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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 $AB = AC = AD = 6.$

 Apr 6, 2023
 #1
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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)

 Apr 6, 2023
 #2
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I ended up getting the answer. So basically I noticed that The points A,B,C, and D are 4 points of a cube with side length 6. The diameter of the sphere is just the space diagonal of the cube. 

 

So sqrt(36+36+36)=6sqrt3. Since thats the diameter, the radius is 3sqrt3

 Apr 6, 2023

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