In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and $\angle BAD={45}^{◦}$. If DC = 5 cm, BC = 4 cm, the area of the trapezium in sq cm is

The correct answer is 28.

To find the area of the trapezium ABCD, let us analyze its geometric properties:

1. AB is parallel to DC:
$AB\parallel DC$

2. BC is perpendicular to DC:
$BC\perp DC$
Since AB is parallel to DC, BC is also perpendicular to AB. This gives:
$\angle BCD={90}^{◦}$
and
$\angle ABC={90}^{◦}$

3. The angle BAD is given as:
$\angle BAD={45}^{◦}$

4. The given lengths are:
$DC=5\text{ cm}$
and
$BC=4\text{ cm}$

Let us draw a perpendicular line from vertex D to the side AB, meeting AB at point E. This construction divides the trapezium into a rectangle EBCD and a right-angled triangle AED (where $\angle AED={90}^{◦}$).

Since EBCD is a rectangle, its opposite sides are equal:
$DE=BC=4\text{ cm}$
and
$EB=DC=5\text{ cm}$

Now, let us find the length of AE in the right-angled triangle AED:
The angle $\angle EAD$ is ${45}^{◦}$.
The angle $\angle ADE$ can be calculated as:
$\angle ADE={180}^{◦}-{90}^{◦}-{45}^{◦}={45}^{◦}$
Since $\angle EAD=\angle ADE={45}^{◦}$, triangle AED is an isosceles right-angled triangle. Thus:
$AE=DE=4\text{ cm}$

We can now calculate the total length of the side AB:
$AB=AE+EB=4\text{ cm}+5\text{ cm}=9\text{ cm}$

The area of a trapezium is given by:
$\text{Area}=\frac{1}{2}\times \left(\text{Sum of parallel sides}\right)\times \left(\text{Perpendicular distance}\right)$
Substituting the side lengths AB and DC, and the height BC:
$\text{Area}=\frac{1}{2}\times (AB+DC)\times BC$
$\text{Area}=\frac{1}{2}\times (9+5)\times 4$
$\text{Area}=\frac{1}{2}\times 14\times 4$
$\text{Area}=28{\text{ cm}}^2$