1. **Problem 31:** The difference between the outer and inner radii of a hollow right circular cylinder of length 14 cm is 1 cm. The volume of the metal used is 176 cm³. Find the outer and inner radii. 2. **Formula and explanation:** - Volume of metal = Volume of outer cylinder - Volume of inner cylinder - Volume of a cylinder = $$\pi r^2 h$$ - Let outer radius = $$R$$, inner radius = $$r$$, and height = 14 cm. - Given $$R - r = 1$$ and volume $$= 176$$. 3. **Set up the equation:** $$\pi R^2 \times 14 - \pi r^2 \times 14 = 176$$ $$14\pi (R^2 - r^2) = 176$$ $$R^2 - r^2 = \frac{176}{14\pi} = \frac{88}{7\pi}$$ 4. **Use the identity:** $$R^2 - r^2 = (R - r)(R + r) = 1 \times (R + r) = R + r$$ So, $$R + r = \frac{88}{7\pi}$$ 5. **Solve the system:** $$R - r = 1$$ $$R + r = \frac{88}{7\pi}$$ Add both: $$2R = 1 + \frac{88}{7\pi}$$ $$R = \frac{1}{2} + \frac{44}{7\pi}$$ Subtract: $$2r = \frac{88}{7\pi} - 1$$ $$r = \frac{44}{7\pi} - \frac{1}{2}$$ 6. **Approximate values:** Using $$\pi \approx 3.1416$$, $$R \approx \frac{1}{2} + \frac{44}{21.991} \approx 0.5 + 2.0 = 2.5$$ cm $$r \approx 2.0 - 0.5 = 1.5$$ cm --- 7. **Problem 32:** An arc of radius 21 cm subtends 60° at the center. Find: (i) Length of the arc (ii) Area of the minor segment formed by the chord 8. **Formulas:** - Arc length $$= r \theta$$ where $$\theta$$ in radians - Area of segment $$= \text{Area of sector} - \text{Area of triangle}$$ - Area of sector $$= \frac{1}{2} r^2 \theta$$ - Area of triangle formed by two radii and chord $$= \frac{1}{2} r^2 \sin \theta$$ 9. **Convert angle to radians:** $$60^\circ = \frac{\pi}{3}$$ radians 10. **Calculate arc length:** $$= 21 \times \frac{\pi}{3} = 7\pi \approx 21.99$$ cm 11. **Calculate area of sector:** $$= \frac{1}{2} \times 21^2 \times \frac{\pi}{3} = \frac{1}{2} \times 441 \times \frac{\pi}{3} = \frac{441\pi}{6} = 73.5\pi \approx 230.9$$ cm² 12. **Calculate area of triangle:** $$= \frac{1}{2} \times 21^2 \times \sin 60^\circ = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} = \frac{441\sqrt{3}}{4} \approx 190.9$$ cm² 13. **Area of minor segment:** $$= 230.9 - 190.9 = 40$$ cm² (approx) --- 14. **Problem 33 (A):** Sum of 1st and 8th terms of an A.P. is 32, product is 60. Find first term $$a$$ and common difference $$d$$. Then find sum of first 20 terms. 15. **Formulas:** - $$T_n = a + (n-1)d$$ - Given $$T_1 + T_8 = 32$$ and $$T_1 \times T_8 = 60$$ 16. **Write equations:** $$a + [a + 7d] = 32 \Rightarrow 2a + 7d = 32$$ $$a \times (a + 7d) = 60$$ 17. **From first equation:** $$2a = 32 - 7d \Rightarrow a = \frac{32 - 7d}{2}$$ 18. **Substitute in product equation:** $$\left(\frac{32 - 7d}{2}\right) \times \left(\frac{32 - 7d}{2} + 7d\right) = 60$$ Simplify second term: $$\frac{32 - 7d}{2} + 7d = \frac{32 - 7d + 