Faces diagonals

If the diagonals of a cuboid are x, y, and z (wall diagonals or three faces) respectively than find the volume of a cuboid.

Solve for x=1.2, y=1.7, z=1.45

Result

V =  0.803

Solution:

$x = 1.2 \ \\ y = 1.7 \ \\ z = 1.45 \ \\ \ \\ x^2 = a^2+b^2 \ \\ y^2 = b^2+c^2 \ \\ z^2 = a^2+c^2 \ \\ \ \\ a^2 = x^2 - b^2 = x^2-y^2+c^2 = x^2-y^2+z^2-a^2 \ \\ 2a^2 = x^2-y^2+z^2 \ \\ \ \\ a = \sqrt{ \dfrac{ x^2-y^2+z^2 }{ 2 } } = \sqrt{ \dfrac{ 1.2^2-1.7^2+1.45^2 }{ 2 } } \doteq 0.5712 \ \\ \ \\ ... \ \\ b = \sqrt{ \dfrac{ x^2+y^2-z^2 }{ 2 } } = \sqrt{ \dfrac{ 1.2^2+1.7^2-1.45^2 }{ 2 } } \doteq 1.0553 \ \\ \ \\ ... \ \\ c = \sqrt{ \dfrac{ -x^2+y^2+z^2 }{ 2 } } = \sqrt{ \dfrac{ -1.2^2+1.7^2+1.45^2 }{ 2 } } \doteq 1.3328 \ \\ \ \\ V = a \cdot \ b \cdot \ c = 0.5712 \cdot \ 1.0553 \cdot \ 1.3328 \doteq 0.8034 = 0.803$

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