1. **State the problem:** We have a cyclic quadrilateral PQRS inscribed in a circle with center O. Given: PR = 17.4 units, QR = 10.7 units, and $\angle QPR = 21.7^\circ$. We need to find: - 9.1 $\angle Q$ - 9.2 $\angle S$ - 9.3 Diameter of the circle - 9.4 Area of quadrilateral PQRS 2. **Recall properties and formulas:** - In a cyclic quadrilateral, opposite angles sum to 180°: $\angle Q + \angle S = 180^\circ$. - Law of Cosines to find side lengths or angles in triangles. - Diameter relates to chord length and subtended angle. - Area of cyclic quadrilateral can be found by dividing into triangles or using Brahmagupta's formula if all sides known. --- ### 9.1 Find $\angle Q$ 3. Consider triangle PQR with sides PR = 17.4, QR = 10.7, and angle $\angle QPR = 21.7^\circ$. 4. Use Law of Cosines to find side PQ: $$PQ^2 = PR^2 + QR^2 - 2 \times PR \times QR \times \cos(21.7^\circ)$$ Calculate: $$PQ^2 = 17.4^2 + 10.7^2 - 2 \times 17.4 \times 10.7 \times \cos(21.7^\circ)$$ 5. Compute values: $$17.4^2 = 302.76, \quad 10.7^2 = 114.49$$ $$2 \times 17.4 \times 10.7 = 372.36$$ $$\cos(21.7^\circ) \approx 0.928$$ So, $$PQ^2 = 302.76 + 114.49 - 372.36 \times 0.928 = 417.25 - 345.68 = 71.57$$ 6. Then, $$PQ = \sqrt{71.57} \approx 8.46$$ 7. Now use Law of Cosines again to find $\angle Q$ opposite side PR: $$\cos(\angle Q) = \frac{PQ^2 + QR^2 - PR^2}{2 \times PQ \times QR}$$ Substitute: $$\cos(\angle Q) = \frac{8.46^2 + 10.7^2 - 17.4^2}{2 \times 8.46 \times 10.7} = \frac{71.57 + 114.49 - 302.76}{2 \times 8.46 \times 10.7} = \frac{-116.7}{181.04} = -0.6447$$ 8. Therefore, $$\angle Q = \cos^{-1}(-0.6447) \approx 129.9^\circ$$ --- ### 9.2 Find $\angle S$ 9. Since PQRS is cyclic, opposite angles sum to 180°: $$\angle Q + \angle S = 180^\circ$$ So, $$\angle S = 180^\circ - 129.9^\circ = 50.1^\circ$$ --- ### 9.3 Find diameter of the circle 10. The chord PR subtends $\angle QPR = 21.7^\circ$ at the circumference. 11. The central angle $\angle POR$ subtending the same chord PR is twice the inscribed angle: $$\angle POR = 2 \times 21.7^\circ = 43.4^\circ$$ 12. Using triangle POR (O is center), PR = 17.4 is chord length, and radius = r. 13. Chord length formula: $$PR = 2r \sin\left(\frac{\angle POR}{2}\right)$$ So, $$17.4 = 2r \sin(21.7^\circ)$$ 14. Calculate $\sin(21.7^\circ) \approx 0.369$, then $$17.4 = 2r \times 0.369 = 0.738r$$ 15. Solve for radius: $$r = \frac{17.4}{0.738} \approx 23.57$$ 16. Diameter is twice radius: $$d = 2r = 2 \times 23.57 = 47.14$$ --- ### 9.4 Find area of quadrilateral PQRS 17. Divide PQRS into triangles PQR and PRS. 18. Area of triangle PQR using formula: $$\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle QPR)$$ Substitute: $$= \frac{1}{2} \times 8.46 \times 10.7 \times \sin(21.7^\circ)$$ Calculate $\sin(21.7^\circ) \approx 0.369$: $$= 0.5 \times 8.46 \times 10.7 \times 0.369 = 16.7$$ 19. To find area of triangle PRS, note that $\angle S = 50.1^\circ$ and PR = 17.4. 20. Use radius $r = 23.57$ to find length PS or RS if needed, or use formula for area of cyclic quadrilateral with known sides. 21. Since full side lengths are not given, approximate area of PRS using formula: $$\text{Area} = \frac{1}{2} \times PR \times r \times \sin(\angle POR)$$ Where $\angle POR = 43.4^\circ$, radius $r=23.57$: $$= 0.5 \times 17.4 \times 23.57 \times \sin(43.4^\circ)$$ Calculate $\sin(43.4^\circ) \approx 0.687$: $$= 0.5 \times 17.4 \times 23.57 \times 0.687 = 140.8$$ 22. Total area of PQRS: $$16.7 + 140.8 = 157.5$$ --- **Final answers:** - 9.1 $\angle Q \approx 129.90^\circ$ - 9.2 $\angle S \approx 50.10^\circ$ - 9.3 Diameter $d \approx 47.14$ units - 9.4 Area of PQRS $\approx 157.5$ square units