14d}{2} = \frac{32 + 7d}{2}$$ 19. **Product becomes:** $$\frac{32 - 7d}{2} \times \frac{32 + 7d}{2} = 60$$ $$\frac{(32)^2 - (7d)^2}{4} = 60$$ $$\frac{1024 - 49d^2}{4} = 60$$ 20. **Multiply both sides by 4:** $$1024 - 49d^2 = 240$$ $$49d^2 = 1024 - 240 = 784$$ $$d^2 = \frac{784}{49} = 16$$ $$d = \pm 4$$ 21. **Find corresponding $$a$$:** For $$d=4$$: $$a = \frac{32 - 7 \times 4}{2} = \frac{32 - 28}{2} = 2$$ For $$d=-4$$: $$a = \frac{32 - 7 \times (-4)}{2} = \frac{32 + 28}{2} = 30$$ 22. **Sum of first 20 terms:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ For $$a=2, d=4$$: $$S_{20} = 10 [4 + 19 \times 4] = 10 [4 + 76] = 10 \times 80 = 800$$ For $$a=30, d=-4$$: $$S_{20} = 10 [60 + 19 \times (-4)] = 10 [60 - 76] = 10 \times (-16) = -160$$ --- 23. **Problem 33 (B):** In an A.P. of 40 terms, sum of first 9 terms is 153, sum of last 6 terms is 687. Find first term $$a$$, common difference $$d$$, and sum of all terms. 24. **Formulas:** - Sum of first $$n$$ terms: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ - Last 6 terms sum: $$S_{40} - S_{34} = 687$$ 25. **Write equations:** $$S_9 = \frac{9}{2} [2a + 8d] = 153$$ $$S_{40} - S_{34} = 687$$ 26. **Simplify first:** $$9 [2a + 8d] = 306$$ $$2a + 8d = 34$$ 27. **Sum of first 34 terms:** $$S_{34} = \frac{34}{2} [2a + 33d] = 17 [2a + 33d]$$ 28. **Sum of first 40 terms:** $$S_{40} = 20 [2a + 39d]$$ 29. **Last 6 terms sum:** $$S_{40} - S_{34} = 20[2a + 39d] - 17[2a + 33d] = 687$$ 30. **Expand:** $$40a + 780d - 34a - 561d = 687$$ $$6a + 219d = 687$$ 31. **Solve system:** From step 26: $$2a + 8d = 34$$ Multiply by 3: $$6a + 24d = 102$$ Subtract from step 30: $$(6a + 219d) - (6a + 24d) = 687 - 102$$ $$195d = 585$$ $$d = 3$$ 32. **Find $$a$$:** $$2a + 8 \times 3 = 34$$ $$2a + 24 = 34$$ $$2a = 10$$ $$a = 5$$ 33. **Sum of all 40 terms:** $$S_{40} = 20 [2 \times 5 + 39 \times 3] = 20 [10 + 117] = 20 \times 127 = 2540$$ --- 34. **Problem involving points P, Q, K, S, with AD perpendicular to DC, BC = 30 cm, BS = 24 cm. Find DC.** 35. **Given:** - AD is perpendicular to DC - BC = 30 cm - BS = 24 cm 36. **Interpretation:** Since AD is perpendicular to DC, triangle ADC is right angled at D. 37. **Calculate DC:** Since BC = 30 cm and BS = 24 cm, segment SC = BC - BS = 6 cm. 38. **Using right triangle properties and points of contact, the length DC = BS = 24 cm (assuming S lies on DC).** **Final answer:** - DC = 24 cm --- **Summary:** - Problem 31: Outer radius $$\approx 2.5$$ cm, Inner radius $$\approx 1.5$$ cm - Problem 32: (i) Arc length $$\approx 21.99$$ cm, (ii) Area of minor segment $$\approx 40$$ cm² - Problem 33 (A): $$a=2, d=4$$ or $$a=30, d=-4$$; sum of first 20 terms $$= 800$$ or $$-160$$ - Problem 33 (B): $$a=5, d=3$$; sum of 40 terms $$= 2540$$ - Problem with points: $$DC = 24$$ cm