Answer: Option A. -> kite
:
A
Here, the ratio is given to be a:b:b:a.
If KLMJis a quadrilateral, the sides are in the ratio a:b:b:a.
Let KL= a, LM= b, MJ= b and JK= a.
Here, JK and LM are equal and LM and MJ are equal.
We know that,kite is a quadrilateral with two pairs of adjacent sides equal in length.
The data given implies that we have two disjoint pairs of adjacent sides which are equal i.e. there are two pairs of adjacent sides which are equal and the opposite sides are not equal.
This is the very definition of a kite.

Answer: Option C. -> ∠A=130∘,∠C=130∘,∠B=50∘
:
C

In a parallelogram, the adjacent sides and anglesare 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.

Answer: Option A. -> True
:
A

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 ABand 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.

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

Answer: Option A. -> Only one angle is known.
:
A
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.

Answer: Option A. -> 150∘
:
A
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
$={360}^{\circ }$-(${40}^{\circ }$+${80}^{\circ }$+${90}^{\circ }$)
$={360}^{\circ }-{210}^{\circ }$
$={150}^{\circ }$

Answer: Option B. -> Rectangle
:
B
We know that,
the opposite angles of a parallelogram areequal.
ie. $\angle A=\angle Cand\angle B=\angle D...\left(i\right)$
If they are supplementary as well, then
$\angle A+\angle C={180}^{\circ }$
from (i),
$\angle A+\angle A={180}^{\circ }$
$2\angle A={180}^{\circ }$
$\angle A={90}^{\circ }$
Similarly,
$\angle C={90}^{\circ }$
Itmeans that each angle will be equal to ${90}^{\circ }.$
A parallelogram whose internal angles are all ${90}^{\circ }$ is a rectangle.Thus, the figure will be a rectangle.

:

SinceBDEF and FDCE are parallelograms,
BD = FE and DC = FE
Hence, we can see that BD is equal to DC and so, D is the mid-point of BC.

Answer: Option D. -> 140∘
:
D
Let the fourthangle be $x^{\circ }$.
By angle sum property, we know that sum of interior angles of a quadrilateral is ${360}^{\circ }$.
$\Rightarrow {37}^{\circ }+{130}^{\circ }+{53}^{\circ }+x={360}^{\circ }$
$\Rightarrow x={360}^{\circ }-({37}^{\circ }+{130}^{\circ }+{53}^{\circ })={360}^{\circ }-{220}^{\circ }={140}^{\circ }$

Answer: Option A. -> True
:
A
The sum of internal angles of a quadrilateral is ${360}^{\circ }$. The sum of external angles of any polygon, irrespective of the number of sides is${360}^{\circ }$. Hence, the given statement is true.