:
False
In a trapezium, only one pair of sides is parallel.

:

We know that the diagonals of a square intersect each other at right angles.
Hence,
$\angle POQ={90}^{\circ }$.

:
Given, an isosceles trapezium, where $AB\parallel DC$, AD = BC and $\angle A={60}^{\circ }$.
Then, $\angle B={60}^{\circ }$.
Draw a line parallel to BC through D which intersects the line AB at E (say).

Then, DEBC is a parallelogram, where
BE = CD = 20 cm and DE = BC = 10 cm
Now, $\angle DEB+\angle CBE={180}^{\circ }$
[adjacent angles are supplementary in parallelogram]
$\Rightarrow \angle DEB={180}^{\circ }-{60}^{\circ }={120}^{\circ }$
$\therefore In\Delta ADE,\angle ADE={60}^{\circ }$ [exterior angle]
Also, $\angle DEA={60}^{\circ }$ [$\because$ AD = DE = 10cm and $\angle DAE={60}^{\circ }$]
Then, $\Delta ADE$ is an equilateral triangle.
$\therefore AE=10cm$
$\Rightarrow AB=AE+EB=10+20=30cm$
Hence, x = 30 cm.

:
Let EC and FC be altitudes and $\angle ECF={30}^{\circ }$.

Let $\angle EDC=x=\angle FBC$
So, $\angle ECD=90-x^{\circ }and\angle BCF={90}^{\circ }-x$
So, by property of the parallelogram,
$\angle ADC+\angle DCB={180}^{\circ }\angle ADC+(\angle ECD+\angle ECF+\angle BCF)={180}^{\circ }\Rightarrow x+{90}^{\circ }-x+{30}^{\circ }+{90}^{\circ }-x={180}^{\circ }\Rightarrow -x={180}^{\circ }-{210}^{\circ }=-{30}^{\circ }\Rightarrow x={30}^{\circ }$
Hence, $\angle DCB={30}^{\circ }+{60}^{\circ }+{60}^{\circ }={150}^{\circ }$.

:
Supplementary
By property of a parallelogram, we know that the adjacent angles of a parallelogram are supplementary.

:
Angles
In a regular polygon, all sides are equal and all angles are equal.

Answer: Option D. -> Kite
:
D
A quadrilateral with two distinct pairs of equal-length adjacent sides is known as akite.

Answer: Option B. -> False
:
B
Let$\angle A=x^{\circ }$
then $\angle D=(180-x{)}^{\circ }$
(Since the adjacent angles of a parallelogram sum up to ${180}^{\circ })$
$\therefore \frac{\angle A}{2}=\frac{x}{2}=\angle DAO$
and$\frac{\angle D}{2}=\frac{(180-x)}{2}=90-\frac{x}{2}=\angle ADO$
In $\Delta AOD,$
$\angle AOD+\angle ADO+\angle DAO={180}^{\circ }$[Angle sum property of a triangle]
$\angle AOD={180}^{\circ }-[\angle ADO+\angle DAO]$
$\angle AOD={180}^{\circ }-[\frac{x}{2}+90-\frac{x}{2}]$
$or\angle AOD=180-90={90}^{\circ }$
$\therefore$ The measure of $\angle AOD$ is ${90}^{\circ }$

Answer: Option B. -> Square
:
B
A regular polygon is a polygon which is equiangular i.e. all the angles are equal and equilateral i.e. all the sides are equal.
The triangle if equilateral is a regular polygon. If not, it is not a regular polygon.
A rhombus has equal sides but not equal angles.
A Rectangle has equal angles but not equal sides. Hence, they are not regular polygons.
A square has equal sides and equal angles. Hence, it is a regular polygon.

Answer: Option C. -> 37.6 cm
:
C
Let ‘a’ be one side of the kite = 7.8 cm
and‘b’ be the other side of the kite = 11 cm
Perimeter of the kite = 2a + 2b units
$=2\times 7.8+2\times 11=15.6+22=37.6cm$