$A$ cone of maximum size is carved out from a cube of edge $14 \, cm$. Find the surface area of the cone and the surface area of the remaining solid left after the cone is carved out.

(N/A) The cone of maximum size carved out from a cube of edge $14 \, cm$ will have a base radius $r = 7 \, cm$ and height $h = 14 \, cm$.
The slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5} \, cm$.
Surface area of the cone $= \pi r l + \pi r^2 = \frac{22}{7} \times 7 \times 7\sqrt{5} + \frac{22}{7} \times 7^2 = 154\sqrt{5} + 154 = 154(\sqrt{5} + 1) \, cm^2$.
Surface area of the cube $= 6 \times (14)^2 = 6 \times 196 = 1176 \, cm^2$.
When the cone is carved out,the circular base of the cone is removed from one face of the cube,but the curved surface area of the cone is added to the total surface area.
Surface area of the remaining solid $= (\text{Total surface area of cube}) - (\text{Area of circular base of cone}) + (\text{Curved surface area of cone})$.
$= 1176 - \pi r^2 + \pi r l = 1176 - 154 + 154\sqrt{5} = (1022 + 154\sqrt{5}) \, cm^2$.