1. **Problem statement:** We have a sector AOB of a circle with center O and radius 16 cm. The angle AOB is $\frac{2\pi}{7}$ radians. Point C lies on OB such that OC = 7.5 cm. AC is a straight line. We need to find: (a) The perimeter of the shaded region (sector AOB minus triangle OCA). (b) The area of the shaded region. 2. **Formulas and rules:** - Length of arc $AB = r \times \theta$ where $r$ is radius and $\theta$ is angle in radians. - Perimeter of shaded region = arc length $AB$ + length $AC$ + length $CB$. - Area of sector = $\frac{1}{2} r^2 \theta$. - Area of triangle OCA can be found using coordinates or law of cosines. 3. **Step (a) Perimeter of shaded region:** - Radius $r = 16$ cm. - Angle $\theta = \frac{2\pi}{7}$. - Length of arc $AB = 16 \times \frac{2\pi}{7} = \frac{32\pi}{7}$ cm. - Length $OB = 16$ cm, point C lies on OB with $OC = 7.5$ cm. - So, $CB = OB - OC = 16 - 7.5 = 8.5$ cm. - To find length $AC$, use law of cosines in triangle $OAC$: - $OA = 16$ cm (radius), $OC = 7.5$ cm. - Angle $AOC = $ same as angle $AOB = \frac{2\pi}{7}$ radians. - Law of cosines: $$AC^2 = OA^2 + OC^2 - 2 \times OA \times OC \times \cos\left(\frac{2\pi}{7}\right)$$ - Substitute values: $$AC^2 = 16^2 + 7.5^2 - 2 \times 16 \times 7.5 \times \cos\left(\frac{2\pi}{7}\right)$$ - Calculate: $$AC^2 = 256 + 56.25 - 240 \times \cos\left(\frac{2\pi}{7}\right)$$ - Approximate $\cos\left(\frac{2\pi}{7}\right) \approx 0.6235$: $$AC^2 = 312.25 - 240 \times 0.6235 = 312.25 - 149.64 = 162.61$$ - So, $$AC = \sqrt{162.61} \approx 12.75 \text{ cm}$$ - Perimeter of shaded region = arc $AB + AC + CB$: $$= \frac{32\pi}{7} + 12.75 + 8.5 \approx 14.37 + 12.75 + 8.5 = 35.62 \text{ cm}$$ 4. **Step (b) Area of shaded region:** - Area of sector AOB: $$= \frac{1}{2} \times 16^2 \times \frac{2\pi}{7} = \frac{1}{2} \times 256 \times \frac{2\pi}{7} = \frac{256\pi}{7} \approx 114.88 \text{ cm}^2$$ - Area of triangle OCA using formula: $$\text{Area} = \frac{1}{2} \times OA \times OC \times \sin\left(\frac{2\pi}{7}\right)$$ - Approximate $\sin\left(\frac{2\pi}{7}\right) \approx 0.7818$: $$= \frac{1}{2} \times 16 \times 7.5 \times 0.7818 = 8 \times 7.5 \times 0.7818 = 60 \times 0.7818 = 46.91 \text{ cm}^2$$ - Area of shaded region = Area of sector - Area of triangle: $$= 114.88 - 46.91 = 67.97 \text{ cm}^2$$ **Final answers:** - (a) Perimeter $\approx 35.62$ cm - (b) Area $\approx 67.97$ cm$^2$