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?

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

...Or 2√6

2sqrt(6) That is correct

Well, if TA, TB, and TC are perpendicular to the other two, then we can say that the face ABC is really happy to have those three perpendicular lines hanging around. It's like having three trusty friends always ready to lend a hand.

Now, let's focus on finding the distance from T to face ABC. What we can do is draw a line from T to the center of the face ABC, and this line will be perpendicular to the face. Let's call this distance x.

Since TB = TA = 12, we can divide the face ABC into two congruent right triangles by drawing a line from A to B. Each of these triangles will have legs TB = TA = 12, and the hypotenuse will be the distance from T to the center of the face ABC, which is x.

Now, we can use the Pythagorean theorem to find x. In each of the right triangles, we have:

x^2 = TA^2 - TC^2
x^2 = 12^2 - 6^2
x^2 = 144 - 36
x^2 = 108
x = √108

So, the distance from T to face ABC is equal to the square root of 108. Now, don't be scared of that square root, it's just a fancy way of expressing the length of the line. But if you want to simplify it further, you can write it as:

x = √(36 x 3)
x = √(6 x 6 x 3)
x = 6√3

Therefore, the distance from T to face ABC is 6√3 units. And with this answer, you've just unlocked the secret to making the face ABC smile even wider!

To find the distance from point T to face ABC, we need to find the length of the perpendicular dropped from T to face ABC.

Given that TA, TB, and TC are perpendicular to the other two sides, we can assume that ABC is a right-angled triangle.

Let's label the point where the perpendicular from T intersects face ABC as P. We need to find the length TP.

Since TP is perpendicular to ABC, we can use similar triangles to find its length.

First, let's consider the triangles TAB and ATP. Both triangles share the side TA, and both angles at T are right angles. So, by congruence, triangle TAB is congruent to triangle ATP.

Similarly, triangles TBC and TCP are congruent. Thus, they also have sides TB and TC in common.

From this, we can conclude that angles TAB and ATP are congruent, and angles TBC and TCP are congruent.

Now, let's focus on triangles TAB and TCP. These triangles are similar because they have congruent angles (TAB and TCP are congruent, and TBA and TPC are congruent right angles). The ratio of corresponding sides in these similar triangles is:

TAB/TCP = TA/TC

Simplifying this equation, we get:

TAB = (TCP * TA)/TC

Substituting the given values, we have:

TAB = (6 * 12)/6 = 12

Since TAB = TP (as they are corresponding sides of similar triangles), we have:

TP = 12 units

Therefore, the distance from T to face ABC is 12 units.

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...