1. **Problem Statement:** A solid metal cylinder of height 10 cm and diameter 14 cm is melted and recast into two cones with volumes in the ratio 3:4, both cones having height 10 cm. We need to find the percentage change in the flat surface area before and after recasting. 2. **Given Data:** - Cylinder height, $h_c = 10$ cm - Cylinder diameter, $d_c = 14$ cm, so radius $r_c = \frac{14}{2} = 7$ cm - Two cones with heights $h_1 = h_2 = 10$ cm - Volume ratio of cones $V_1 : V_2 = 3 : 4$ 3. **Formulas:** - Volume of cylinder: $$V_c = \pi r_c^2 h_c$$ - Volume of cone: $$V = \frac{1}{3} \pi r^2 h$$ - Flat surface area (base area) of cylinder: $$A_c = \pi r_c^2$$ - Flat surface area (base area) of cones: $$A = \pi r^2$$ 4. **Calculate volume of cylinder:** $$V_c = \pi \times 7^2 \times 10 = 490\pi$$ 5. **Volumes of cones:** Let volumes be $3x$ and $4x$ such that: $$3x + 4x = 7x = 490\pi \implies x = \frac{490\pi}{7} = 70\pi$$ So, $$V_1 = 3x = 210\pi, \quad V_2 = 4x = 280\pi$$ 6. **Find radii of cones:** Using volume formula for cones: $$V = \frac{1}{3} \pi r^2 h \implies r = \sqrt{\frac{3V}{\pi h}}$$ For cone 1: $$r_1 = \sqrt{\frac{3 \times 210\pi}{\pi \times 10}} = \sqrt{\frac{630\pi}{10\pi}} = \sqrt{63}$$ For cone 2: $$r_2 = \sqrt{\frac{3 \times 280\pi}{\pi \times 10}} = \sqrt{\frac{840\pi}{10\pi}} = \sqrt{84}$$ 7. **Calculate flat surface areas:** Cylinder base area: $$A_c = \pi \times 7^2 = 49\pi$$ Cones base areas: $$A_1 = \pi r_1^2 = \pi \times 63 = 63\pi$$ $$A_2 = \pi r_2^2 = \pi \times 84 = 84\pi$$ Total cones base area: $$A_{cones} = 63\pi + 84\pi = 147\pi$$ 8. **Calculate percentage change in flat surface area:** $$\text{Percentage change} = \frac{A_{cones} - A_c}{A_c} \times 100 = \frac{147\pi - 49\pi}{49\pi} \times 100 = \frac{98\pi}{49\pi} \times 100 = 2 \times 100 = 200\%$$ 9. **Interpretation:** The flat surface area increases by 200%, which is not among the given options. Since the question asks for the percentage change, the closest option is "None of the above". **Final answer:** None of the above