1. **Stating the problem:** We need to find the area of the sector ALB of a circle with center L, where AL = 14 cm, angle LAC = 25°, and AC = BC. 2. **Understanding the problem:** Since AC = BC, triangle ABC is isosceles with AC = BC. Angle LAC = 25° is given, and AL is the radius of the circle (14 cm). 3. **Find the central angle of sector ALB:** Since AC = BC, angle LAC = angle LBC = 25°. Therefore, angle ALB = angle LAC + angle LBC = 25° + 25° = 50°. 4. **Formula for the area of a sector:** $$\text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2$$ where $\theta$ is the central angle in degrees and $r$ is the radius. 5. **Calculate the area:** $$\text{Area} = \frac{50}{360} \times \pi \times 14^2 = \frac{50}{360} \times \pi \times 196$$ 6. **Simplify the fraction:** $$\frac{50}{360} = \frac{5}{36}$$ 7. **Calculate the area numerically:** $$\text{Area} = \frac{5}{36} \times 196 \times \pi = \frac{5 \times 196}{36} \times \pi = \frac{980}{36} \times \pi = \frac{245}{9} \times \pi$$ 8. **Approximate the value:** Using $\pi \approx 3.1416$, $$\text{Area} \approx \frac{245}{9} \times 3.1416 = 27.222 \times 3.1416 \approx 85.5 \text{ cm}^2$$ 9. **Check the options:** None of the options match 85.5 cm² exactly, so re-examine the problem. 10. **Reconsider the angle ALB:** Since AC = BC and angle LAC = 25°, triangle ABC is isosceles with vertex angle at C. Angle ALB is the central angle subtending arc AB. Because AC = BC, angle ACB = 180° - 2*25° = 130°. 11. **Central angle ALB is twice angle ACB:** In a circle, the central angle is twice the inscribed angle subtending the same arc. Therefore, angle ALB = 2 * 130° = 260°. 12. **Calculate the sector area with corrected angle:** $$\text{Area} = \frac{260}{360} \times \pi \times 14^2 = \frac{260}{360} \times \pi \times 196$$ 13. **Simplify the fraction:** $$\frac{260}{360} = \frac{13}{18}$$ 14. **Calculate the area numerically:** $$\text{Area} = \frac{13}{18} \times 196 \times \pi = \frac{2548}{18} \times \pi = 141.56 \times \pi$$ 15. **Approximate the value:** $$141.56 \times 3.1416 \approx 444.5 \text{ cm}^2$$ This is too large compared to options, so the previous step is incorrect. 16. **Alternative approach:** Since AC = BC and angle LAC = 25°, triangle ALC is isosceles with AL = LC = 14 cm. Angle LAC = 25°, so angle ALC = 25°. Therefore, angle ACB = 180° - 2*25° = 130°. 17. **Angle ALB is the central angle subtending arc AB, which equals 2 * angle ACB = 2 * 130° = 260°. But sector angle cannot be more than 180° for minor sector. 18. **Assuming minor sector angle is 100° (360° - 260°):** Calculate area with $\theta = 100^\circ$: $$\text{Area} = \frac{100}{360} \times \pi \times 14^2 = \frac{5}{18} \times \pi \times 196 = \frac{980}{18} \times \pi = 54.44 \times \pi$$ 19. **Approximate:** $$54.44 \times 3.1416 \approx 171.1 \text{ cm}^2$$ 20. **Final answer:** The area of sector ALB is approximately 171.1 cm². **Answer: E. 171,1 cm²**