Finding the Height of a TetrahedronDate: 05/03/98 at 11:57:47 From: Andrew Jones Subject: Volume of a Tetrahedron Hi Dr. Math, I am trying to find out the height of a tetrahedron, so I can work out its volume using the formula 1/3*(base area)*(perpendicular height). The tetrahedron's sides are all length x, and I want to know its volume relative to x. Please could you tell me how to work out the height of this tetrahedron, and therefore its area? Thank you very much. Yours mathematically, Andrew Jones Date: 05/03/98 at 13:53:37 From: Doctor Sam Subject: Re: Volume of a Tetrahedron Andrew, Great question! I wish I could draw you a picture . . . it is hard enough to talk about two dimensions, but you have jumped into the third. Imagine your tetrahedron sitting on one of its equilateral triangles as its base. You want to find its height, the perpendicular distance from the vertex in space to the triangle base. Try drawing this, and draw in the height h in the middle of the tetrahedron. You want to find h and you know x, the side of the tetrahedron. Since h is perpendicular to the base, there is a right angle at the foot of this line. If you connect the foot of the perpendicualar to one of the corners of the tetrahedron's base, you will have a right triangle that stands up in space with only one of its edges in the plane: /| / | x/ |h / | / | ------- If we could find the length of the third side of this triangle (in terms of x), we could use the Pythagorean Theorem to find h. So now look closely at this third side. This is inside the equilateral triangle that is the tetrahedron's base. In fact, if you join the foot of h to all three of the vertices of the base, you will have a picture of an equilateral triangle with three lines meeting in the center of the triangle. If you extend these lines, you will have a picture of an equilateral triangle with three lines drawn from the vertices to the opposite bases, all meeting in the center of the triangle. In an equilateral triangle, these lines are medians (they divide the opposite side in half) and also angle bisectors (they divide the angles in half) and also altitudes (they are perpendicular to the sides of the equilateral triangle). Are you familiar with properties of these lines? There are two properties that will let you figure out the length you want in terms of x. The first is that medians divide the opposite side in half. Therefore, in the base triangle, each median cuts the equilateral triangle into two right triangles. The hypotenuse of these triangles is x, the side of the tetrahedron. The edge of the triangle on the base is x/2. Together, these values will let you calculate the third side of the right triangle, the median/altitude/angle bisector line. The second important property is that the medians of any triangle intersect at a point that divides the median into the ratio of 2 to 1. That is, the length that you want (the base of the vertical triangle) is two-thirds of the length of the median (in the base triangle). I hope you were able to follow all that . . . it is difficult without a figure to look at. So draw a good one for yourself! Good luck, Andrew. I hope that helped. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/