91Ó°ÊÓ

The labelled diagram is:

From the given diagram it can be noticed that:

$AD=AE+ED$.

Therefore, it can be obtained that:

$\begin{array}{c}AD=AE+ED\\ =2+3\\ =5\end{array}$

Therefore, it can be noticed that:

$AD=AB=BC=CD=5$.

Therefore, $ABCD$is a square.

Therefore, all the angles in the square $ABCD$will be of measure $90°$each.

Consider the measure of the angle $∠DEH$as x.

In the triangle $\mathrm{Δ}DEH$, it can be noticed that:

$\begin{array}{c}∠DEH+∠EHD+∠EDH=180\\ x+∠EHD+90=180\\ ∠EHD=180−90−x\\ ∠EHD=90−x\end{array}$

It is given that the shaded figure is a square that implies $EFGH$is a square.

Therefore, all the angles in the square $EFGH$will be of measure $90°$each.

Now, from the given diagram it can also be noticed that the angles $∠DEH$, $∠HEF$and $∠FEA$forms the linear pair.

Therefore, it can be obtained that:

$\begin{array}{c}∠DEH+∠HEF+∠FEA=180\\ x+90+∠FEA=180\\ ∠FEA=180−90−x\\ ∠FEA=90−x\end{array}$

Now as in the triangles $\mathrm{Δ}DEH$and $\mathrm{Δ}AFE$, it can be noticed that:

$\begin{array}{l}∠EHD=FEA=90−x\\ ∠EDH=FAE=90\end{array}$

Therefore, by AA similarity, the triangles $\mathrm{Δ}DEH$and $\mathrm{Δ}AFE$are similar.

Therefore, $∠DEH=AFE=x$.

Similarly, it can be obtained that all the triangles $\mathrm{Δ}DEH$, $\mathrm{Δ}AFE$, $\mathrm{Δ}BGF$and $\mathrm{Δ}CHG$are similar.

Therefore, the diagram is:

Now it can be noticed that one side which is the hypotenuse of each triangle are also equal as they are the sides of the square $EFGH$.

Therefore, the triangles are congruent.

Therefore, it can be said that:

$\mathrm{Δ}DEHâ‰\dots \mathrm{Δ}AFEâ‰\dots \mathrm{Δ}BGFâ‰\dots \mathrm{Δ}CHG$.

Therefore, by corresponding parts of congruent triangles it can be noticed that:

$AF=BG=CH=DE=3$ and $EA=FB=GC=HD=2$

Therefore, the diagram is:

Now, in the triangle $\mathrm{Δ}GCH$, it can be noticed that:

Therefore, the measure of the side $GH$ is $\sqrt{13}$.

The area of the square is given by:

$area=side×side$

Therefore, the area of the square $EFGH$ having measure of each side as $\sqrt{13}$ is given by:

Therefore, the area of the shaded square is 13 units.