Step 1: Determine the angle of the sector. The figure shows a triangle OAB where OA and OB are radii of the circle, so $OA = OB = 5.00 \text{ cm}$. The length of the chord AB is given as $8.66 \text{ cm}$. We can use the Cosine Rule to find the angle $\theta = \angle AOB$. The Cosine Rule states $c^2 = a^2 + b^2 - 2ab \cos C$. In triangle OAB: $$AB^2 = OA^2 + OB^2 - 2(OA)(OB) \cos(\theta)$$ $$(8.66)^2 = (5)^2 + (5)^2 - 2(5)(5) \cos(\theta)$$ $$74.9956 = 25 + 25 - 50 \cos(\theta)$$ $$74.9956 = 50 - 50 \cos(\theta)$$ $$74.9956 - 50 = -50 \cos(\theta)$$ $$24.9956 = -50 \cos(\theta)$$ $$\cos(\theta) = \frac{24.9956}{-50} = -0.499912$$ This value is very close to $-0.5$. Therefore, $\theta = \arccos(-0.5) = 120^\circ$. To use in the sector area formula, convert to radians: $$\theta = 120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} \text{ radians}$$ Step 2: Calculate the area of the sector OAB. The area of a sector is given by the formula $A_{\text{sector}} = \frac{1}{2} r^2 \theta$ (where $\theta$ is in radians) or $A_{\text{sector}} = \frac{\theta}{360^\circ} \pi r^2$ (where $\theta$ is in degrees). Using degrees: $$A_{\text{sector}} = \frac{120^\circ}{360^\circ} \times \pi (5)^2$$ $$A_{\text{sector}} = \frac{1}{3} \times 25\pi$$ $$A_{\text{sector}} = \frac{25\pi}{3} \text{ cm}^2$$ $$A_{\text{sector}} \approx \frac{25 \times 3.14159265}{3} \approx 26.1799 \text{ cm}^2$$ Step 3: Calculate the area of the triangle OAB. The area of a triangle can be calculated using the formula $A_{\text{triangle}} = \frac{1}{2} ab \sin C$. $$A_{\text{triangle}} = \frac{1}{2} (OA)(OB) \sin(\theta)$$ $$A_{\text{triangle}} = \frac{1}{2} (5)(5) \sin(120^\circ)$$ $$A_{\text{triangle}} = \frac{25}{2} \times \frac{\sqrt{3}}{2}$$ $$A_{\text{triangle}} = \frac{25\sqrt{3}}{4} \text{ cm}^2$$ $$A_{\text{triangle}} \approx \frac{25 \times 1.73205081}{4} \approx 10.8253 \text{ cm}^2$$ Step 4: Calculate the area of the shaded region. The shaded region is a segment of the circle, which is the area of the sector minus the area of the triangle. $$A_{\text{segment}} = A_{\text{sector}} - A_{\text{triangle}}$$ $$A_{\text{segment}} = \frac{25\pi}{3} - \frac{25\sqrt{3}}{4}$$ $$A_{\text{segment}} = 25 \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)$$ $$A_{\text{segment}} \approx 26.1799 - 10.8253$$ $$A_{\text{segment}} \approx 15.3546 \text{ cm}^2$$ Rounding to two decimal places: $$A_{\text{segment}} \approx 15.35 \text{ cm}^2$$ The final answer is $\boxed{15.35 \text{ cm}^2}$. 3 done, 2 left today. You're making progress.