Ans: A
For the right circular cylinder,
For the right circular cylinder,
$\begin{array}{rcl}
a & = & 2\pi r \times \dfrac{r}{2} \\
& = & \pi r^2
\end{array}$
For the hemisphere,
$\begin{array}{rcl}
b & = & \dfrac{1}{2} \times 4\pi r^2 \\
& = & 2\pi r^2
\end{array}$
For the right circular cone,
$\begin{array}{rcl}
c & = & \pi r \times \sqrt{(2r)^2+r^2} \\
& = & \sqrt{5}\pi r^2
\end{array}$
Therefore, $a < b < c$.
