Polygons in squares!

Let us work on the 5th figure and logically generalize for the $n^{th}$ squared figure.

Consider the figure to be divided into four sections. The first section is $\triangle JOI$ . The second section is $\triangle JOU$ , the third section is $\triangle UOT$ and the fourth section is $\triangle TOI$ .

To find the number of triangles in this figure is quite simple.
One triangle embedded in the figure is $\triangle OAB$ .
To find the number of such triangles, consider the point $O$ as one vertex of the triangle. Now, the third side can be either one of $AB$ , $DC$ , $EF$ , $GH$ or $IJ$ . So, it makes $\color{#D61F06}{5}$ of them. There are similar triangles in the four sections, which makes $5 \times 4 = \color{#D61F06}{20}$ of them.
Logically thinking, if we had the $n^{th}$ squared figure we would have $\color{#3D99F6}{4n}$ of them. $\longrightarrow \boxed{1}$ .
Another type of triangles is of the type $\triangle ABK$ . Observe that if the vertex $B$ is changed to $D$ then we get $\triangle DNC$ . Similarly, change the vertices to $F$ , $H$ and $J$ to get three more of such triangles. This makes $\color{#D61F06}{5}$ of them. Here we have consider the sections 1 and 2, there are five such triangles each in section(2 and 3), section(3 and 4), section(4 and 1). This makes $5 \times 4 = \color{#D61F06}{20}$ of them.
Logically thinking, if we had the $n^{th}$ squared figure we would have $\color{#3D99F6}{4n}$ of them. $\longrightarrow \boxed{2}$ .
From $\boxed{1}$ and $\boxed{2}$ we would get $4n$ + $4n$ = $\color{#20A900}{8n}$ triangles in the $n^{th}$ squared figure.

To find the number of quadrilaterals.
One type is $\square ABCD$ . Fix $AB$ as one side of the trapezium. The opposite side can be $CD$ , $EF$ , $HG$ or $JI$ . This makes $\color{#D61F06}{4}$ of them. Now if we take the base as $CD$ and start selecting the opposite side, observe that we have already counted the one is which we would be taking $AB$ as the opposite side, so the choices for the opposite side to be selected decreases to three. Then for base $EF$ we have two choices for the opposite side and for base $GH$ only one. So the number of such quadrilaterals are $4 + 3 + 2 + 1 \ = \ \color{#D61F06}{10}$ . Now, there would be similar trapezium in all the four sections making it $10 \times 4 = \color{#D61F06}{40}$ of them.
Logically thinking, if we had the $n^{th}$ square figure we would have $4\left[(n-1) \ + \ (n-2) \ + \ \cdots \ + \ 2+1\right] \ = \ \color{#3D99F6}{2n(n-1)}$ of them. $\longrightarrow \boxed{3}$ .

Another type of quadrilaterals embedded here are the concentric squares visible . Here there are $\color{#D61F06}{5}$ of them. Obviously, the $n^{th}$ square figure would have $\color{#3D99F6}{n}$ of them. $\longrightarrow \boxed{4}$

Another type of quadrilaterals is of the form $JBKI$ . Here we consider one base line in section one and one perpendicular line in section 2. Suppose we consider $JI$ as the base line then we can form such quadrilaterals by choosing the base line as $HR$ , $FP$ , $DN$ or $BK$ . Now, if we change the base line to line $GH$ , we would have only 3 options for the perpendicular line. If the base line was $EF$ , then the options would decrease to two and for base line $CD$ only $CDBK$ is possible. This makes $4 + 3 + 2 + 1 = \ \color{#D61F06}{10}$ of them. But we can also consider the base line to be in section 2 and the perpendicular line to be in section 1. This makes $10 + 10 = \color{#D61F06}{20}$ in each two consecutive sections. Now there are our such consecutive sections, this makes $20 \times 4 = \color{#D61F06}{80}$ such quadrilaterals. Logically thinking, if we had the $n^{th}$ square figure we would have $2\left( 4\left[(n-1) \ + \ (n-2) \ + \ \cdots \ + \ 2+1\right]\right) \ = \ \color{#3D99F6}{4(n-1)n}$ of them. $\longrightarrow \boxed{5}$ .
From $\boxed{3}$ , $\boxed{4}$ and $\boxed{5}$ , we get the total number of quadrilaterals in the $n^{th}$ squared figure as $2n(n-1) \ + \ n \ + \ 4(n-1)n \ = \color{#20A900}{n(6n - 5)}$ of them.

Now, to find the number of pentagons is quite simple.
There is only one type of pentagons in this figure. $ABOKL$ is one of them, here the base line is $AL$ , there are four more such pentagons which can be obtained by changing the base line to $CM$ , $EQ$ , $GS$ or $IT$ . This makes $\color{#D61F06}{5}$ of them. Now, there are four such sections, making a total of $5 \times 4 = \ \color{#D61F06}{20}$ of them.
Logically thinking, if we had the $n^{th}$ squared figure we would have $\color{#20A900}{4n}$ of them.

