Inscribed circle

A circle is inscribed at the bottom wall of the cube with an edge (a = 1). What is the radius of the spherical surface that contains this circle and one of the vertex of the top cube base?

Result

r =  0.8

Solution:

a=1 r1=a/2=1/2=12=0.5  u=2 a=2 1=21.4142 r2=u/2=1.4142/20.7071  r2=x2+r12 r2=(ax)2+r22  x2+r12=(ax)2+r22 r12=2xa+a2+r22  x=a2+r22r122 a=12+0.707120.522 1=58=0.625  r3=x2+r12=0.6252+0.520.8004  r3=r r=(xa)2+r22=(0.6251)2+0.707120.8004=0.8a = 1 \ \\ r_{ 1 } = a/2 = 1/2 = \dfrac{ 1 }{ 2 } = 0.5 \ \\ \ \\ u = \sqrt{ 2 } \cdot \ a = \sqrt{ 2 } \cdot \ 1 = \sqrt{ 2 } \doteq 1.4142 \ \\ r_{ 2 } = u/2 = 1.4142/2 \doteq 0.7071 \ \\ \ \\ r^2 = x^2 + r_{ 1 }^2 \ \\ r^2 = (a-x)^2 + r_{ 2 }^2 \ \\ \ \\ x^2 + r_{ 1 }^2 = (a-x)^2 + r_{ 2 }^2 \ \\ r_{ 1 }^2 = -2xa + a^2 + r_{ 2 }^2 \ \\ \ \\ x = \dfrac{ a^2 + r_{ 2 }^2-r_{ 1 }^2 }{ 2 \cdot \ a } = \dfrac{ 1^2 + 0.7071^2-0.5^2 }{ 2 \cdot \ 1 } = \dfrac{ 5 }{ 8 } = 0.625 \ \\ \ \\ r_{ 3 } = \sqrt{ x^2 + r_{ 1 }^2 } = \sqrt{ 0.625^2 + 0.5^2 } \doteq 0.8004 \ \\ \ \\ r_{ 3 } = r \ \\ r = \sqrt{ (x-a)^2 + r_{ 2 }^2 } = \sqrt{ (0.625-1)^2 + 0.7071^2 } \doteq 0.8004 = 0.8



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See also our trigonometric triangle calculator. Pythagorean theorem is the base for the right triangle calculator.

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