1. **Problem statement:** We have two circles with centers P (radius 6 cm) and Q (radius 5 cm), intersecting with common chord AB of length 8 cm. 2. **Find the length of PQ.** - Let O be the midpoint of chord AB. Since AB = 8 cm, AO = OB = 4 cm. - Triangles PAO and QAO are right triangles (because the radius is perpendicular to the chord's midpoint). - For circle P: $$PA = 6, AO = 4 \Rightarrow PO = \sqrt{6^2 -4^2} = \sqrt{36 -16} = \sqrt{20} = 2\sqrt{5}$$ - For circle Q: $$QA = 5, AO=4 \Rightarrow QO = \sqrt{5^2 -4^2} = \sqrt{25 -16} = \sqrt{9} = 3$$ - Since P, O, and Q are collinear, the length $$PQ = PO + QO = 2\sqrt{5} + 3\approx 4.47 +3 = 7.47 \text{ cm}$$ 3. **Find angle APB.** - Triangle APB is isosceles with PA = PB = 6 cm, and base AB = 8 cm. - Use the cosine rule: $$\cos(\angle APB) = \frac{PA^2 + PB^2 - AB^2}{2 \times PA \times PB} = \frac{6^2 + 6^2 - 8^2}{2 \times 6 \times 6} = \frac{36 + 36 - 64}{72} = \frac{8}{72} = \frac{1}{9}$$ - So, $$\angle APB = \cos^{-1}(\frac{1}{9}) \approx 83.62^\circ$$ 4. **Find angle AQB.** - Triangle AQB is isosceles with QA = QB = 5 cm, base AB = 8 cm. - Using cosine rule: $$\cos(\angle AQB) = \frac{5^2 + 5^2 - 8^2}{2 \times 5 \times 5} = \frac{25 + 25 - 64}{50} = \frac{-14}{50} = -0.28$$ - So, $$\angle AQB = \cos^{-1}(-0.28) \approx 106.26^\circ$$ 5. **Find the area of the shaded region (the overlapping lens shape).** - The shaded region equals the sum of the segments of each circle cut off by chord AB. - Segment area formula for circle radius $$r$$ and chord half-length $$c$$: $$\text{Segment area} = r^2 \cos^{-1}\left(\frac{d}{r}\right) - d \sqrt{r^2 - d^2}$$ where $$d$$ is the distance from the center to chord midpoint. - For circle P (radius 6 cm, $$d=PO = 2\sqrt{5} \approx 4.47$$): $$\theta_P = 2 \cos^{-1}\left(\frac{d}{r}\right) = 2 \cos^{-1}\left(\frac{4.47}{6}\right) = 2 \cos^{-1}(0.745) \approx 2 \times 41.8^\circ = 83.6^\circ = 1.459 \text{ rad}$$ - Segment area P: $$ = \frac{r^2}{2}(\theta - \sin\theta) = \frac{36}{2}(1.459 - \sin 1.459) = 18 (1.459 - 0.993) = 18 \times 0.466 = 8.39 \text{ cm}^2$$ - For circle Q (radius 5 cm, $$d=QO=3$$): $$\theta_Q = 2 \cos^{-1}\left(\frac{3}{5}\right) = 2 \cos^{-1}(0.6) = 2 \times 53.13^\circ = 106.26^\circ = 1.855 \text{ rad}$$ - Segment area Q: $$= \frac{25}{2}(1.855 - \sin 1.855) = 12.5 (1.855 - 0.960) = 12.5 \times 0.895 = 11.19 \text{ cm}^2$$ - Total shaded area: $$8.39 + 11.19 = 19.58 \text{ cm}^2$$ **Final answers:** - (a) $$PQ \approx 7.47$$ cm - (b)(i) $$\angle APB \approx 83.62^\circ$$ - (b)(ii) $$\angle AQB \approx 106.26^\circ$$ - (c) Shaded area $$\approx 19.58$$ cm$^2$