1. **State the problem:** We need to find the area of the center section of a circular table when the drop-leaves are down. The table has a diameter of 44 in., so the radius is 22 in. Two chords of length 38 in. represent the drop-leaves. 2. **Identify known values:** - Diameter of circle = 44 in. - Radius $r = \frac{44}{2} = 22$ in. - Chord length $AB = 38$ in. 3. **Find $OA$:** Since $O$ is the center and $A$ is on the circle, $OA$ is the radius. $$OA = 22 \text{ in.}$$ 4. **Drop perpendicular $OC$ from center $O$ to chord $AB$:** Since $OC$ is perpendicular to $AB$, it bisects $AB$. $$AC = BC = \frac{38}{2} = 19 \text{ in.}$$ 5. **Use right triangle $OCA$ to find angle $\theta$:** By definition of sine, $$\sin \theta = \frac{AC}{OA} = \frac{19}{22} \approx 0.8636$$ Calculate $\theta$: $$\theta = \sin^{-1}(0.8636) \approx 59.5^\circ$$ 6. **Find central angle $\angle AOB$:** Since $\theta$ is half of $\angle AOB$, $$\angle AOB = 2 \times 59.5^\circ = 119^\circ$$ Convert to radians for area calculations: $$119^\circ = \frac{119 \pi}{180} \approx 2.077 \text{ radians}$$ 7. **Calculate area of sector $AOB$:** Formula for sector area: $$\text{Area}_{sector} = \frac{1}{2} r^2 \theta$$ Substitute values: $$= \frac{1}{2} \times 22^2 \times 2.077 \approx \frac{1}{2} \times 484 \times 2.077 = 502.7 \text{ in}^2$$ 8. **Calculate area of triangle $AOB$:** Area of triangle with two sides and included angle: $$\text{Area}_{\triangle} = \frac{1}{2} r^2 \sin(\angle AOB)$$ Calculate $\sin(119^\circ)$: $$\sin(119^\circ) = \sin(180^\circ - 119^\circ) = \sin(61^\circ) \approx 0.8746$$ Area: $$= \frac{1}{2} \times 22^2 \times 0.8746 = \frac{1}{2} \times 484 \times 0.8746 = 211.7 \text{ in}^2$$ 9. **Calculate area of segment $AOB$ (sector minus triangle):** $$\text{Area}_{segment} = 502.7 - 211.7 = 291.0 \text{ in}^2$$ 10. **Find area of center section:** The center section is the circle minus the two drop-leaf segments. Area of full circle: $$\pi r^2 = \pi \times 22^2 = 1520.53 \text{ in}^2$$ Area of two segments: $$2 \times 291.0 = 582.0 \text{ in}^2$$ Area of center section: $$1520.53 - 582.0 = 938.53 \text{ in}^2$$ **Final answers:** - $OA = 22$ in. - $\theta \approx 59.5^\circ$ - Area of sector $AOB \approx 502.7$ in$^2$ - Area of center section $\approx 938.5$ in$^2$