# In a regular quadrangular prism, the side of the base is 1. Height √2. M – middle of AA1.

**In a regular quadrangular prism, the side of the base is 1. Height √2. M – middle of AA1. Find the distance from M to DA1C1.**

Let us determine the lengths of the diagonals DA1 and DС1 of the side faces.

DA1 ^ 2 = DC1 ^ 2 = A1D1 ^ 2 + DD1 ^ 2 = 1 + 4 = 5.

DА1 = DС1 = √5 cm.

Let us determine the length of the A1C1 diagonal.

A1C1 ^ 2 = A1B1 ^ 2 + B1C1 ^ 2 = 1 + 1 = 2 cm. A1C1 = √2 cm.

Let’s draw the height of the BP of the isosceles triangle DA1C1.

Then DH ^ 2 = DA1 ^ 2 – A1H ^ 2 = 5 – 1/2 = 9/2.

DН = 3 / √2 cm.

Then Sda1c1 = A1C1 * DH / 2 = √2 * (3 / √2) / 2 = 3/2 cm2.

Let’s draw a segment MC1 and consider a triangular pyramid C1A1MD with a base A1MD.

Determine the area of the triangle A1MD.

Saa1d = Saa1d1d / 2 = 2 * 1/2 = 1 cm2.

DM is the median of the triangle АА1D, then Sa1md = Saa1d / 2 = 1/2 cm2.

Then the volume of the S1A1MD pyramid is: V = Saa1d * S1D1 / 3 = (1/2) * 1/3 = 1/6 cm3.

Consider the same pyramid C1A1MD, taking triangle DA1C1 as the base.

Then V = Sda1s1 * MK / 3 = (3/2) * MK / 3.

1/6 = 1/2 * MK.

MK = (1/6) / (1/2) = 1/3.

Answer: The distance from M to DA1C1 is 1/3 cm.