Circular sector inscribed in a square

$\cos \alpha = \dfrac{5}{10}$

$\alpha = 60^\circ$
 

$\theta = 180^\circ - 2\alpha$

$\theta = 180^\circ - 2(60)$

$\theta = 60^\circ$
 

$\dfrac{A_\text{sector}}{\theta_\text{degree}} = \dfrac{\pi r^2}{360^\circ}$

$\dfrac{A_\text{sector}}{60^\circ} = \dfrac{\pi(10^2)}{360^\circ}$

$A_\text{sector} = 52.36 ~ \text{cm}^2$           answer