A solid is in the form of a cone mounted on a hemisphere. The radius and height of the cone are 3 m and 4 m. Find the surface area of the given solid.

Given: Radius of the cone ($r$) = 3 m

Height of the cone ($h$) = 4 m

The slant height of the cone ($l$) = $\sqrt{h^2+r^2}$

= $\sqrt{4^2+3^2}$

= $\sqrt{25}$

= 5 m

The surface area of the solid = C.S.A of the cone + C.S.A of the hemisphere

As per the question, the radius of the cone = Radius of the hemisphere

Therefore, the surface area of the solid

$=\mathrm{\pi }rl+2\mathrm{\pi }{r}^2$

$=(3.14\times 3\times 5)+(2\times 3.14\times 3^2)$

$=47.1+56.52$

$=103.62 {\mathrm{m}}^2$