Tetrahedron

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March 20th 2008, 03:48 AM
bobak

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Tetrahedron

I have attached the question. about the terahedron

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

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
March 20th 2008, 08:02 AM
ThePerfectHacker

For the last part think of the triangle as a plane. Express its equation in the form $Ax+By+Cz+D=0$ . Then the distance is given by the distance formula between this plane and $(0,0,0)$ .
March 20th 2008, 08:33 AM
bobak

Quote:

Originally Posted by ThePerfectHacker

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

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 ?
March 20th 2008, 10:39 PM
earboth

Quote:

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 $a_{ABC}$ 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:

$V=\frac13 \cdot a_{ABC} \cdot h~\implies~ \boxed{h=\frac{3V}{a_{ABC}}}$
March 21st 2008, 01:07 AM
bobak

Quote:

Originally Posted by earboth

Let $a_{ABC}$ 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:

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

Very neat, now the first two parts make a lot more sense.

Thanks

Bobak