1. The problem provides definitions: radius = line from circle center to a point on circle; chord = line connecting two points on circle; diameter = chord through center with length twice radius. 2. Given radius $r = 12$ cm and chord length $c = 16$ cm, find the length of the segment joining the midpoints of arcs AB and CB. 3. The chord of length 16 cm creates two arcs AB and CB. The segment joining the midpoints of these arcs equals the length of the radius because both arcs share the radius line. 4. Using the right triangle formed by half the chord and the distance from center to chord midpoint, calculate half chord: $\frac{c}{2} = 8$ cm. 5. Let $d$ be the distance from center to chord, then by Pythagoras: $$r^2 = d^2 + \left(\frac{c}{2}\right)^2$$ Substitute knowns: $$12^2 = d^2 + 8^2$$ $$144 = d^2 + 64$$ $$d^2 = 80$$ $$d = \sqrt{80} = 4\sqrt{5} \approx 8.94\,\text{cm}$$ 6. Segment joining midpoints of arcs AB and CB equals twice this distance: $$2d = 2 \times 8.94 = 17.88\,\text{cm}$$ 7. For problem with chord 15 cm and perpendicular distance (height) 9 cm from center, find radius. 8. Use Pythagoras: $$r^2 = 9^2 + \left(\frac{15}{2}\right)^2 = 81 + 56.25 = 137.25$$ $$r = \sqrt{137.25} \approx 11.72\,\text{cm}$$ 9. For Activity 5, matching major and minor arcs with given points: - (A) AB = Minor arc: B - (B) AC = Major arc: A - (C) BC = Minor arc: C - (D) AD = Major arc: D 10. Measuring major arcs: - Given radius 10 cm, circumference $= 2\pi r = 20\pi$ - Arc DF: $m(DF) = 86^\circ$, length $= 5.27 \pi$ which matches $20\pi \times \frac{86}{360} \approx 5.27 \pi$ - Similarly for arcs FB and BD, the given degree measures and lengths correspond appropriately. Final Answers: 1) Radius = 12 cm 2) Chord = 16 cm 3) Half chord = 8 cm 4) Twice radius = 24 cm 5) Length joining midpoints = 17.88 cm 6) Radius (from second problem) ≈ 11.72 cm 7) Arcs matched as above 8) Major arcs measured accordingly