$EB = (r - 8) + r = 2r - 8$

From the Theorem of Intersecting Chords:

$AE \times EB = CE \times DE$

$8(2r - 8) = 12(20)$

$r = 19 ~ \text{cm}$

$r - 8 = 11 ~ \text{cm}$

Apply Cosine Law to Triangle *EOC*

$12^2 = r^2 + (r - 8)^2 - 2r(r - 8) \cos \theta$

$12^2 = 19^2 + 11^2 - 2(19)(11) \cos \theta$

$\cos \theta = \dfrac{19^2 + 11^2 - 12^2}{2(19)(11)}$

$\theta = 36.04^\circ$

Area of *AEC* = Area of Sector *AOC* – Area of Triangle *EOC*

$A_{AEC} = \frac{1}{2}r^2 \theta_\text{rad} - \frac{1}{2}r(r - 8) \sin \theta_\text{deg}$

$A_{AEC} = \frac{1}{2}(19^2)(36.04^\circ)\left( \dfrac{\pi}{180^\circ} \right) - \frac{1}{2}(19)(11)\sin 36.04^\circ$

$A_{AEC} = 52.05 ~\text{cm}^2$ ← Answer: [ B ]