In a rectangular parallelepiped AB = 1, AD = 2, AA1 = 1. Find the distance from point D to line A1B

Since, according to the condition, АА1 = AB = 1 cm, then the two lateral faces of the straight parallelepiped are squares, then the face АА1D1 is equal to the faces ABСD, which means that its diagonals DА1 and DВ are equal.

Then triangle ВDА1 is isosceles.

By the Pythagorean theorem, we determine the length DВ.

DB ^ 2 = BC ^ 2 + CD ^ 2 = 4 + 1 = 5.

DВ = √5 cm.Then A1D = √5 cm

A1B ^ 2 = AB ^ 2 + AA1 ^ 2 = 1 + 1 = 2.

A1B = √2 cm.

The desired distance DH is the height and median of the isosceles triangle BDA1, then BH = A1H = A1H = A1B / 2 = √2 / 2 cm.

Then DH ^ 2 = DB ^ 2 – BH ^ 2 = 5 – 2/4 = 9/2.

DН = 3 / √2 = 3 * √2 / 2 cm.

Answer: From point D to straight line A1B 3 * √2 / 2.