Tetrahedron help! (1 Viewer)

shved · Nov 30, 2012

Hey all,
(This is my first thread, so um yeah don't kill me)

So i was just casually reading through my maths textbook (I'm a nerd?) , and i found this question that I've been trying to figure out for the past hour. Here it is:

9. Calculate the volume of a regular tetrahedron that has all its edges 6cm in length using trig + pythagoras and the formula v=1/3ah

I worked out that one face would be 9root3 cm^2 and now im stuck, any help would be appreciated

Thanks

SpiralFlex · Nov 30, 2012

First we need to consider what a tetrahedron is. It contains 4 equilateral triangles stuck to form a 3D shape. What do we know about equilateral triangles? They have interior angles of 60 degrees each. Now we need to apply this fact using our basic knowledge.

Let the edge be a.

We need to find the base area, that's simple it's just,

$A=\frac{1}{2}a^2\sin 60$

$=\frac{\sqrt{3}}{4}a^2$

Now, we need to find the altitude of a face, call it h

$h=a\sin 60$

$h=\frac{a\sqrt{3}}{2}$

Now let the space height be H, [Note: We are considering the triangle within the shape standing up. By Pyth.

$(\frac{2}{3}h)^2+H^2=a^2$

$\frac{4}{9}[\frac{a\sqrt{3}}{2}]^2+H^2=a^2$

$H=\frac{\sqrt{6}a}{3}$

$V=\frac{1}{3}Ah$

$V=\frac{1}{3}(\frac{a^2\sqrt{3}}{4})*(\frac{\sqrt{6}a}{3})$

$V=\frac{a^3\sqrt{2}}{12}$

Let a = 6,

$\therefore V=18\sqrt{2}$

shved · Nov 30, 2012

Thanks mate, looks like i didn't know about the properties (2/3 and 1/3). Oh well, more revision for me in the holidays.

Tetrahedron help! (1 Viewer)

shved

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SpiralFlex

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shved

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