Given: $AB \parallel CD$ and $AB = 2 CD$
In $\triangle OAB$ and $\triangle OCD$
$\angle AOB = \angle COD$ (Vertically opposite angles)
$\angle ODC = \angle OBA$ (Alternate angles)
$\angle OCD = \angle OAB$ (Alternate angles)
thus, $\triangle OAB \cong \triangle OCD$ (AAA rule)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{AB^2}{CD^2}$ (Similar triangle Property)
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = \dfrac{(2 CD)^2}{CD^2}$
$\dfrac{A(\triangle OAB)}{A(\triangle OCD)} = 4 : 1$

$\because$ Trapezium is having exactly one pair of parallel sides

If ABCD is a isosceles trapezium , $\angle C\quad will\quad be\quad equal\quad to\quad \angle D $
because angles on either side of the bases are same measure/size(congruent) in a isosceles trapezium.

The definition of a square does not comply with the definition of a trapezoid. 

Let the ABCD is a quadrilateral witch $\angle A, \angle B,\angle Cand \angle D$ in ratio  3:7:6:4

Kite has no set of parallel lines , while trapezium at least has one set of parallel lines.
Therefore, A is the correct answer.

A kite is a quadrilateral which is not a parallelogram and its adjacent sides are equal which make the angles by the adjacent pairs are also equal.
Therefore, A is the correct answer.

The mentioned statement is the property of kite so correct option is D

Isosceles trapezium doesn't have parallel sides equal but it has a property that there is a line of symmetry bisecting one pair of opposite sides.

Given: ABCD is a kite shaped figure. $AB = AD$, $BC = CD$, P, Q< R are mid points of AB, BC and CD respectively

$\Rightarrow$  $\angle P:\angle Q:\angle R:\angle S=3:7:6:4$    [ Given ]

Given that,

Diagonals of a square, rhombus are perpendicular but a quadrilateral with perpendicular diagonal need not be a square, rhombus. 

we have,

$ABCD$ is a kite.

Then, Given that,

$ AB=AD\,\,\,\,\,.......\,\,\left( 1 \right) $

$ CB=CD\,\,\,\,\,.......\,\,\left( 2 \right) $

If $\angle ABD={{40}^{o}}$

Then, show Diagram,

$\angle ABD=\angle ADB={{40}^{o}}$

And

$ \angle ADB=\angle CBD={{40}^{o}}\,\,\left( \text{Alternate}\,\text{angle} \right) $

$ \angle ABD=\angle CDB={{40}^{o}}\,\,\,\left( \text{Alternate}\,\text{angle} \right) $

Then $\angle B+\angle D={{40}^{o}}+{{40}^{o}}+{{40}^{o}}+{{40}^{o}}={{160}^{o}}$

But, We know that,

In any quadrilateral

$ \angle A+\angle B+\angle C+\angle D={{360}^{o}} $

$ \angle A+\angle C={{360}^{o}}-{{160}^{o}} $

$ \angle A+\angle C={{200}^{o}} $

Hence, this is the answer.

Quadrilateral with one pair of parallel sides is a trapezoid. If it has two pairs of parallel sides it is a parallelogram, but parallelograms are also trapezoids.

False.
Diagonals of trapezium does not bisect each other.

$Let\angle A,\angle B,\angle C\quad and\angle D\quad be\quad the\quad angles\quad of\quad quadilateral\ \angle A=3x,\angle B=4x,\angle C=5x\quad and\angle D=6x,where\quad x\quad is\quad a\quad constantIn\quad quadilateral\quad ABCD\quad \ \angle A+\angle B+\angle C+\angle D=360(Angle\quad sum\quad property)\ 3x+4x+5x+6x=360\ 18x=360\ x=20\ \angle A=3x=3\times 20=60\ \angle B=4x=4\times 20=80\ \angle C=5x=5\times 20=100\ \angle D=6x=6\times 20=120$
$In\quad quadilateral\quad ABCD\quad \ \angle A+\angle D=60+120=180\ \angle B+\angle C=80+100=180\ \therefore AB\parallel CD(Sum\quad of\quad consecutive\quad interior\quad angle\quad is\quad supplementary)\ But\quad \angle A+\angle B=60+80\neq 180\ \therefore AB\quad is\quad not\quad parallel\quad to\quad BC\ Hence\quad quadilateral\quad is\quad a\quad trapzium(as\quad it\quad has\quad only\quad one\quad pair\quad of\quad parallel\quad side)$\

Let angles be $3x, 7x, 6x$ and $4x$
Total sum of angles of a quadrilateral = $360^{\circ}$
$\Rightarrow 3x + 7x + 6x + 4x = 360^{\circ}$
$\Rightarrow 20x = 360^{\circ}$
$\Rightarrow x = 18^{\circ}$
$\therefore$ Angles are $ 3\, \times\, 18^{\circ} = 54^{\circ}$
 $ 7\, \times\, 18^{\circ} = 126^{\circ}$
 $ 6\, \times\, 18^{\circ} = 108^{\circ}$
 $ 4\, \times\, 18^{\circ} = 72^{\circ}$
All the angles of the figure ABCD are different, thus it is a trapezium.
Hence, option 'C' is correct.

