Melting Cylinder into Cones

For the cylinder: Radius $r = 7\text{ cm}$, height $h = 10\text{ cm}$. Volume $V = \pi r^2 h = \pi (7^2)(10) = 490\pi$. Flat surface area before melting $= 2\pi r^2 = 2\pi (7^2) = 98\pi$.

Volumes of the two cones are in $3:4$ ratio: $V_1 = \frac{3}{7} \times 490\pi = 210\pi$ $V_2 = \frac{4}{7} \times 490\pi = 280\pi$

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 combined $= \pi r_1^2 + \pi r_2^2 = \pi(63 + 84) = 147\pi$.

Percentage change in flat surface area: $\frac{147\pi - 98\pi}{98\pi} \times 100\% = \frac{49}{98} \times 100\% = 50\%$