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Find the area of the blue shaded region of the square in the following image:

How to solve this question?

Draw the red line as shown:

$hspace5cm$

Note:
$$begincases1=2+3\
4=3+6endcases Rightarrow 1+6=2+4 Rightarrow S=frac12 cdot 4.8cdot 6=14.4.$$
To prove $4=3+6$, note that the triangles $3-5-6$ and $4-5$ are congruent, because corresponding one side and three angles are equal.

Please see the following picture.

Hint:

Consider a unit square with the bottom left corner at the origin and the point $(t,1)$.

The oblique from the bottom left corner has equation

$$y=fracxt$$

and the other oblique is perpendicular, from the bottom right corner, hence

$$y=-t(x-1).$$

The intersection point is

$$left(fract^2t^2+1,fractt^2+1right).$$

To determine $t$, we express the value of the ratio of the known sides,

$$frac4.86=fracleft(dfract^2t^2+1,dfractt^2+1right)-(1,0)right.$$

All the rest will follow.

Let us start with a square $ABCD$ and construct on the sides $AB$, $BC$, $CD$, $DA$ points $E,F,G,H$ so that $AE=BF=CG=DH$:

Let $X$ be the intersection $X=AGcap BH$. Similar points $Y,Z,W$ obtained by rotation around the center of the square were also drawn. This realizes the situation from the given problem. We want to become independent of the given values for $XG$ and $XB$, and show in general:

The blue area together is equal to the area of triangle $Delta BXG$.

(To have some symmetry, and some common part with $Delta BXG$ some green triangles have been also drawn.)

(Note that because of $DH=CG$ the area of the blue triangle $Delta DHB$ is equal to the area of the blue triangle in the OP, $Delta BGC$.)

Proof:
$$
beginaligned
2operatornameArea(Delta BXG)
&=
2operatornameArea(Delta YXW)
+
2operatornameArea(Delta YWB)
+
2operatornameArea(Delta GWB)
\
&=
operatornameArea(square XYZW)
+
YBcdot XW
+
WGcdot XB
\
&=
operatornameArea(square XYZW)
+
AXcdot XY
+
YEcdot XB
\
&=
operatornameArea(square XYZW)
+
2operatornameArea(Delta AXE)
+
2operatornameArea(Delta XEB)
\
&=text"Blue area plus green area"
\
&=
2operatornameArea(AXH) +
2operatornameArea(DHB)
,
endaligned
$$
and the result follows.

$square$

I tried to show "the square in the middle", and use in the proof the full symmetry.