Find the area of the blue shaded region of the square in the following image:
[Added by Jack:]
The area of the triangle in the middle of the square is given by $$ 4.8\times 6=28.8\ (cm^2) $$
Other than this, it seems difficult to go further with the given information. It seems that one has yet to use the assumption of "square".
How can one solve this problem?
On
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
$$y=\frac{x}{t}$$
and the other oblique is perpendicular, from the bottom right corner, hence
$$y=-t(x-1).$$
The intersection point is
$$\left(\frac{t^2}{t^2+1},\frac{t}{t^2+1}\right).$$
To determine $t$, we express the value of the ratio of the known sides,
$$\frac{4.8}{6}=\frac{\left\|\left(\dfrac{t^2}{t^2+1},\dfrac{t}{t^2+1}\right)-(t,1)\right\|}{\left\|\left(\dfrac{t^2}{t^2+1},\dfrac{t}{t^2+1}\right)-(1,0)\right\|}.$$
All the rest will follow.
On
Draw the red line as shown:
$\hspace{5cm}$
Note: $$\begin{cases}1=2+3\\ 4=3+6\end{cases} \Rightarrow 1+6=2+4 \Rightarrow S=\frac12 \cdot 4.8\cdot 6=14.4.$$ 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.
On
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 $X=AG\cap BH$. 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: $$ \begin{aligned} 2\operatorname{Area}(\Delta BXG) &= 2\operatorname{Area}(\Delta YXW) + 2\operatorname{Area}(\Delta YWB) + 2\operatorname{Area}(\Delta GWB) \\ &= \operatorname{Area}(\square XYZW) + YB\cdot XW + WG\cdot XB \\ &= \operatorname{Area}(\square XYZW) + AX\cdot XY + YE\cdot XB \\ &= \operatorname{Area}(\square XYZW) + 2\operatorname{Area}(\Delta AXE) + 2\operatorname{Area}(\Delta XEB) \\ &=\text{"Blue area plus green area"} \\ &= 2\operatorname{Area}(AXH) + 2\operatorname{Area}(DHB) \ , \end{aligned} $$ and the result follows.
$\square$
I tried to show "the square in the middle", and use in the proof the full symmetry.
Later edit: Here is a further picture that illustrates the way to split the triangle $\Delta BXG$ in three triangles, that can be pasted elsewhere with equivalent area:
$\Delta BXG$ in three parts, one part is half of a square, two other parts add to a side triangle.">
Please see the following picture.