The pentagon is made of 5 triangles exactly the same as $\mathrm{AOB}$, and the pentangle is made of 5 shapes exactly the same as $\mathrm{AOBC}$.

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{\frac{\mathrm{Area}(\mathrm{pentangle})}{\mathrm{Area}(\mathrm{pentagon})}}& =& \frac{\mathrm{Area}(\mathrm{AOBC})}{\mathrm{Area}(\mathrm{AOB})}\\ \multicolumn{1}{c}{}& =& 1-\frac{\mathrm{Area}(\mathrm{ACB})}{\mathrm{Area}(\mathrm{AOB})}\\ \multicolumn{1}{c}{}& =& 1-\frac{\mathrm{Area}(\mathrm{DCB})}{\mathrm{Area}(\mathrm{DOB})}\\ \multicolumn{1}{c}{}& =& 1-\frac{\mathrm{DC}}{\mathrm{DO}}\\ \multicolumn{1}{c}{}& =& \frac{\mathrm{CO}}{\mathrm{DO}}<\frac{1}{2}.\end{array}}$

In order to calculate this ratio exactly we first find the angles and then use trigonometry.

${\displaystyle \angle \mathrm{AOB}={72}^{\circ }}$

${\displaystyle \angle \mathrm{DOB}=\frac{1}{2}\angle \mathrm{AOB}={36}^{\circ }}$

As $\mathrm{AC}$ is parallel to $\mathrm{PM}$ and $\mathrm{CB}$ is parallel to $\mathrm{MK}$,

${\displaystyle \angle \mathrm{ACB}=\angle \mathrm{PMK}={108}^{\circ }.}$

${\displaystyle \angle \mathrm{DCB}=\frac{1}{2}\angle \mathrm{ACB}={54}^{\circ }.}$

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{\frac{\mathrm{Area}(\mathrm{pentangle})}{\mathrm{Area}(\mathrm{pentagon})}}& =& 1-\frac{\mathrm{DB}\tan 36}{\mathrm{DB}\tan 54}\\ \multicolumn{1}{c}{}& =& 1-\frac{\tan 36}{\tan 54}\\ \multicolumn{1}{c}{}& =& 1-\tan {}^{2}36\\ \multicolumn{1}{c}{}& =& 1-(5-2\sqrt{5})\\ \multicolumn{1}{c}{}& =& 2\sqrt{5}-4\\ \multicolumn{1}{c}{}& =& 0.47\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{approx}.\end{array}}$

This ratio is just less than 0.5 meaning the pentangle is a bit smaller than half the pentagon.

Footnote: You don't need a calculator, from the diagram it is possible to calculate exact values for the trig. ratios for ${18}^{\circ }$, ${36}^{\circ }$, ${54}^{\circ }$ and ${72}^{\circ }$. All the angles marked with a spot can be shown to be ${36}^{\circ }$ using simple properties of triangles. Let $\mathrm{CA}=\mathrm{CB}=x$. The triangle $\mathrm{PAC}$ is an isosceles triangle with base angles of ${72}^{\circ }$. If $\mathrm{PA}=\mathrm{PC}=1$ then $\mathrm{PB}=1+x$. Triangles $\mathrm{PAB}$ and $\mathrm{ACB}$ are similar, hence

${\displaystyle \frac{x}{1}=\frac{1}{1+x}}$

This gives a quadratic equation which can be solved to give $x=\frac{1}{2}(\sqrt{5}-1).$

Now we have $\mathrm{AD}=\frac{1}{2}$ and so (using Pythagoras Theorem to find $\mathrm{CD}$):

${\displaystyle \tan {}^2{36}^{\circ }=4{\mathrm{CD}}^2=4(x^2-\frac{1}{4})=4x^2-1=5-2\sqrt{5}.}$