Question 1209484
To find the area of trapezoid \( ABCD \), let’s analyze the problem:

### Step 1: Divide the trapezoid into triangles
The diagonals of the trapezoid intersect at point \( E \), dividing it into four triangles: \( \triangle ABE \), \( \triangle CDE \), \( \triangle ADE \), and \( \triangle BCE \).

The given:
- The area of \( \triangle CDE \) is \( n^2 \).
- The bases of the trapezoid are \( AB = m \) and \( DC = n \).

### Step 2: Property of the diagonals
In a trapezoid, the diagonals divide each other proportionally. Let the ratio of the division of each diagonal at \( E \) be \( k : 1 \), where \( k \) is the ratio of the bases:
\[
k = \frac{AB}{DC} = \frac{m}{n}.
\]

### Step 3: Areas of the triangles
The areas of triangles divided by the diagonals are proportional to the lengths of the corresponding bases:
1. Area of \( \triangle EDC \) is \( n^2 \) (given).
2. Area of \( \triangle ABE \) is proportional to \( k^2 \), so:
   \[
   \text{Area of } \triangle ABE = k^2 \cdot n^2 = \left( \frac{m}{n} \right)^2 \cdot n^2 = m^2.
   \]
3. Area of \( \triangle ADE \) and \( \triangle BCE \) are each proportional to \( k \), and their total area equals twice the area of \( \triangle CDE \) (since diagonals divide the trapezoid evenly).

### Step 4: Total area of the trapezoid
The area of trapezoid \( ABCD \) is the sum of the areas of the four triangles:
\[
\text{Area of } ABCD = \text{Area of } \triangle ABE + \text{Area of } \triangle CDE + 2 \cdot \text{Area of } \triangle CDE.
\]
\[
\text{Area of } ABCD = m^2 + n^2 + 2 \cdot n^2.
\]
\[
\text{Area of } ABCD = m^2 + 3n^2.
\]

### Final Answer:
The area of trapezoid \( ABCD \) is:
\[
\boxed{m^2 + 3n^2}.
\]