Find the radius of the sphere which passes through points $A,$ $B,$ $C,$ and $D$ in space if $\angle BAC = \angle BAD = \angle CAD = 90^\circ$ and $AB = AC = AD = 6.$
The four points A, B, C, and D form a regular tetrahedron. The radius of the sphere that passes through these points is called the circumradius of the tetrahedron.
The circumradius of a tetrahedron can be calculated using the following formula:
r = s / √2
where s is the semiperimeter of the tetrahedron, which is equal to:
s = (AB + AC + AD) / 2
In this case, we have:
s = (6 + 6 + 6) / 2 = 9
Therefore, the circumradius of the tetrahedron is:
r = 9 / sqrt(2) = 9/2*sqrt(2)
