# Commits

committed 3ec26c0

Convert the Dodecahedron page to MathJax.

• Participants
• Parent commits 6748fcd

# File t2/MathVentures/dodeca.html.wml

 #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" />
 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 180&deg;*3/5=108 degrees, then the angle CDA
-is equal to (180&deg;-108&deg;)/2=36&deg; degrees.</p>
+pentagon is equal to:
+</p>

-<p>Since angle ACD is a right angle, we find that CA (and CB) is
+<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 AE/CA = AB/CA/2 = AD/CA/2 = 1
-/ (CA/AD) / 2 = 1 / (2 * sin(36&deg;)). 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&deg;))).
-(approximately 58.28&deg;).</p>
+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>
 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 OD is OA *
-tan(arcsin(1/(2*sin(36&deg;)))) = 1/2 * tan(54&deg;) *
-tan(arcsin(1/(2*sin(36&deg;)))) * <b>a</b>. The volume of the
+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>
+divided by 3 and so it is equal to:
+</p>

-<p>5/4 * tan(54&deg;) * <b>a</b>^2 *<br />
-1/2 * tan(54&deg;) * tan(arcsin(1/(2*sin(36&deg;)))) * <b>a</b>
-*<br />
-1/3 =<br />
-5/24 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) *
-<b>a</b>^3</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>5/2 * tan(54&deg;)^2 * tan(arcsin(1/(2*sin(36&deg;)))) *
-<b>a</b>^3 = ~7.66 * <b>a</b>^3.</p>
+<p>
+$ +\frac{5}{2} \cdot \tan{54&deg;}^2 \cdot <get-var mytan /> \cdot +{\bf a}^3 = ~7.66 * {\bf a}^3. +$
+</p>