In the adjoining figure, three equal squares are placed. The squares are unit squares The area of the shaded region in $$(incm^2)$$ is

Let us name the points and draw the dotted line as shown in the figure below:

Since all the squares are unit squares, each side of all 3 squares is $$1$$ cm long

Hence,$$ AE=AD+DE=1+1=2$$ cm

Now, $$EH=AH=\sqrt{DH^2+(AD=DE=1)^2}=\sqrt{5}$$ cm

Now, in triangle $$AEH$$:

Semiperimeter, $$S=1+\sqrt5$$ cm

Hence, Area of triangle AEH$$=\dfrac{ \sqrt{s(s-2)(s-\sqrt5)(s-\sqrt5)}}{2}$$ square cm

Also, Area of triangle AEH$$=\dfrac{AJ*EH}{2}=1$$

So, $$AJ=\dfrac{2}{\sqrt5}$$ cm

Since we are given that $$\angle AJE=90^\circ$$

So, in triangle $$AJE$$

$$AJ^2+EJ^2=AE^2$$

$$EJ=\sqrt{4-\dfrac{4}{5}}=\dfrac{4}{\sqrt{5}}$$ cm

Finally, Area of shaded region=Area of triangle $$AJE=\dfrac{AJ*EJ}{2}=\dfrac{4}{5}$$ square cm