It is known that in the cube abcd-a1b1c1d1, the edge length is 1 and the area of triangle a1bc is calculated

It is known that in the cube abcd-a1b1c1d1, the edge length is 1 and the area of triangle a1bc is calculated


BC ⊥ plane abb1a1,
A1B ∈ plane abb1a1,
BC⊥A1B,
Δ a1bc is RT, A1B = √ 2,
S△A1BC=A1B*BC/2=√2*1/2=√2/2.



In the cube abcd-a1b1c1d1, if the surface area of the tetrahedron with vertices a, C, B1 and D1 is 43, the edge length of the cube ()
A. 2B. 2C. 4D. 22


The surface area of a tetrahedron with vertices a, C, B1 and D1 is 43, so the area of one side is: 3, the edge length of a tetrahedron is: A, from 34a2 = 3, the solution is a = 2, the edge length of a tetrahedron is the diagonal of a cube, so the edge length of a cube is: X, 2x2 = 4, the solution is x = 2