Points A, B, C, and T are in space such that TA, TB, and TC are perpendicular to the other two. If TA = TB = 12 and TC = 6, then what is the distance from T to face ABC?

Answer 1

I found out that the volume of the pyramid is $144$, so all I have to do is find the base of ABC and divide that by 144 to get the height which is the distance from T to face ABC. I tried to find the area of the base using Heron's Formula, but it got too messy. I am pretty sure that there is a better way...

Answer 2

Using the intercept form for the plane ABC, its equation is
x/12 + y/12 + z/6 = 1
x+y+2z = 12
Now use the distance formula from (x,y,z) to the plane and you get
12/√6

check, using Heron's formula. The area of the base is found using
s = (6√5+6√5+12√2)/2 = 6(√5+√2)
a = √((6(√5+√2))(6(√5+√2)-6√5)(6(√5+√2)-6√5)(6(√5+√2)-12√2))
= √(6(√5+√2)(6√2)(6√2)(6√5-6√2)
= √(6*6*6*6(√5+√2)(√2)(√2)(√5-√2))
= 36√6
Now since V = 1/3 Bh, and the volume of the pyramid is 144,
1/3 * 36√6 h = 144
h = 12/√6

Answer 3

...Or 2√6

Answer 4

2sqrt(6) That is correct