1. **State the problem:** We have trapezium ABCD with AD parallel to BC. Diagonals AC and BD intersect at E. Given: - BE = \(\frac{4}{5}\) of BD - AE = \(\frac{1}{5}\) of AC - Area of triangle ABE = 20 cm² We need to find the area of trapezium ABCD. 2. **Analyze the ratio relationship in the diagonals:** Since E divides the diagonals such that \(AE : EC = 1 : 4\) and \(BE : ED = 4 : 1\), point E divides AC and BD in these ratios. 3. **Use properties of trapezium and intersecting diagonals:** The ratios of division imply similar triangles formed inside. Area of triangle ABE relates to area of trapezium ABCD via these segment ratios. 4. **Relate areas:** The diagonals divide the trapezium into four triangles. Using the ratios, the ratio of areas of triangles formed by the diagonals is equal to the ratio of products of the segments. 5. **Calculate the area of trapezium:** Since triangle ABE has area 20 and BE = \(\frac{4}{5}\) BD, and AE = \(\frac{1}{5}\) AC, the whole trapezium area is 25 times the area of triangle ABE. 6. **Final calculation:** $$\text{Area of trapezium ABCD} = 20 \times 25 = 500 \text{ cm}^2$$ **Answer:** The area of trapezium ABCD is \(500\) square centimeters.