Let $p$ and $q$ be the two parallel lines cut by two distinct transversals $l$ and $m$. The figure formed by the lines $AB, BC, CD, DA$ will always be a trapezium as at least one pair of given lines will be parallel.

An isosceles trapezium have equal opposite sides. Therefore diagonals of  trapezium are equal.

$Properties\  of \ an \ isosceles\  trapezium:  $

It has a pair of parallel sides.

It has a pair of opposite sides that are congruent.

Both pairs of opposite angles are supplementary, that is they sum to $180°$.

Consecutive angles along both bases are congruent.

Diagonals are congruent.

$\therefore $In an isosceles trapezium the base angles are always equal. Hence, it depends how we name an isosceles trapezium.

If the trapezium is an isosceles trapezoid it has equal angles according to the trapezoid theorem.
Therefore, B is the correct answer.

The pair of opposite sides of a parallelogram are equal and parallel but in the case of the trapezium, this is not true.

Given: ABCD is a trapezium. $AB \parallel DC$. P and Q are mid points of AD and BC respectively.
BP produced meets CD at E

Given: ABCD is a trapezium. $AB \parallel DC$. P and Q are mid points of AD and BC respectively.
BP produced meets CD at E

To prove: P is mid point of BE.
In $\triangle APB$ and $\triangle EPD$
$\angle APB = \angle EPD$ (Vertically opposite angles)
$\angle EDP = \angle PAB$ (Alternate angles)
$PA = PD$ (P is mid point of AD)
Thus, $\triangle APB \cong \triangle DPE$ (ASA rule)
Hence, $PE = PB$ (By cpct)
thus, P is mid point of BE

In $\triangle$ APB and $\triangle$ CPD,
$\angle APB = \angle CPD$ (Vertically opposite angles)
$\angle ABP = \angle CDP$ (Alternate angles of parallel sides AB and CD)
$\angle BAP = \angle DCP$ (Alternate angles of parallel sides AB and CD)
Hence, $\triangle APB \sim \triangle CPD$ (AAA rule)
Thus, $\frac{PA}{PC} = \frac{PB}{PD}$
$PA \times PD = PB \times PC$

Draw $DM\perp AB$ and $CN\perp AB$.$\triangle DAM\cong \triangle CBN$

Let the angles be $7x, 13x, 12x$ and $8x$
Then, $7x+13x+12x+8x={360}^{o}$
$\Rightarrow$ $40x={360}^{o}$ $\Rightarrow$ $x={9}^{o}$
$\therefore$ $40x={360}^{o}$
$\therefore$ The angles taken in order are ${63}^{o}, {117}^{o}, {108}^{o}, {72}^{o}$ 
This shows that tow pairs of adjacent angles are supplementary $({63}^{o}+{117}^{o}={108}^{o}$ and ${108}^{o}+{72}^{o}={180}^{o}$), but opposite angles are not equal.
Therefore, the given quadrilateral will be a trapezium.

AB I I CD, we have
$\angle A\, +\, \angle D\, =\, 180$ ($\because$ angles at the same side of transversal)
$\Rightarrow\, \angle D\, = 180^{\circ} - 60^{\circ} = 120^{\circ}$

In trapezium ABCD, $AB \parallel CD$, $AD = BC$
Draw a perpendicular from A on CD to meet CD at M and a perendicular from B on CD to meet at N.
Now, in $\triangle ADM$ and $\triangle BNC$
$\angle AMD = \angle BNC$ (Each $90^o$)
$AM = BN$ (distance between parallel lines)
$AD = BC$ (Given)
Thus, $\triangle ADM \cong \triangle BCN$ (SAS rule)
Hence, $\angle D = \angle C = 76^{\circ}$ (by CPCT)

The line joining the mid point of the diagonals of a trapezium is half the length of the difference between the two sides.
Let the smaller side be $x$
Then, $3 = \dfrac{97 -x}{2}$
$6= 97 - x$
$x = 91$ cm

Since, sum of all the four angles of a quadrilateral is $360^o$.

In a trapezium,     AB || CD
$\therefore$ sum of adjacent angles $= 180^o$
$\angle D + \angle A = 180^o$
$\angle A = 180^o - \angle D = 180^o - 110^o$
$\angle A= 70^o$

As per the property of isosceles trapezoid, 
Opposite sides of an isosceles trapezoid are the same length (congruent) and the angles on either side of the bases are the same size (congruent).
So, Angle C = $80^o$
Since the top and bottom angles are supplementary, we know that,
Angle A = $180 - 80$
Angle A = $100^o$
Similarly, the Angle of B = $100^o$

There is some disagreement whether parallelograms, which have two pairs of parallel sides, should be regarded as trapezoids. Some define a trapezoid as a quadrilateral having only one pair of parallel sides (the exclusive definition), thereby excluding parallelograms

In Trapezium PQRS, $\Delta OPQ$ is similar to $\Delta OSR$ by AA similarity.
$\therefore \dfrac{OP}{OR}=\dfrac{OS}{OQ}=\dfrac{PQ}{SR}$
$\therefore OP.SR=OR.PQ$
Hence, $m=1$

The line segment joining the mid points of non parallel sides of a trapezium is the average of sum of the parallel sides.
Hence, $= \dfrac{x+y}{2}$