Surface Areas and Volumes: CBSE Class 10 Mathematics practice questions

C. $214.5\ \text{cm}^2$

The radius is $r = 3.5$ and the cone height is $15.5 - 3.5 = 12\ \text{cm}$, so the slant height is $l = \sqrt{3.5^2 + 12^2} = 12.5\ \text{cm}$. Area $= 2\pi r^2 + \pi r l = \pi(24.5 + 43.75) = 68.25\pi = 214.5\ \text{cm}^2$. Traps: adding the full surface areas of both solids gives 291.5 (the hidden flat faces are counted), using $15.5$ as the cone height gives 251.8, and painting only the cone gives 137.5.

D. $62.8\ \text{cm}^2$

The cone has $r = 3$, $h = 4$ so $l = \sqrt{9 + 16} = 5\ \text{cm}$. Orange area $= \pi r l + \pi(r^2 - r_c^2) = \pi(15 + 9 - 4) = 20\pi = 62.8\ \text{cm}^2$ with $r_c = 2$. Leaving out the ring subtraction gives 75.36, painting only the curved cone gives 47.1, and adding the cylinder base instead of subtracting gives 87.92.

A. $638.5\ \text{cm}^2$

Surface area $=$ TSA of cube $-$ base of hemisphere $+$ CSA of hemisphere $= 600 - \pi r^2 + 2\pi r^2 = 600 + \pi r^2 = 600 + 38.5 = 638.5\ \text{cm}^2$ with $r = 3.5$. Adding $2\pi r^2$ without removing the covered circle gives 677, only removing the circle gives 561.5, and adding the hemisphere's total surface area $3\pi r^2$ without removing the covered circle gives 715.5.