Tic Tac Toe

It is clear from the figure above that the perimeter of the polygon is:

$\begin{aligned} p & = 4 \times [\text{the perimeter of }4\times 16 \text{ rectangle}] - 4 \times [\text{the perimeter of }4\times 4 \text{ square}] \\ & = 4 \times 40 - 4 \times 16 = \boxed{96} \end{aligned}$

Note: It must be stated in the problem that the figure is not drawn to scale. The length of $x$ is $\dfrac{16-2(4)}{3}=\dfrac{8}{3}$ .

The length of the $\color{#D61F06}red$ line is $2(16)=$ $\color{#D61F06}32$ .

The length of the $\color{#3D99F6}blue$ line is $2(16)=$ $\color{#3D99F6}32$ .

The length of the $\color{#20A900}green$ line is $3\left(\dfrac{8}{3}\right)(2)=$ $\color{#20A900}16$ .

The length of the $\color{#69047E}purple$ line is $3\left(\dfrac{8}{3}\right)(2)=$ $\color{#69047E}16$ .

So the perimeter of the figure above is $\color{#D61F06}32$ $+$ $\color{#3D99F6}32$ $+$ $\color{#20A900}16$ $+$ $\color{#69047E}16$ $=$ $\color{#624F41}\large \boxed{96}$ $\large \boxed{answer}$

Note:

I am not sure that the shape should be called a polygon, and it is not clear from the wording of the problem that the "perimeter" includes the perimeter of the internal square. So some changes in wording might be in order. Solution:

As long as the two vertical pieces are exactly vertical and intersect the two horizontal pieces, which are exactly horizontal, the exact position of the pieces does not influence the final answer.

The three red lines will add up to $16-2\times4=8$ . There are $4$ such groups of three, each forming an interrupted line, in a vertical direction, and $4$ in horizontal direction, adding up to $8\times8=64$ .

The only parts left are the ends, line the one shown in green. There are $4$ vertical ones and $4$ horizontal ones making a total of $8$ and contributing $8\times4=32$ to the perimeter.

Total perimeter of the figure is $64+32=\boxed{96}$ .