} = \frac{\sqrt{13}}{6} \), \( NL = \frac{\sqrt{10}}{3} \).

In trapezoid \( AKNL \), draw the line \( KF \parallel LN \) and the height of the trapezoid \( KP \). Since \( KL = \frac{1}{2}AN \), it follows that \( AF = FN \) and \( LK \).

By the law of cosines, we have \( KF^2 = (AK)^2 + (AF)^2 - 2 \cdot AK \cdot AF \cdot \cos \hat{A} \), thus \( \cos \hat{A} = \frac{3}{\sqrt{65}} \) hence \( \sin \hat{A} = \sqrt{\frac{56}{65}} = \frac{2\sqrt{14}}{\sqrt{65}} \). Consequently, \( KP = \sqrt{\frac{14}{13}} \) and the area \(S\) of the trapezoid \( \text{trapezoid } = \frac{1}{2} KP \cdot \frac{\sqrt{13}}{2} = \frac{\sqrt{14}}{4} \).

Present the answer in LaTex format: \boxed{Your answer}
user