# Rectangles ABCD and ABMC lie in different planes. The sum of their perimeters is 46 cm.

**Rectangles ABCD and ABMC lie in different planes. The sum of their perimeters is 46 cm. AK = 6 cm, BC = 5 cm. Calculate the distance between the straight lines AK and BC.**

Let’s construct rectangles ABCD in which BC = 5 cm and ABMK in which AK = 6 cm, they have a common side AB, it is perpendicular to the sides BC and AK – this means that this will be the distance between the straight lines BC and AK.

Perimeter of rectangle ABCD:

p1 = 2AB + 2BC.

The perimeter of the AВMK rectangle:

p2 = 2AB + 2MK.

Perimeter sum:

p = p1 + p2 = 2AB + 2BC + 2AB + 2MK = 4AB + 2BC + 2MK = 46.

Let’s substitute the numerical values:

4AB + 2 * 5 + 2 * 6 = 46.

Let’s solve the equation:

4AB = 46-10-12.

4AB = 24 cm.

AB = 6 cm.

Answer: the distance between the straight lines AK and BC is 6 cm.