Pentagons and circles

Consider the diagram above.

Let $s$ be the area of one lune and $x$ be the length of one side of the pentagon.

$A_{sector}=\dfrac{\theta}{360} \pi r^2=\dfrac{72}{360}\left(\dfrac{22}{7}\right)(10^2)=62.857$

$A_{segment}=A_{sector}-A_{triangle}=62.857-\dfrac{1}{2}(10^2)(\sin~72)=15.304$

By cosine law, $x^2=10^2+10^2-2(10)(10)(\cos~72)$ , from here $x=11.756$

$s=area~of~small~semicrcle-A_{segment}=\dfrac{1}{2}\left(\dfrac{22}{7}\right)\left(\dfrac{11.756}{2}\right)^2-15.304=38.99$

Finally,

$5s=5(38.99)\approx$ $\boxed{195}$