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.