1. ## Tetrahedron

I have attached the question. about the terahedron

part i) I had no problem with i got the volume as $\displaystyle \frac{1}{6} abc$

for part ii) i use Pythagorus theorem to get the lengths of sides CA CA and AB then just used the cosine rule.

I am just stuck on the last part about the perpendicular distance.

Thanks, Bobak

2. For the last part think of the triangle as a plane. Express its equation in the form $\displaystyle Ax+By+Cz+D=0$. Then the distance is given by the distance formula between this plane and $\displaystyle (0,0,0)$.

3. Originally Posted by ThePerfectHacker
For the last part think of the triangle as a plane. Express its equation in the form $\displaystyle Ax+By+Cz+D=0$. Then the distance is given by the distance formula between this plane and $\displaystyle (0,0,0)$.
That worked a treat, I used the formula outlined here Equation of a plane

however my only concern is that the question is aimed at students who are not familiar with plane geometry. I worried i missed something, can this be done another way ?

4. Originally Posted by bobak
That worked a treat, I used the formula outlined here Equation of a plane

however my only concern is that the question is aimed at students who are not familiar with plane geometry. I worried i missed something, can this be done another way ?
Let $\displaystyle a_{ABC}$ denote the area of the base triangle.
Use the hint: If you already know the volume of the tetrahedron then you can calculate the height by:

$\displaystyle V=\frac13 \cdot a_{ABC} \cdot h~\implies~ \boxed{h=\frac{3V}{a_{ABC}}}$

5. Originally Posted by earboth
Let $\displaystyle a_{ABC}$ denote the area of the base triangle.
Use the hint: If you already know the volume of the tetrahedron then you can calculate the height by:

$\displaystyle V=\frac13 \cdot a_{ABC} \cdot h~\implies~ \boxed{h=\frac{3V}{a_{ABC}}}$
Very neat, now the first two parts make a lot more sense.

Thanks

Bobak