Given that the area of trapezoid $ABCD$ is $140 \, \text{cm}^2$, $AB = 20 \, \text{cm}$, and $CD = 8 \, \text{cm}$, we are asked to find the area of triangle $\triangle AND$ where $DN \parallel BC$.

Step 1: Area of the Trapezoid

The formula for the area of a trapezoid is: $\text{Area of } ABCD = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$ Here, $\text{Base}_1 = AB = 20 \, \text{cm}$ and $\text{Base}_2 = CD = 8 \, \text{cm}$.

Let the height of the trapezoid be $h$. Then, $140 = \frac{1}{2} \times (20 + 8) \times h$ $140 = \frac{1}{2} \times 28 \times h$ $140 = 14h$ $h = \frac{140}{14} = 10 \, \text{cm}$ So, the height of the trapezoid $ABCD$ is $10 \, \text{cm}$.

Step 2: Understanding Triangle $\triangle AND$

Since $DN \parallel BC$, triangle $\triangle AND$ and trapezoid $ABCD$ share the same height from vertex $D$ to line $AB$. The height of triangle $\triangle AND$ is thus also $10 \, \text{cm}$.

Let the length of segment $AN = x \, \text{cm}$.

Step 3: Ratio of Similarity

Since $DN \parallel BC$, triangles $\triangle AND$ and $\triangle BCD$ are similar. The ratio of similarity is given by: $\text{Ratio} = \frac{AN}{AB} = \frac{x}{20}$ The length of $DN$ corresponds to the length of $CD = 8 \, \text{cm}$.

Step 4: Area of Triangle $\triangle AND$

The area of triangle $\triangle AND$ is given by: $\text{Area of } \triangle AND = \frac{1}{2} \times \text{Base} \times \text{Height}$ $\text{Area of } \triangle AND = \frac{1}{2} \times x \times 10 = 5x \, \text{cm}^2$

Step 5: Relationship Between Areas

The area of the trapezoid $ABCD$ is composed of the area of $\triangle AND$ and the area of $\triangle BCD$. Because $\triangle AND$ and $\triangle BCD$ are similar, the areas of these triangles are proportional to the squares of the corresponding sides: $\frac{\text{Area of } \triangle AND}{\text{Area of Trapezoid } ABCD} = \left(\frac{x}{20}\right)^2$ Let $k = \frac{x}{20}$, then: $\text{Area of } \triangle AND = k^2 \times 140$

Since $k^2$ represents the proportion of the total area that belongs to $\triangle AND$ and also equals the ratio $\frac{8}{20}$, we can calculate: $k = \frac{8}{20} = 0.4$ Thus: $\text{Area of } \triangle AND = (0.4) \times 140 \, \text{cm}^2 = 56 \, \text{cm}^2$

The correct area calculation is thus: $\boxed{56}$ cm².