+#include '../template.wml'

+<subject "What's the Volume of a Dodecahedron?" />

+<img src="dodeca.gif" alt="A Dodecahedron" /> <br />

+The picture above shows a dodecahedron. It's a solid body which has

+12 perfect pentagons of the same size as its sides. I wondered what

+is the volume of such a body.

+<td>What is the volume of a dodecahedron with an edge-length of

+ The solution can be found some space below:

+Since a dodecahedron is a perfect solid with 12 identical sides at

+the same distance from its center, we can divide it into twelve

+identical pyramids. The corner of each pyramid is found at the

+dodecahedron's center, and the base is one of the sides.

+<p>The volume of a pyramid is the area of the base multiplied by

+its height divided by 3. We can calculate the height by the angle

+between the base and one of the surface sides. Since two pyramids

+are adjacent at every edge of the dodecahedron, we can find it by

+taking the angle between two adjacent sides of the dodecahedron and

+<p>To find that, let's take a corner of the dodecahedron, and form

+a triangle at the points which are at a certain distance along the

+edges. We get the following picture:</p>

+<p><img src="dodeca_corner.gif"

+alt="Corner of a dodecahedron from two different views" /><br />

+From point C, which is found somewhere along the edge, let's lower

+two perpendiculars to the edge, down to the edges of the base

+triangle. We get the triangle CAB. Now since the surface sides of

+this pyramid are isoscles triangles, and the angle of a perfect

+pentagon is equal to 180°*3/5=108 degrees, then the angle CDA

+is equal to (180°-108°)/2=36° degrees.</p>

+<p>Since angle ACD is a right angle, we find that CA (and CB) is

+equal to AD * sin(36°), and since the base triangle is perfect,

+AB is equal to AD. Thus, we know that AE/CA = AB/CA/2 = AD/CA/2 = 1

+/ (CA/AD) / 2 = 1 / (2 * sin(36°)). Thus, the angle ACE, which

+is half the angle between two sides of the dodecahedron and the

+angle we seek, is equal to arcsin(1/(2*sin(36°))).

+(approximately 58.28°).</p>

+<p>Now, let's take the look at a side pyramid of a dodecahedron and

+calculate its volume:</p>

+<p><img src="dodeca_side_pyramid.gif"

+alt="A Dodecahedron Side Pyramid" /><br />

+Since the angle of a perfect pentagon is equal to 108°, angle

+OBA which is half of it is equal to 54°. Since AB is

+<b>a</b>/2, OA is equal to tan(54°)*<b>a</b>/2. Now, the base

+of the pyramid is made of 5 equal triangles each of which has a

+base of length <b>a</b> and a height of OA. Thus, the area of the

+base is 5*(OA*<b>a</b>/2) = 5*tan(54°)*<b>a</b>/2*<b>a</b>/2 =

+5/4*tan(54°)*<b>a</b>^2.</p>

+<p>In the previous section we found out that the angle OAD is equal

+to arcsin(1/(2*sin(36°))), and therefore OD is OA *

+tan(arcsin(1/(2*sin(36°)))) = 1/2 * tan(54°) *

+tan(arcsin(1/(2*sin(36°)))) * <b>a</b>. The volume of the

+pyramid is the area of its base multiplied by its height (OD)

+divided by 3 and so it is equal to:</p>

+<p>5/4 * tan(54°) * <b>a</b>^2 *<br />

+1/2 * tan(54°) * tan(arcsin(1/(2*sin(36°)))) * <b>a</b>

+5/24 * tan(54°)^2 * tan(arcsin(1/(2*sin(36°)))) *

+<p>Since there are 12 such pyramids in a dodecahedron, its volume

+is equal to this volume multiplied by 12. Thus, we get that the

+volume of a dodecahedron is:</p>

+<p>5/2 * tan(54°)^2 * tan(arcsin(1/(2*sin(36°)))) *

+<b>a</b>^3 = ~7.66 * <b>a</b>^3.</p>

+<h3><a href="./">Back to the Math-Ventures Web page</a></h3>