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  x2=a2+b2 y2=b2+c2 z2=a2+c2  a2=x2b2=x2y2+c2=x2y2+z2a2 2a2=x2y2+z2  a=x2y2+z22=1.221.72+1.45220.5712  ... b=x2+y2z22=1.22+1.721.45221.0553  ... c=x2+y2+z22=1.22+1.72+1.45221.3328  V=a b c=0.5712 1.0553 1.33280.8034=0.803x = 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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