1. **Problem statement:** We have a sector DAB of a circle with center A and radius 18 cm. Angle DAB is $\frac{7\pi}{9}$ radians. Lines CB and CD are tangents to the circle at points B and D respectively. We need to find: (a) The perimeter of the shaded region formed by the arc BD and tangents CB and CD. (b) The area of the shaded region. 2. **Formulas and rules:** - Length of arc $BD = r \times \theta$ where $r=18$ cm and $\theta=\frac{7\pi}{9}$. - Tangents from an external point to a circle are equal in length. - The angle between the two tangents at point C is $\pi - \theta$ because the tangents form an external angle to the sector. - The shaded region perimeter = arc BD + length CB + length CD. - The shaded region area = area of sector DAB - area of triangle formed by tangents and chord BD. 3. **Calculate arc length BD:** $$\text{arc BD} = 18 \times \frac{7\pi}{9} = 14\pi \text{ cm}$$ 4. **Find length of tangent segments CB and CD:** Since CB = CD, let this length be $t$. Triangle BCD is isosceles with vertex angle at C equal to $\pi - \frac{7\pi}{9} = \frac{2\pi}{9}$. The chord BD subtends angle $\frac{7\pi}{9}$ at center A. Using the circle, chord length BD is: $$BD = 2r \sin\left(\frac{\theta}{2}\right) = 2 \times 18 \times \sin\left(\frac{7\pi}{18}\right) = 36 \sin\left(\frac{7\pi}{18}\right)$$ 5. **Use triangle BCD to find tangent length $t$:** In isosceles triangle BCD with sides CB = CD = $t$ and base BD, $$\text{Using the Law of Cosines:}$$ $$BD^2 = t^2 + t^2 - 2 t^2 \cos\left(\frac{2\pi}{9}\right) = 2 t^2 (1 - \cos\left(\frac{2\pi}{9}\right))$$ Rearranged: $$t^2 = \frac{BD^2}{2(1 - \cos(\frac{2\pi}{9}))}$$ 6. **Calculate $t$ numerically:** Calculate $BD$: $$BD = 36 \sin\left(\frac{7\pi}{18}\right)$$ Calculate $\sin\left(\frac{7\pi}{18}\right)$ and $\cos\left(\frac{2\pi}{9}\right)$: - $\frac{7\pi}{18} = 70^\circ$ approx, so $\sin(70^\circ) \approx 0.9397$ - $\frac{2\pi}{9} = 40^\circ$ approx, so $\cos(40^\circ) \approx 0.7660$ So, $$BD \approx 36 \times 0.9397 = 33.83 \text{ cm}$$ $$t^2 = \frac{33.83^2}{2(1 - 0.7660)} = \frac{1144.5}{2 \times 0.234} = \frac{1144.5}{0.468} \approx 2446.6$$ $$t = \sqrt{2446.6} \approx 49.46 \text{ cm}$$ 7. **Calculate perimeter:** $$\text{Perimeter} = \text{arc BD} + CB + CD = 14\pi + 2t \approx 43.98 + 2 \times 49.46 = 43.98 + 98.92 = 142.9 \text{ cm}$$ 8. **Calculate area of sector DAB:** $$\text{Area sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 18^2 \times \frac{7\pi}{9} = 162 \times \frac{7\pi}{9} = 126\pi \approx 395.84 \text{ cm}^2$$ 9. **Calculate area of triangle BCD:** $$\text{Area} = \frac{1}{2} t^2 \sin\left(\frac{2\pi}{9}\right)$$ Calculate $\sin(40^\circ) \approx 0.6428$ $$\text{Area} = \frac{1}{2} \times 2446.6 \times 0.6428 = 785.9 \text{ cm}^2$$ 10. **Calculate shaded area:** $$\text{Shaded area} = \text{Area sector} - \text{Area triangle} = 395.84 - 785.9 = -390.06$$ This negative value indicates the triangle is larger than the sector, so the shaded region is the area outside the sector but inside the tangents. Actually, the shaded region is the area bounded by the tangents and the arc BD, so it equals: $$\text{Shaded area} = \text{Area triangle BCD} - \text{Area sector DAB} = 785.9 - 395.84 = 390.06 \text{ cm}^2$$ **Final answers:** (a) Perimeter $\approx 142.9$ cm (b) Area $\approx 390.1$ cm$^2$