Cylinder Melted into Cones Surface Area Change

Radius of cylinder $r = 7\text{ cm}$, height $h = 10\text{ cm}$. Flat surface area of cylinder $= 2 \pi r^2 = 2 \pi (7^2) = 98\pi$.

Volume of cylinder $= \pi r^2 h = 490\pi$. Volumes of two cones are $210\pi$ and $280\pi$. For cone 1: $\frac{1}{3} \pi r_1^2 (10) = 210\pi \implies r_1^2 = 63$. For cone 2: $\frac{1}{3} \pi r_2^2 (10) = 280\pi \implies r_2^2 = 84$. Flat surface area of the two cones $= \pi r_1^2 + \pi r_2^2 = \pi(63 + 84) = 147\pi$.

Percentage change $= \frac{147\pi - 98\pi}{98\pi} \times 100\% = \frac{49}{98} \times 100\% = 50\%$.