# Cube cut

In the ABCDA'B'C'D'cube, it is guided by the edge of the CC' a plane witch dividing the cube into two perpendicular four-sided and triangular prisms, whose volumes are 3:2. Determine in which ratio the edge AB is divided by this plane.

Result

r =  1:4

#### Solution:

$V_{ 1 }:V_{ 2 } = 3:2 \ \\ V_{ 1 } = S_{ 1 } \ a \ \\ V_{ 2 } = S_{ 2 } \ a \ \\ S_{ 1 }:S_{ 2 } = 3:2 \ \\ \ \\ S_{ 2 } = \dfrac{ ax }{ 2 } \ \\ S_{ 1 } = a^2 - S_{ 2 } = a^2 - \dfrac{ ax }{ 2 } \ \\ \ \\ (a^2 - \dfrac{ ax }{ 2 } ) / \dfrac{ ax }{ 2 } = 3:2 \ \\ (2a^2-ax) / ax = 3:2 \ \\ (2a-x) / x = 3:2 \ \\ 2a-x = 1.5 \ x \ \\ \ \\ 2a = 2.5 \ x \ \\ \ \\ r = (a-x)/x = (2.5x/2 - x) /x \ \\ \ \\ r = 2.5/2 - 1 = \dfrac{ 1 }{ 4 } = 0.25 = 1:4$

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