In the rectangular parallelepiped ABCDA1B1C1D1, edges 1 AB = 4, AD = 7, AA = 6 are known.

| April 25, 2021 | education

In the rectangular parallelepiped ABCDA1B1C1D1, edges 1 AB = 4, AD = 7, AA = 6 are known. Points K and P are the midpoints of edges AA1 and DD1, respectively. Find the cross-sectional area of the parallelepiped by the plane passing through the straight line KP and the vertex B1.

Points K and P are the middle of the lateral sides, then A1K = AK = AA1 / 2 = 6/2 = 3 cm.

D1P = DP = DD1 / 2 = 6/2 = 3 cm.

In the quadrangle ADPK, the diagonals are equal, since the angles at the base are straight, and AK = DP, then the quadrilateral ADPK is a rectangle, and KP is parallel to AD, and therefore parallel to B1C1.

Segments KB1 and PC1 are equal and parallel.

Then the section of the rectangular parallelepiped is the KRS1V1 rectangle.

From the right-angled triangle KA1B1, according to the Pythagorean theorem, we determine the length of the hypotenuse KB1.

KB1 ^ 2 = A1K ^ 2 + A1B1 ^ 2 = 9 + 15 = 25.

KB1 = PC1 = 5 cm.

Then Ssec = KA1 * KR = 5 * 7 = 35 cm2.

Answer: The cross-sectional area is 35 cm2.



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