Solid Geometry

[ZyzzBrah]

Given a sphere of radius r. Find the volume of the regular tetrahedron that circumscribes the sphere.
Choices:
a. $\displaystyle 6\sqrt{3}r^3$

b. $\displaystyle 12\sqrt{3}r^3$

c. $\displaystyle 8\sqrt{3}r^3$

d. $\displaystyle 10\sqrt{3}r^3$

Relevant Equations
the equation for the volume of regular tetrahedron is $\displaystyle V = \frac{1}{12}e^3\sqrt{2}$
where e is the edge

To attempt a solution:
i let radius r pass from the center to the vertices of equilateral triangle with side e

then cut a portion of it and a smaller triangle results which is an isosceles triangle of sides r,r and e

law of sine
sin(30)/r = sin(120)/e
$\displaystyle e = \sqrt{3}r$
substitute:
$\displaystyle V = \frac{1}{12}(\sqrt{3}r)^3\sqrt{2}$

$\displaystyle V = \frac{1}{4}\sqrt{6}r$
help something seems to be wrong it isnt in the choices

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[Unknown008]

Given a sphere of radius r. Find the volume of the regular tetrahedron that circumscribes the sphere.
Choices:
a. $\displaystyle 6\sqrt{3}r^3$

b. $\displaystyle 12\sqrt{3}r^3$

c. $\displaystyle 8\sqrt{3}r^3$

d. $\displaystyle 10\sqrt{3}r^3$

Relevant Equations
the equation for the volume of regular tetrahedron is $\displaystyle V = \frac{1}{12}e^3\sqrt{2}$
where e is the edge

To attempt a solution:
i let radius r pass from the center to the vertices of equilateral triangle with side e

then cut a portion of it and a smaller triangle results which is an isosceles triangle of sides r,r and e

law of sine
sin(30)/r = sin(120)/e
$\displaystyle e = \sqrt{3}r$
substitute:
$\displaystyle V = \frac{1}{12}(\sqrt{3}r)^3\sqrt{2}$

$\displaystyle V = \frac{1}{4}\sqrt{6}r$
help something seems to be wrong it isnt in the choices

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The tetrahedron circumscribes the sphere, which means the sphere is inside the tetrahedron

 

[ZyzzBrah]

my bad
nvmd

 

[Deleted member 4993]

Here the planes of tetrahedra are tangent to the sphere - edges are not.

I would GUESS that the centroids of the triangles(faces) would be the tangent points of the system.