In the adjoining figure, $$(\triangle ABC)$$ is an isosceles right-angled triangle. $$(BE)$$ is perpendicular to $$(AD)$$. If $$(BE) = 1\text{ cm}$$, then the area in $$\text{cm}^2$$ of quadrilateral $$(ABCD)$$ is

Place $$A = (0,0)$$, let $$AD$$ be horizontal and write $$B = (u,1)$$ because $$BE = 1$$. Since $$\triangle ABC$$ is right isosceles at $$B$$, a rotation of $$\overrightarrow{BA}$$ through $$90^\circ$$ gives $$C = (u+1,1-u)$$, and hence $$D = (u+1,0)$$. The shoelace formula gives the area as $$\frac{1}{2}\left|u(1-u)-(u+1)-(1-u)(u+1)\right| = 1$$.

Alternative solution

Lets extend CD to F such that FD = BE and $$\angle BFC = 90\degree$$. 

$$\triangle BAC$$ is a right angled isosceles triangle. 

$$ \angle BAC = \angle BCA = 45\degree$$

Lets assume $$\angle CAD = \theta$$

In $$\triangle BAE$$ we get $$ \angle ABE = 180\degree - 90\degree - (45\degree + \theta) = 45 -\theta$$

Now we know that $$\angle EBF = 90\degree$$ (By construction)

$$\angle ABC = 90\degree$$

Equating the two we get

$$\angle ABE + \angle EBC = \angle EBC + \angle CBF$$

$$ \implies \angle ABE = \angle CBF$$

Now, in $$\triangle ABE, \triangle CBF$$

$$AB = BC, \angle ABE = \angle CBF , \angle BFC = \angle BEA$$. Thus, the triangles are congruent to each other.

This implies $$ BE = BF = 1$$

Since all the angles are $$90\degree$$ and the adjacent sides are equal, BEDF is a square.

Now, we need to find the area of ABCD = Area of BAE + Area of BEDC.

Since triangle BAE is congruent to triangle BCF, their areas are equal.

Now we can replace 

Area of ABCD = Area of BCF + Area of BEDC = Area of square BEDF = 1