According to the angle sum property of a quadrilateral, the sum of all the angles of a quadrilateral is $={360}^{\circ }$

$\therefore$The fourth angle
$={360}^{\circ }$-(${40}^{\circ }$+${80}^{\circ }$+${90}^{\circ }$)
$={360}^{\circ }-{210}^{\circ }$
$={150}^{\circ }$

A regular quadrilateral is a shape that has four equal sides with all the interior angles equal.
Among the given options, only square has all its four sides  equal with each angle being ${90}^{\circ }$.

Trapezium is a quadrilateral in which only one pair of opposite sides is parallel.

A parallelogram is made up of two triangles, each having an area equal to the half of the product of base and corresponding height. Adding these two, we find the given statement to be true.

If one angle of a parallelogram is known, the other angles can be found out by applying the following two properties:
1. The adjacent angles are supplementary.
2. The opposite angles are equal.

In a parallelogram, the opposite angles are equal. Hence, the non-adjacent angle is ${80}^{\circ }$.

Suppose ABCD is a parallelogram and BD is the diagonal.
There are two triangles - $\Delta$ ABD and $\Delta$ CDB
In $\Delta$ ABD and $\Delta$ CDB,
AD = BC   (opposite sides of a parallelogram are equal)
AB = CD   (opposite sides of a parallelogram are equal)
BD is common
$\therefore$ By SSS criterion of congruency,
$\Delta$ ABD $\cong$ $\Delta$ CDB
Hence, the given statement is true.

In a parallelogram, the adjacent sides and angles are not equal and no sides are perpendicular to other.
$\angle A+\angle B={130}^{\circ }+{50}^{\circ }={180}^{\circ }$
Here, $\angle A$ and $\angle B$ are co-interior angles and their sum is ${180}^{\circ }$. 
By the converse of co-interior angles theorem, AD || BC and AB is the transversal.
$\angle D+\angle A={50}^{\circ }+{130}^{\circ }={180}^{\circ }$
Here, $\angle D$ and $\angle A$ are co-interior angles and their sum is ${180}^{\circ }$. 
By the converse of co-interior angles theorem, AB || DC and AD is the transversal. 

In the given parallelogram, AD || BC and AB || DC
Consider the diagonal BD which acts as a transversal for BC and AD.
$\angle ADB=\angle DBC=x^{\circ }$ ( alternate angles)

Consider the diagonal BD which acts as a transversal for AB and BC.
$\angle ABD=\angle BDC=y^{\circ }$ ( alternate angles)

We can see that
$\angle B=\angle ABD+\angle DBC=x^{\circ }+y^{\circ }$
Also, $\angle D=\angle ADB+\angle BDC=x^{\circ }+y^{\circ }$
Thus, $\angle B=\angle D$.

Hence, opposite angles of a parallelogram are equal.

Let the fourth angle be $x^{\circ }$.

By angle sum property, we know that sum of interior angles of a quadrilateral is ${360}^{\circ }$.

$\Rightarrow {37}^{\circ }+{130}^{\circ }+{53}^{\circ }+x={360}^{\circ }$

$\Rightarrow x={360}^{\circ }-({37}^{\circ }+{130}^{\circ }+{53}^{\circ })={360}^{\circ }-{220}^{\circ }={140}^{\circ }$