1. **State the problem:** We have a sector OAB of a circle with radius $20$ cm and central angle $\theta$ radians. The perimeter of the sector is $65$ cm. Lines CA and CB are tangents to the circle at A and B, forming an isosceles triangle ABC with $AC = CB$. We need to find the area of the shaded region between the tangents and the arc AB. 2. **Given:** - Radius $r = 20$ cm - Perimeter of sector $= 65$ cm - $AC = CB$ (isosceles triangle) - Central angle $\angle AOB = \theta$ 3. **Find $\theta$ using the perimeter:** The perimeter of the sector is the sum of two radii and the arc length: $$\text{Perimeter} = OA + OB + \text{arc } AB = 2r + r\theta = 65$$ Substitute $r=20$: $$2 \times 20 + 20\theta = 65$$ $$40 + 20\theta = 65$$ Subtract 40: $$20\theta = 25$$ Divide both sides by 20: $$\theta = \frac{25}{20} = \frac{5}{4} = 1.25 \text{ radians}$$ 4. **Find length of chord AB:** Chord length $AB = 2r \sin\left(\frac{\theta}{2}\right)$ $$AB = 2 \times 20 \times \sin\left(\frac{1.25}{2}\right) = 40 \sin(0.625)$$ Calculate $\sin(0.625) \approx 0.5868$: $$AB \approx 40 \times 0.5868 = 23.47 \text{ cm}$$ 5. **Find length of tangent segments CA and CB:** Since CA and CB are tangents from point C to the circle, and triangle ABC is isosceles with $AC=CB$, point C lies on the perpendicular bisector of AB. The length of each tangent from C to the circle is $t = \sqrt{d^2 - r^2}$ where $d$ is the distance from O to C. 6. **Find distance OC:** Triangle ABC is isosceles with base AB and equal sides CA and CB tangent to the circle. The angle at O is $\theta = 1.25$ radians. The angle at C is $\pi - \theta$ because the tangents form an angle supplementary to the central angle. Using the Law of Cosines in triangle ABC: $$AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos(\angle C)$$ But $AC=BC$, so triangle ABC is isosceles with base AB and equal sides AC. Alternatively, the length of tangent $t$ can be found by: $$t = r \tan\left(\frac{\theta}{2}\right)$$ Calculate: $$t = 20 \times \tan\left(\frac{1.25}{2}\right) = 20 \times \tan(0.625)$$ Calculate $\tan(0.625) \approx 0.7265$: $$t \approx 20 \times 0.7265 = 14.53 \text{ cm}$$ 7. **Find area of sector OAB:** $$\text{Area}_{sector} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 20^2 \times 1.25 = 200 \times 0.625 = 125 \text{ cm}^2$$ 8. **Find area of triangle ABC:** The base is $AB = 23.47$ cm. The height is the length of tangent $t = 14.53$ cm. Area of triangle: $$\text{Area}_{triangle} = \frac{1}{2} \times AB \times t = \frac{1}{2} \times 23.47 \times 14.53$$ Calculate: $$\approx 0.5 \times 23.47 \times 14.53 = 170.5 \text{ cm}^2$$ 9. **Find area of shaded region:** The shaded region is the area of triangle ABC minus the area of sector OAB: $$\text{Area}_{shaded} = \text{Area}_{triangle} - \text{Area}_{sector} = 170.5 - 125 = 45.5 \text{ cm}^2$$ **Final answer:** $$\boxed{45.5 \text{ cm}^2}$$