I used 25 alphabets

The area is made up of 8 triangles, all congruent to $VLK\dots(1)$

Consider $VL=\frac{1}{4}$ to be the base of the triangle $\dots(2)$

Let $V$ be the centre of a coordinate system, then the height of the triangle is the x-coordinate of $K$ . $K$ lies on the intersection of $BD$ and $ZC$ , in other words its coordinates simultaneously satisfy

$y=x\text{ and }y=\frac{1}{2}-2x$

$\implies x=\frac{1}{6}\dots(3)$

Combining these three observations gives the total area as

$8\times \frac{1}{2}\times \frac{1}{4} \times \frac{1}{6}=\frac{1}{6}$

and so

$a=\frac{1}{\frac{1}{6}}=\boxed{6}$

I am a good guesser hehehe

We have 8 congruent triangles.

angle LVK = 45 degree, VL = (1/4)ZX = 1/4 units, VK = (1/2)GK, GK = (1/3)BD,

BD = 2^(1/2) units. So, VK = (1/6)*(2^0.5)

area of triangle LVK =(1/2)* VL* VK* Sin(pi/4) = 1/48 sq. units.

AREA OF OCTAGON = 1/6 sq. units
a=6 Ans.

we have same analysis,that's nice

Note that the Area of WVL is given by $\frac{1}{2} (\frac{1}{2}) (\frac{1}{4}) = \frac{1}{16} \\$ Note that WFE and FVE have the same base and height and hence, area so triangle WVL is composed of thee triangles with the same area (2 blue and one white). As such, the Blue region FVLE is $\frac{2}{3}(\frac{1}{16}) = \frac{1}{24}\\$ The Octogon is composed of 4 quadrants with that area giving its total area as $4(\frac{1}{24}) = \frac{1}{6}\\$ Hence $a=\boxed{6}$

Not the most mathematical method, but one that is simple to understand and fits the common man on the street.

This is considered cheating to some but here we go!

We zoom in on $\triangle VLD$ .

The area of $\triangle VLD$ is $\frac { 1 }{ 16 }$ that of the area of the unit square $ABCD$ .

By measurement, we find, with a ruler on the screen, that the ratio of the length of $VK$ to the length of $KD$ is $1:2$ .

This implies that the area of $\triangle VLK$ is $\frac { 1 }{ 3 }$ that of the area of $\triangle VLD$ , as both triangles have a common perpendicular height.

This means that the area of $\triangle VLK$ is $\frac { 1 }{ 48 }$ that of the area of the unit square $ABCD$ .

Multiply that by 8 and you get the area of the octagon: $\frac { 1 }{ 6 } { units }^{ 2 }$

$\therefore a=\boxed{6}$

area of blue octagon = 8*(area of triangle LVK)

first we find area of triangle LVK

observe Line segment is divided into four parts in figure i.e., SL=LV=VH=HX

BD is divided into three parts i.e, BG = GK = KD

and also GV = VK since AYCW is parallelogram

this implies VK = VD/3 and VL = ZV/2

this implies area of triangle LVK = 1/6(area of ZVD)

but area of ZVD is 1/8(area of square)

so area of triangle of LVK = 1/8 * 1/6 *(area of square)

area of octagon = 8* (area of LVK )

area of octagon= 8*1/8 * 1/6 *(area of square)

area of octagon= 1/6 *(area of square)

therefore a =6

by putting a line connecting Z and Y we can consider that triangle HZY(1) and HIV(2) are similar triangles.... considering that VL =1/4, we can also consider that the dimension of triangle HZY(1) are Base1(HZ) = 3/4 and Height1(VY) = 1/2... knowing that HZY(1) and HIV(2) are similar, therefore their dimensions are in proportion.... B1(HZ):H1(VY) = B2(VH): H2(?)....... 3/4:1/2 = 1/4:( ?)...... and by solving the ratio and proportion, we can have H2 = 1/6.....

Area for triangle HIV = 1/2 x B x H..... then Base = 1/4 and Height = 1/6.......

A= 1/2 x 1/4 x1/6 = 1/48...... to solve for the entire octagon Area of HIV would be multiplied into 8..... 1/48 x 8 = 8/48 = 1/6......

for A= 1/a............ a =6

Even if we are rounding of the answer should be 5.