Difference between revisions of "Regular tetrahedron/Introductory problem"
(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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Line 14: | Line 14: | ||
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 .
Solution
Now we find the lateral height:
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: