In the rectangular parallelepiped ABCDA1B1C1D1, the edges 1 AB = 3, AD = 3, AA = 2 are known. Find the area

In the rectangular parallelepiped ABCDA1B1C1D1, the edges 1 AB = 3, AD = 3, AA = 2 are known. Find the area of section of the parallelepiped by the plane passing through the edge AD and the center P of the upper base.

The center of the upper base, point P, is the point of intersection of the diagonals A1C1 and B1D1.

Then the required section is a rectangle KMDA, whose side KM = AD = 3 cm.

Points K and M divide the faces A1B1 and C1D1 in half, then A1K = A1B1 / 2 = AB / 2 = 3/2 = 1.5 cm.

In a right-angled triangle AA1K, we determine the length of the hypotenuse AK.

AK ^ 2 = AA1 ^ 2 + A1K ^ 2 = 4 + 2.25 = 6.25.

AK = 2.5 cm.

Then the cross-sectional area is equal to:

Ssection = AK * AD = 2.5 * 3 = 7.5 cm2.

Answer: The cross-sectional area is 7.5 cm2.