Sector of a circle

In general the perimeter of a sector is $2r + r \theta$ and its area is $\dfrac{1}{2} r^{2} \theta$ . We are then given that

$2r + r \theta = 12$ and $\dfrac{1}{2} r^{2} \theta = 9 \Longrightarrow \theta = \dfrac{18}{r^{2}}$ . Upon substitution we then have that

$2r + r \theta = 2r + r \times \dfrac{18}{r^{2}} = 12 \Longrightarrow 2r + \dfrac{18}{r} = 12 \Longrightarrow r + \dfrac{9}{r} = 6 \Longrightarrow r^{2} - 6r + 9 = 0 \Longrightarrow (r - 3)^{2} = 0$ .

Thus $r = 3, \theta = \dfrac{18}{r^{2}} = 2$ and $r + \theta = \boxed{5}$ .