2 pentagons inside a decagon

Let the side length of decagon be 'a'.

Area of a regular decagon is $a^2$ × $\frac{5(\sqrt{5+2\sqrt{5}})}{2}$ .
Area of a regular pentagon is $a^2$ × $\frac{\sqrt{5(5+2\sqrt{5})}}{4}$ .

Now the shaded region can be split into 2 pentagons and a isosceles triangle.

Now all the angles of pentagon are $108^{\circ}$ . One can easily see that the isosceles $\triangle$ is also a part of a pentagon.

$\therefore$ the vertical angle of triangle is $108^{\circ}$ .

$\begin{aligned} \triangle (Area)&= \frac{1}{2}× a^2×sin(108)\\ &= \frac{1}{2}× a^2× \frac{\sqrt{5+\sqrt{5}}}{2\sqrt{2}}\\ &= a^2× \frac{\sqrt{5+\sqrt{5}}}{4\sqrt{2}} \end{aligned}$

$\therefore$ Area of shaded region is

= $\normalsize 2×(Pentagon) + Triangle$
= $\normalsize 2×a^2×\frac{\sqrt{5(5+2\sqrt{5})}}{4} + a^2× \frac{\sqrt{5+\sqrt{5}}}{4\sqrt{2}}$

Ratio of shaded region to area of decagon
$\begin{aligned} &= \frac{2×a^2×\frac{\sqrt{5(5+2\sqrt{5})}}{4} + a^2× \frac{\sqrt{5+\sqrt{5}}}{4\sqrt{2}}}{a^2× \frac{5(\sqrt{5+2\sqrt{5}})}{2}}\\ &= \frac{\frac{\sqrt{5(5+2\sqrt{5})}}{2} + \frac{\sqrt{5+\sqrt{5}}}{4\sqrt{2}}}{\frac{5(\sqrt{5+2\sqrt{5}})}{2}}\\ &= \frac{5\sqrt{5}-1}{20}\\ &= 0.509.... \end{aligned}$

Therefore the shaded region has greater area since its value is more than 0.5.

If we rearranged the parts, the whole thing boiled down to whether one unshaded isosceles triangle with two same sides and 36° as the angle between the two is bigger or smaller in area compared to two congruent shaded isosceles triangles which have a base (we make them have left-right balance) of the same length with the unshaded's same sides and 108° as the angle opposite it.

Then, if we 'tangramed' the two shaded triangles it's easy to see that together they covered up the unshaded's area nicely.

Shaded : 2 obtuse isosceles ones with angle 36°-108°-36°, different side as base with length x.
Unshaded : 1 acute isosceles triangle with angle 72°-36°-72°, same sides with length x.