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To solve the problem in the second question, we are given:

- ABCD is a rectangle.
- DF = 20 cm, CE = 30 cm.
- We need to find the area of the rectangle ABCD.

### Key Observations:
1. \( \triangle DFE \) is a right-angled triangle because angle \( \angle DFE = 90^\circ \).
2. DF = 20 cm is one leg of the right triangle.
3. CE = 30 cm is the other leg of the triangle, which is equivalent to BE (since \( \triangle BEC \) is congruent to \( \triangle DFE \) as both are right-angled triangles).

### Step-by-Step Solution:
- First, apply the Pythagorean theorem in triangle \( \triangle DFE \):
  \[
  DE^2 = DF^2 + FE^2
  \]
  Here, \( DE \) is the diagonal of the rectangle ABCD, and DF = 20 cm.

  We are also given that \( BE = 30 \text{ cm} \), but notice that BE = FE (since they are parts of the diagonals of the rectangle).

Now we know two sides of right triangle DEF. Let's calculate the area of the rectangle, knowing that the length and width correspond to FE

ABCD is a rectangle. DF = 20 cm, CE = 30 cm. What is the area of rectangle ABCD?

This geometry problem involves finding the area of rectangle ABCD where DF = 20 cm and CE = 30 cm using the Pythagorean theorem. Key concepts include right-angled triangles and area calculation. Suitable for Grades 8-10 students.

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