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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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
[FONT=verdana, geneva, lucida, 'lucida grande', arial, helvetica, sans-serif][/FONT]
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