In parallelogram ABCD, side AD = 20, BE is the height lowered from vertex B

In parallelogram ABCD, side AD = 20, BE is the height lowered from vertex B (point E belongs to side AD), AE = 5, AB = √2 AE. Find the area of the parallelogram ABCD.

The area ABCD is equal to two areas of the triangle ABD or 2 * (AB * BE) / 20 = AB * BE. Determine the value of BE from the right-angled triangle ABE: BE ^ 2 = AB ^ 2 = AE ^ 2;

BE ^ 2 = (5√2) ^ 2 – 5 ^ 2 = 2 * 25 – 25 = 50 – 25 = 25. Whence the value BE = √25 = 5.

The area of the parallelogram ABCD is equal to AB * BE = 20 * 5 = 1009 (square units).

Answer: area ABCD 100 (sq. Units).