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Find the volume of a shpere circumscribed around a regular terahedron with side = 8cm

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  • 1 decade ago
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    Assume one of the vertices of the tetrahedron is on the north pole of the sphere. A regular tetrahedron has 4 faces that are all equilateral triangles. Therefore the the other 3 vertices of the tetrahedron are on a single latitude line. We gotta find which line that is. Let's give the spherical coordinates of the 4 vertices. Then to find the latitude line we will use the fact that the average of the 4 vertices must be the origin.

    The coordintes are ordered as (radius, theta, phi) where phi is the angle off from the ray going straight up, and theta is the angle that the projection onto the xy-plane is off from the ray going in the positive x direction.

    (r, 0, 0) = north pole

    (r, 0, phi), (r, 120, phi), (r, 240, phi) the 3 vertices on the latitude line that is 30 degree south of the equator.

    We will get the same answer for the latitude (phi) if we assume for the moment that r=1.

    Convert these to (x,y,z) coordinates using

    (x,y,z) = (r*cos(theta)*sin(phi), r*sin(theta)*sin(phi), r*cos(phi))

    (0, 0, 1),

    (cos(0)*sin(phi), sin(0)*sin(phi), cos(phi)),

    (cos(120)*sin(phi), sin(120)*sin(phi), cos(phi)),

    (cos(240)*sin(phi), sin(240)*sin(phi), cos(phi)).

    or

    (0,0,1),

    (sin(phi), 0, cos(phi)),

    ((-1/2)*sin(phi), (sqrt(3)/2)*sin(phi), cos(phi))

    ((-1/2)*sin(phi), (-sqrt(3)/2)*sin(phi), cos(phi))

    The sum of these 4 points is:

    (0, 0, 1 + 3cos(phi))

    So to make the average be the origin, we need 1+3cos(phi)=0.

    so cos(phi) = -1/3 (so phi = arccos(-1/3) though we won't be needing this fact)

    After constructing the right triangle with adjacent side 1 and hypoteneuse 3 (which implies the opposite side must be 2sqrt(2)), we get sin(phi) = (2sqrt(2)/3) (we know it must be positive because 0<phi<180, so sin(phi)>0 ).

    So the (x,y,z) coordinates of the vertices are:

    (0,0,r),

    (r*(2sqrt(2)/3), 0, -r/3),

    (r*(-1/2)*(2sqrt(2)/3), r*(sqrt(3)/2)*(2sqrt(2)/3), -r/3)

    (r*(-1/2)*(2sqrt(2)/3), r*(-sqrt(3)/2)*(2sqrt(2)/3), -r/3)

    To find the radius r, we'll use the assumption that the distance between any 2 of these is 8cm.

    So ||(0,0,r)-(r*(2sqrt(2)/3), 0, -r/3)|| = 8cm

    or, sqrt((r^2)*(8/9) + 0 + (16/9)*r^2) = 8*cm^2

    so (24/9)r^2=64*cm^2

    so r = sqrt(24)cm

    so volume = (4/3)*pi*r^3

    = (4/3)*pi*(24sqrt(24)cm^3) = 64*sqrt(6)*pi*cm^3

  • locuaz
    Lv 7
    1 decade ago

    the volume is:

    64sqrt(6)pi

    Source(s): h
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