Through vertex A of rhombus ABCD, a perpendicular SA is drawn to the plane of the rhombus.

| May 2, 2021 | education

Through vertex A of rhombus ABCD, a perpendicular SA is drawn to the plane of the rhombus. Find the distance between straight lines SA and BC, if AC = 2cm, DB = 2√3cm.

Since the diagonals of the rhombus at the intersection point are divided in half, BO = BD / 2 = 2 * √3 / 2 = √3 cm.

AO = AC / 2 = 2/2 = 1 cm.

The diagonals of the rhombus, at the point of intersection, form a right angle, then, in a right-angled triangle AOB, according to the Pythagorean theorem, AB ^ 2 = AO ^ 2 + BO ^ 2 = 12 + (√3) 2 = 4.

AB = BC = CD = AD = 2 cm.

Let’s define the area of a rhombus in two ways.

Savsd = АС * ВD / 2 = 2 * 2 * √3 / 2 = 2 * √3 cm2.

Savsd = ВС * АН = 2 * АН.

Then: 2 * √3 = 2 * AN.

AH = √3 cm.

Answer: The distance from the straight line SA to the side of the BC is √3 cm.



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