2016-II-19

I is true. Let $\theta$ be the angle of the sector $OAB$. Consider the area of the shaded region $ABDC$.

$\begin{array}{rcl}\pi (OC{)}^2\times {\displaystyle \frac{\theta }{{360}^{\circ }}}–\pi (OA{)}^2\times {\displaystyle \frac{\theta }{{360}^{\circ }}}& =& 72\pi \\ \pi (39{)}^2\times {\displaystyle \frac{\theta }{{360}^{\circ }}}–\pi (33{)}^2\times {\displaystyle \frac{\theta }{{360}^{\circ }}}& =& 72\pi \\ 432\theta & =& {25920}^{\circ }\\ \theta & =& {60}^{\circ }\end{array}$

II is not true. By the result of I, $\theta ={60}^{\circ }$. Therefore, the area of the sector $OAB$

$\begin{array}{cl}=& \pi (33{)}^2\times {\displaystyle \frac{{60}^{\circ }}{{360}^{\circ }}}\\ =& 181.5\pi {\text{ cm}}^2\end{array}$

III is not true. By the result of I, $\theta ={60}^{\circ }$. Therefore, the perimeter of the sector $OCD$

$\begin{array}{cl}=& 2\pi (39)\times {\displaystyle \frac{{60}^{\circ }}{{360}^{\circ }}}+2(39)\\ =& (13\pi +78)\text{ cm}\end{array}$