+0

# I need help

0
21
2

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
0

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
0

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