1. **State the problem:** We have a circle with area 144 cm². A chord is drawn 6 cm from the center, dividing the circle into two segments. We need to find the area of the smaller segment. 2. **Find the radius of the circle:** The area of a circle is given by $$A = \pi r^2$$. Given $$A = 144$$, we solve for $$r$$: $$r^2 = \frac{144}{\pi}$$ $$r = \sqrt{\frac{144}{\pi}} = \frac{12}{\sqrt{\pi}}$$ 3. **Identify the distance from center to chord:** The chord is 6 cm from the center, so the perpendicular distance from the center to the chord is $$d = 6$$ cm. 4. **Calculate the half-length of the chord:** Using the right triangle formed by the radius, half-chord, and distance from center to chord: $$\text{half-chord} = \sqrt{r^2 - d^2} = \sqrt{\left(\frac{12}{\sqrt{\pi}}\right)^2 - 6^2} = \sqrt{\frac{144}{\pi} - 36}$$ 5. **Calculate the central angle $$\theta$$ in radians subtended by the chord:** $$\cos\left(\frac{\theta}{2}\right) = \frac{d}{r} = \frac{6}{\frac{12}{\sqrt{\pi}}} = \frac{6 \sqrt{\pi}}{12} = \frac{\sqrt{\pi}}{2}$$ So, $$\frac{\theta}{2} = \arccos\left(\frac{\sqrt{\pi}}{2}\right)$$ $$\theta = 2 \arccos\left(\frac{\sqrt{\pi}}{2}\right)$$ 6. **Calculate the area of the segment:** The area of a segment is given by: $$\text{Segment area} = \frac{r^2}{2} (\theta - \sin\theta)$$ Substitute values: $$= \frac{\frac{144}{\pi}}{2} \left(2 \arccos\left(\frac{\sqrt{\pi}}{2}\right) - \sin\left(2 \arccos\left(\frac{\sqrt{\pi}}{2}\right)\right)\right)$$ Simplify: $$= \frac{72}{\pi} \left(2 \arccos\left(\frac{\sqrt{\pi}}{2}\right) - \sin\left(2 \arccos\left(\frac{\sqrt{\pi}}{2}\right)\right)\right)$$ 7. **Numerical approximation:** Calculate $$\arccos\left(\frac{\sqrt{\pi}}{2}\right)$$: $$\sqrt{\pi} \approx 1.77245$$ $$\frac{\sqrt{\pi}}{2} \approx 0.88623$$ $$\arccos(0.88623) \approx 0.485$$ radians Then, $$\theta = 2 \times 0.485 = 0.97$$ radians Calculate $$\sin(\theta) = \sin(0.97) \approx 0.824$$ Finally, $$\text{Segment area} \approx \frac{72}{3.1416} (0.97 - 0.824) = 22.918 \times 0.146 = 3.35$$ cm² **Answer:** The area of the smaller segment is approximately **3.35 cm²**.