1. **State the problem:** We have a semicircle with center O and radius $r$. Points P and Q lie on the horizontal diameter, with P at the right endpoint and O at the center. Point R lies on the circumference forming an arc PQR with a total central angle of 150°, split into two arcs: PQ with 90° and QR with 60°. The length of arc PQR is 15 units, and we want to find the length of arc QR. 2. **Formula for arc length:** The length $L$ of an arc subtending an angle $\theta$ (in degrees) in a circle of radius $r$ is given by: $$L = \frac{\theta}{360} \times 2\pi r$$ 3. **Find the radius $r$ using arc PQR:** The total angle for arc PQR is 150°, and its length is 15 units. $$15 = \frac{150}{360} \times 2\pi r$$ Simplify the fraction: $$\frac{150}{360} = \frac{5}{12}$$ So, $$15 = \frac{5}{12} \times 2\pi r = \frac{5}{12} \times 2\pi r$$ Multiply both sides by 12 to clear denominator: $$15 \times 12 = 5 \times 2\pi r$$ $$180 = 10\pi r$$ Divide both sides by $10\pi$: $$r = \frac{180}{10\pi} = \frac{18}{\pi}$$ 4. **Find the length of arc QR:** Arc QR subtends an angle of 60°. Using the arc length formula: $$L_{QR} = \frac{60}{360} \times 2\pi r = \frac{1}{6} \times 2\pi r = \frac{2\pi r}{6} = \frac{\pi r}{3}$$ Substitute $r = \frac{18}{\pi}$: $$L_{QR} = \frac{\pi}{3} \times \frac{18}{\pi} = \frac{18}{3} = 6$$ **Final answer:** The length of arc QR is **6 units**.