In the cube ABCDA1B1C1D1, point K is the midpoint of edge AD, point L belongs to CD and CL: LD = 1: 2.

| June 2, 2021 | education

In the cube ABCDA1B1C1D1, point K is the midpoint of edge AD, point L belongs to CD and CL: LD = 1: 2. A plane is drawn through points K, L and D1. Find the angle between the planes KLD1 and ABC, as well as the area of the resulting section if the edge of the cube is equal to a.

Point K is the middle of AD, then DK = a / 2 cm, CL / LD = 1/2, then DL = 2 * a / 3 cm.

The triangle KDL is rectangular, in which, according to the Pythagorean theorem, KL ^ 2 = DK ^ 2 + DL ^ 2 = a ^ 2/4 + 4 * a ^ 2/9 = (9 * a ^ 2 + 16 * a ^ 2) / 36 = 25 * a ^ 2/36.

CL = 5 * a / 6 cm.

The area of ​​the KDL triangle is equal to: Skdl = KD * DL / 2 = (a / 2) * (2 * a / 3) / 2 = a ^ 2/6 cm2.

Also Skdl = KL * DН / 2.

DH = 2 * Skdl / CL = 2 * (a ^ 2/6) / (5 * a / 6) = 2 * a / 5 cm.

Triangle DD1H is rectangular, then HD1 ^ 2 = DH ^ 2 + DD1 ^ 2 = (4 * a ^ 2/25) + a ^ 2 = 29 * a ^ 2/25.

НД1 = a * √29 / 5 cm.Then Ssec = CL * НD1 / 2 = (5 * a / 6) * (a * √29 / 5) / 2 = a ^ 2 * √29 / 12 cm2.

In a right-angled triangle DD1H, tgH = DD1 / DH = a / (2 * a / 5) = 5/2 = 2.5.

Angle H = arctg2.5.

Answer: The cross-sectional area is a ^ 2 * √29 / 12 cm2, the angle is arctg2.5.



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