# In rhombus ABCD, angle A is 60, the side of the rhombus is 4cm. Line AE is perpendicular to the plane of the rhombus

**In rhombus ABCD, angle A is 60, the side of the rhombus is 4cm. Line AE is perpendicular to the plane of the rhombus. The distance from point E to line CD is 4 cm. Find the distance from point E to the plane of the rhombus and from point A to the plane (EDC).**

Knowing the length of the side of the rhombus and the angle between them, we determine its area.

Savsd = AB * AD * Sin60 = 16 * √3 / 2 = 8 * √3 cm.

Let us construct the height of AK on the continuation of CD, then AK = Savsd / CD = 8 * √3 / 4 = 2 * √3 cm.

The triangle AEK is rectangular, then, by the Pythagorean theorem, AE ^ 2 = EK ^ 2 – AK ^ 2 = 16 – 12 = 4. AE = 2 cm.

The area of the triangle AEK is equal to: Saek = AE * AK / 2 = 2 * 2 * √3 / 2 = 2 * √3 cm.

Then AH = 2 * Saek / EK = 2 * 2 * √3 / 4 = √3 cm.

Answer: The length of the segment AE is 2 cm, from point A to the plane ЕDC √3 cm.