# In rectangular parallelepiped ABCDA1B1C1D1 edge BC = 4, edge, edge BB1 = 2. Point K is the midpoint of edge CC1

In rectangular parallelepiped ABCDA1B1C1D1 edge BC = 4, edge, edge BB1 = 2. Point K is the midpoint of edge CC1. Find the area of the section through points B1, A1 and K.

From point K, the middle of CC1, construct a segment KN parallel to CD. Section А1В1СD is a rectangle, then КН = СD = AB = 2 * √5 cm.

From point K we construct a segment KM parallel to BC, then KM = BC = 4 cm.

Since point K is the middle of CC1, then KC1 = MB1 = CC1 / 2 = 4/2 = 2 cm.

In a right-angled triangle КМВ1, according to the Pythagorean theorem, КВ1 ^ 2 = КМ ^ 2 + MB1 ^ 2 = 16 + 4 = 20. КВ1 = 2 * √5 cm.

Then Ssec = КВ1 * КН = 2 * √5 * 2 * √5 = 20 cm2.

Answer: The cross-sectional area is 20 cm2. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.