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#include '../template.wml'

#include "mathjax.wml"

<latemp_subject "What’s the Volume of a Dodecahedron?" />

<latemp_more_keywords "volume, dodecahedron, pythagoran solid, solid, pythagoran, perfect, pentagons, 12, 5, geometry, mathematics, maths, math, spatial geometry" />

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

<p>
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.
</p>

<table border="1">
<tr>
<td>What is the volume of a dodecahedron with an edge-length of
a?</td>
</tr>
</table>

<br />
<p>
 The solution can be found some space below:
</p>

<longblank />

<h2>Solution:</h2>

<p>
Since a dodecahedron is a perfect solid with 12 identical sides at
the same distance from its centre, we can divide it into twelve
identical pyramids. The corner of each pyramid is found at the
dodecahedron’s centre, and the base is one of the sides.
</p>

<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
dividing it by 2.</p>

<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 isosceles triangles, and the angle of a perfect
pentagon is equal to:
</p>

<p>
$$ \frac{180&deg; \cdot 3}{5} = 108&deg; $$
</p>

<p>
Then the angle CDA
is equal to:
</p>

<p>
$$ \frac{180&deg;-108&deg;}{2}=36&deg; $$
</p>

<p>
Since angle ACD is a right angle, we find that CA (and CB) is
equal to AD * sin(36&deg;), and since the base triangle is perfect,
AB is equal to AD. Thus, we know that:
</p>

<p>
\[

\frac{AE}{CA} = \frac{\frac{AB}{CA}}{2} = \frac{\frac{AD}{CA}}{2} =
\frac{\frac{1}{\frac{CA}{AD}}}{2} =
\frac{1}{2\sin{36&deg;}}
\]
</p>

<p>
Thus, the angle ACE, which is half the angle between two sides of the
dodecahedron and the angle we seek, is equal to:
</p>

<p>
$$ \arcsin{\frac{1}{2 \sin{36&deg;}}} $$
</p>

<p>Approximately 58.28&deg;.</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&deg;, angle
OBA which is half of it is equal to 54&deg;. Since AB is
<b>a</b>/2, OA is equal to tan(54&deg;)*<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&deg;)*<b>a</b>/2*<b>a</b>/2 =
5/4*tan(54&deg;)*<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&deg;))), and therefore:
</p>

<set-var mytan="\\tan{\\left[\\arcsin{\\left(\\frac{1}{2 \\sin{36&deg;}}\\right)}\\right]}" />
<p>
\[
OD = OA \cdot <get-var mytan />
= \frac{1}{2} \tan{54&deg;} <get-var mytan /> \cdot
{\bf a}
\]
</p>

<p>
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>
\[
\frac{5}{4} \cdot \tan{54&deg;} \cdot {\bf a}^2 \cdot \\*
\frac{1}{2} \cdot tan{54&deg;} \cdot <get-var mytan /> \cdot {\bf a} \cdot \frac{1}{3} = \\*
\frac{5}{24} \cdot \tan^2{54&deg;} \cdot <get-var mytan />  \cdot {\bf a}^3
\]
</p>

<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>
\[
\frac{5}{2} \cdot \tan{54&deg;}^2 \cdot <get-var mytan /> \cdot
{\bf a}^3 = ~7.66 * {\bf a}^3.
\]
</p>