Complicated geometry question...?

Question

What is volume of a regular tetrahedron in terms of the radius of a sphere inscribed within it?

Answer 1

Hmm, good question.

The volume for a tetrahedron of side length "s" is (s^3)√2 / 12. If we can get s in terms of "r", the radius of the inscribed sphere, then we can just substitute.

Imagine the very center of the tetrahedron, which would also be the center of the sphere inside of it. The sphere touches each of the four faces in the center of each face. So the line from the center of the sphere and perpendicular to each face is a radius of the sphere.

Now imagine four lines coming from the center of the sphere to each of the tetrahedron's four verticies. This splits the tetrahedron up into four identical pyramids. The height of each of these pyramids is r.

Each of these little pyramids has a volume of (1/3)Ar, where "A" is the area of one face of the tetrahedron. But the combined volume of these four little pyramids has to be the volume of the entire tetrahedron, which is (1/3)Ah where h is the height of the tetrahedron. So:
4(1/3)Ar = (1/3)Ah, and 4r = h.

The height of a tetrahedron (which you can prove on the side) is s√6/3. So s√6/3 = 4r, and s = 12r / √6.

Plug this into (s^3)√2 / 12 and we get:
V = r^3 (12 / √6)^3 (√2 / 12)
V = r^3 (12^2 / 6√6) (√2)
V = r^3 (24√2 / √6)
V = r^3 (24√2√6 / 6)
V = r^3 (4√12)
V = r^3 (4√(4*3))
V = (r^3) (8√3)

Answer 2

The radius of the inscribed sphere ("r") is sqrt(6)/12 times the edge length (which we'll call "a").

The volume of the tetrahedron is sqrt(2)/12 times the cube of the edge length.

So...

r = sqrt(6)/12 * a
V = sqrt(2)/12 * a^3

Solve the first equation for a:

r = sqrt(6)/12 * a
a = 12/sqrt(6) * r
a = 2 * sqrt(6) * r

Now plug that into the volume equation:

V = sqrt(2)/12 * a^3
V = sqrt(2)/12 * (2*sqrt(6)*r)^3
V = sqrt(2)/12 * 2^3 * sqrt(6)^3 * r^3
V = sqrt(2)/12 * 8 * 6 * sqrt(6) * r^3
V = sqrt(2) * 4 * sqrt(6) * r^3
V = 8 * sqrt(3) * r^3

Answer 3

quantity = length x width x top, so in case you narrow back 2 of the variables via 20%, the quantity may be .8 x .8 = .sixty 4 cases smaller if the top remained an identical. to maintain the quantity consistent, you will possibly decide to enhance the quantity via one million/.sixty 4 = one million.5625, that's comparable to announcing fifty six.25% greater.