In the regular decagon shown, there are two regular pentagons inscribed. Which has the greater area?

Unshaded region
Both are equal
Shaded region

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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.