My Math Forum How do I find the base height of a Tetrahedron?

 Algebra Pre-Algebra and Basic Algebra Math Forum

 January 28th, 2013, 10:57 AM #1 Member   Joined: Feb 2012 Posts: 78 Thanks: 0 How do I find the base height of a Tetrahedron? What I am trying to do is find the volume of a tetrahedron. However, I need the base height of the tetrahedron, which I have no idea how to figure out. I know that the edges of the tetrahedron = 1 and the surface area is 4 times the square root of 3. Any help would be greatly appreciated.
January 28th, 2013, 11:43 AM   #2
Senior Member

Joined: Feb 2010

Posts: 706
Thanks: 140

Re: How do I find the base height of a Tetrahedron?

Quote:
 Originally Posted by swm06 What I am trying to do is find the volume of a tetrahedron. However, I need the base height of the tetrahedron which I have no idea how to figure out. I know that the edges of the tetrahedron = 1 and the surface area is 4 times the square root of 3. Any help would be greatly appreciated.
The surface area of a regular tetrahedron with edge 1 is $\sqrt{3}$

 January 28th, 2013, 04:09 PM #3 Member   Joined: Feb 2012 Posts: 78 Thanks: 0 Re: How do I find the base height of a Tetrahedron? Multiplied by 4 because a tetrahedron has 4 sides. But how do you find its base height?
January 28th, 2013, 05:08 PM   #4
Senior Member

Joined: Feb 2010

Posts: 706
Thanks: 140

Re: How do I find the base height of a Tetrahedron?

Quote:
 Originally Posted by swm06 Multiplied by 4 because a tetrahedron has 4 sides. But how do you find its base height?
The area of an equilateral triangle with side s is given by

$\dfrac{s^2\sqrt{3}}{4}$

Since the edge of the tetrahedron is s = 1, then the area of one face is

$\dfrac{\sqrt{3}}{4}$

There are four faces so multiplying by 4 gives the total surface area equal to $\sqrt{3}$.

By the way, the height is $\dfrac{\sqrt{6}}{3}$

 January 29th, 2013, 01:44 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,636 Thanks: 2080 Each face of the tetrahedron has inradius of $\frac{\sqrt{\small3}}{\small6}$ and altitude of length $\frac{\sqrt{\small3}}{\small2}$. Hence the height of the tetrahedron is $\sqrt{(\sqrt3/2)^{\small2}\,-\,(\sqrt3/6)^{\small2}}\,=\,\sqrt{\frac{\small2}{\small3}}\, =\,\frac{\sqrt{\small6}}{\small3}$.

 Tags base, find, height, tetrahedron

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# tetrahedron height

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post r-soy Physics 4 April 23rd, 2012 08:27 AM bhtp Algebra 1 June 20th, 2010 11:14 AM manich44 Algebra 9 October 27th, 2009 02:15 AM symmetry Algebra 2 September 11th, 2007 07:20 PM symmetry Algebra 3 July 29th, 2007 09:04 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top