Art of Problem Solving

Regular tetrahedron/Introductory problem: Difference between revisions

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New page: ==Problem== Find the volume of a tetrahedron whose sides all have length <math>2</math>. ==Solution== We find the area of the base: <math>\mathrm {A} =\dfrac{4\sqrt{3}}{4}=\sqrt...
 
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Now we can find the height of the tetrahedron:
Now we can find the height of the tetrahedron:


<math>\mathrm {h}=\sqrt{\sqrt{3})^2-(\dfrac{1}{3}*\sqrt{3})^2}=\sqrt{\dfrac{8}{3}</math>
<math>\mathrm {h}=\sqrt{(\sqrt{3})^2-(\dfrac{1}{3}*\sqrt{3})^2}=\sqrt{\dfrac{8}{3}</math>


Now the volume of the tetrahedron is:
Now the volume of the tetrahedron is:

Latest revision as of 19:30, 30 December 2012

Problem

Find the volume of a tetrahedron whose sides all have length $2$.

Solution

We find the area of the base:

$\mathrm {A} =\dfrac{4\sqrt{3}}{4}=\sqrt{3}$

Now we find the lateral height:

$\mathrm {lh}=\sqrt{2^2-1^2}=\sqrt{3}$

Now we can find the height of the tetrahedron:

$\mathrm {h}=\sqrt{(\sqrt{3})^2-(\dfrac{1}{3}*\sqrt{3})^2}=\sqrt{\dfrac{8}{3}$ (Error compiling LaTeX. Unknown error_msg)

Now the volume of the tetrahedron is:

$\dfrac{1}{3}*\sqrt{3}*\sqrt{\dfrac{8}{3}}=\dfrac{\sqrt{8}}{3}$


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