Smallest Part From The Circle That Was Divided Into Four Parts By Perpendicular Chords

$\theta = 90^\circ - 2\alpha$

$\theta = 90^\circ - 2\arcsin \left( \dfrac{5}{13} \right)$

$\theta = 44.76^\circ$
 

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

$A_\text{sector} = \dfrac{\pi (13^2)(44.76^\circ)}{360^\circ}$

$A_\text{sector} = 66.01 ~ \text{cm}^2$
 

$x = \sqrt{13^2 - 5^2}$

$x = 12 ~ \text{cm}$
 

$b = x - 5 = 12 - 5$

$b = 7 ~ \text{cm}$
 

$A_\text{triangle} = \frac{1}{2}(7)(5)$

$A_\text{triangle} = 17.5 ~ \text{cm}^2$
 

Required Area:
$A = A_\text{sector} - 2A_\text{triangle}$

$A = 66.01 - 2(17.5)$

$A = 31.01 ~ \text{cm}^2$   ←   answer

$y_U = \sqrt{169 - x^2}$
 

$x_2 = \sqrt{13^2 - 5^2}$

$x_2 = 12 ~ \text{cm}$
 

$\displaystyle A = \int_{x_1}^{x_2} (y_U - y_L) \, dx$

$\displaystyle A = \int_5^{12} \left( \sqrt{169 - x^2} - 5 \right) \, dx$

$A = 31.01 ~ \text{unit}^2$   ←   answer