Regular dodecahedron: Surface area and volume

Example

To find the area of edge a of a dodecahedron $a=10m$.

To calculate the area of edge $a$ of the regular dodecahedron it will be necessary to first find the area of side $a$ of a regular pentagon.

We need to use trigonometry to find the area of the pentagon from $a$. This can be very useful, since $a$ will not be a regular occurrence in problems like finding the area of a pentagon, or a dodecahedron.

The five triangles with height $ap$ that compose the pentagon are equal, $$\begin{gathered} A_{pentagon}=5\cdot ({\displaystyle \frac{a\cdot ap}{2}}) \\ {b}^2=a{p}^2+({\displaystyle \frac{a}{2}}{)}^2 \end{gathered}$$ We have to, first, find $b$. As, $\beta ={\displaystyle \frac{{360}^{\circ }}{5}}={72}^{\circ }$ and using the definition of the sinus of the triangle, $$\begin{gathered} \sin {\displaystyle \frac{\beta }{2}}=\sin {36}^{\circ }={\displaystyle \frac{opposite}{hypotenuse}}={\displaystyle \frac{{\displaystyle \frac{a}{2}}}{b}} \\ b={\displaystyle \frac{5}{\sin {36}^{\circ }}}=8,5m \\ 8,5^2=a{p}^2+5^2 \\ ap=6,9m \end{gathered}$$

Then, $$\begin{gathered} A_{pentagon}=5\cdot {\displaystyle \frac{10\cdot 6,9}{2}}=172m^2 \\ {A}_{dodecahedron}=2063,5m^2 \end{gathered}$$

Finally, the following expressions allow us to find the area and volume of the dodecahedron of edge $a$: $$\begin{gathered} A=30\cdot a\cdot ap \\ V={\displaystyle \frac{1}{4}}(15+7\sqrt{5})a^3 \end{gathered}$$