1. **Problem statement:** We have a sector ABC of a circle with radius $a$ and central angle $\angle BAC = \frac{\pi}{6}$. Points D and E lie on AB and AC such that $AD = AE = ka$ with $k < 1$. The line DE divides the sector into two equal-area regions. We need to find $k$ to 4 significant figures. 2. **Formula for area of sector:** The area of sector ABC is given by $$\text{Area}_{ABC} = \frac{1}{2} a^2 \theta$$ where $\theta = \frac{\pi}{6}$. 3. **Area of sector ABC:** $$\text{Area}_{ABC} = \frac{1}{2} a^2 \times \frac{\pi}{6} = \frac{\pi a^2}{12}$$ 4. **Area of smaller sector ADE:** Since D and E lie on AB and AC at distance $ka$ from A, the smaller sector ADE has radius $ka$ and the same angle $\frac{\pi}{6}$. So, $$\text{Area}_{ADE} = \frac{1}{2} (ka)^2 \times \frac{\pi}{6} = \frac{\pi k^2 a^2}{12}$$ 5. **Area of trapezium DEBC:** The line DE divides the sector into two equal areas, so $$\text{Area}_{ADE} = \text{Area}_{DEBC} = \frac{1}{2} \times \text{Area}_{ABC} = \frac{\pi a^2}{24}$$ 6. **Equate areas:** $$\frac{\pi k^2 a^2}{12} = \frac{\pi a^2}{24}$$ 7. **Simplify equation:** Cancel $\pi a^2$ from both sides: $$\frac{k^2}{12} = \frac{1}{24}$$ Multiply both sides by 24: $$2 k^2 = 1$$ 8. **Solve for $k$:** $$k^2 = \frac{1}{2}$$ $$k = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.7071$$ **Final answer:** $$k = 0.7071$$ (to 4 significant figures)