Now, to find the number of hexagons.
One type of hexagons is $DBALMC$ . Here the “opposite vertices” are $A$ and $C$ . So, the number of hexagons in this section can be obtained by finding the total number of pairs of opposite vertices which can be selected from the lot of $A$ , $C$ , $F$ , $G$ and $I$ . This can be given by $\dbinom{5}{2} = \ \color{#D61F06}{10}$ . Now there are four such consecutive sections, so a total of $10 \times 4 = \color{#D61F06}{40}$ such hexagons can be obtained. Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \dbinom{n}{2} \ = \color{#3D99F6}{2n(n-1)}$ of them. $\longrightarrow \boxed{6}$ .
Another type of hexagons is $JHRUTI$ . Here, $J$ and $H$ can be considered as the “edge points” of this hexagon. Now, the number of such hexagons can be obtained by choosing a pair of the possible edge points $J$ , $H$ , $F$ , $D$ , $B$ which can be given by $\dbinom{5}{2}$ . Now there are four such sections possible so the number of such hexagons is $4 \times \dbinom{5}{2} \ = \color{#D61F06}{40}$ . . Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \dbinom{n}{2} \ = \color{#3D99F6}{2n(n-1)}$ of them. $\longrightarrow \boxed{7}$ .
Another type of hexagons is $JHRSTI$ . Here, $J$ and $H$ can be considered as the “edge points” of this hexagon. Now, the number of such hexagons can be obtained by choosing a pair of the possible edge points $J$ , $H$ , $F$ , $D$ , $B$ which can be given by $\dbinom{5}{2}$ . Now there are four such sections possible so the number of such hexagons is $4 \times \dbinom{5}{2} \ = \color{#D61F06}{40}$ . . Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \dbinom{n}{2} \ = \color{#3D99F6}{2n(n-1)}$ of them. $\longrightarrow \boxed{8}$ .
Another type of hexagons is $JHRSGI$ . Here, $J$ and $H$ can be considered as the “edge points” of this hexagon. Now, the number of such hexagons can be obtained by choosing a pair of the possible edge points $J$ , $H$ , $F$ , $D$ , $B$ which can be given by $\dbinom{5}{2}$ . Now there are four such sections possible so the number of such hexagons is $4 \times \dbinom{5}{2} \ = \color{#D61F06}{40}$ . . Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \dbinom{n}{2} \ = \color{#3D99F6}{2n(n-1)}$ of them. $\longrightarrow \boxed{9}$ .
Another type of hexagons is $JHROTI$ . Here, $J$ and $H$ can be considered as the “edge points” of this hexagon. Now, the number of such quadrilaterals can be obtained by choosing a pair of the possible edge points $J$ , $H$ , $F$ , $D$ , $B$ which can be given by $\dbinom{5}{2}$ . Also, this hexagon is not symmetrical as the previous hexagons. So, for the two consecutive sections, the number of such hexagons is $2\dbinom{5}{2} = \ \color{#D61F06}{20}$ . Now there are four such sections possible so the number of such hexagons is $4 \times 20 \ = \color{#D61F06}{80}$ .. Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \left( 2 \times \dbinom{n}{2} \right) \ = \color{#3D99F6}{4n(n-1)}$ of them. $\longrightarrow \boxed{10}$ .
Another type of hexagons is $JHRSOI$ . Here, $J$ and $H$ can be considered as the “edge points” of this hexagon. Now, the number of such quadrilaterals can be obtained by choosing a pair of the possible edge points $J$ , $H$ , $F$ , $D$ , $B$ which can be given by $\dbinom{5}{2}$ . Also, this hexagon is not symmetrical as the previous hexagons. So, for the two consecutive sections, the number of such hexagons is $2\dbinom{5}{2} = \ \color{#D61F06}{20}$ . Now there are four such sections possible so the number of such hexagons is $4 \times 20 \ = \color{#D61F06}{80}$ .. Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \left( 2 \times \dbinom{n}{2} \right) \ = \color{#3D99F6}{4n(n-1)}$ of them. $\longrightarrow \boxed{11}$ .

Another type of hexagons is $JOGSTI$ , the number of such hexagons is $4 \times 20 \ = \color{#D61F06}{80}$ .. Logically thinking, if we had the $n^{th}$ squared figure we would have $4 \times \left( 2 \times \dbinom{n}{2} \right) \ = \color{#3D99F6}{4n(n-1)}$ of them. $\longrightarrow \boxed{12}$ .

So, from $\boxed{6}$ , $\boxed{7}$ , $\boxed{8}$ , $\boxed{9}$ , $\boxed{10}$ , $\boxed{11}$ , $\boxed {12}$ the total number of hexagons in the figure is $\color{#20A900}{20n(n-1)}$ .

Therefore, $\delta_{n} = 8n$ , $\beta_{n} = n(6n - 5)$ , $\gamma_{n} = 4n$ , $\delta_{n} = 20n(n-1)$ .

$\begin{aligned} \lim_{n \to \infty} \dfrac{\delta_{n} \times \gamma_{n}}{\alpha_{n} \times \beta_{n}} & = \dfrac{20n(n-1) \times 4n}{8n \times n(6n-5)}\\ \\ & = \dfrac{80}{48}\\ \\ & = \dfrac{5}{3}\\ \\ \therefore a + b = 5 + 3 & = \color{#69047E}{\boxed{8}} \end{aligned}$