Answer: Option C. -> 70∘
:
C

Given: In rectangle ABCD,
$\angle BAO=35°$
To find: Angle between the diagonals $\angle AODor\angle BOC$
In $\Delta$AOB
OA = OB
$\because$ Diagonals of a rectangle bisecteach other
$\Rightarrow \angle OBA=\angle OAB={35}^{\circ }$(As the angles opposite to equal sides are equal)
Applying exterior angle property in$\Delta$AOB
$\angle AOD=\angle OBA+\angle OAB$
$\angle AOD=35°+35°$
$\angle AOD=70°$

Answer: Option C. -> diagonals of PQRS are perpendicular and bisect each other
:
C
Let ABCD be a rectangle such as AB = CD and BC = DA. P,Q,R and S are the midpoints of the sides AB, BC, CD and DA respectively.

Let us join AC and BD
In $\Delta$ABC,
P and Q are the mid-points of AB and BC respectively.
$\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC (Midpoint theorem)...........(1)
Similarly in $\Delta$ ADC,
SR || AC and SR=$\frac{1}{2}$AC (Midpoint theorem)..........(2)
Clearly, PQ || SR and PQ = SR
Since, in quadrilateral PQRS, one pair of opposite sideis equal and parallel to each other, it is parallelogram
$\therefore$ PS || QR and PS = QR (opposite sides of parallelogram).........(3)
In $\Delta$ BCD, Q and R are the mid-points of sides BC and CD respectively.
$\therefore$ QR || BD andQR=$\frac{1}{2}$BD(Midpoint theorem)..........(4)
However, the diagonals of a rectangle are equal.
$\therefore$ AC = BD ...........(5)
By using equation (1), (2), (3), (4), (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus and hence, the given statement is true.

Answer: Option B. -> False
:
B
In a trapezium only one pair of opposite sides areparallel.
Ina parallelogram, both pairs of opposite sides are equal and parallel.
Therefore a trapezium is not a parallelogram.

:
By joining the mid points of the sides of a quadrilateral, another quadrilateral is formed whose opposite sides will be equal and parallel. Hence the quadrilateral formed is aparallelogram.

Answer: Option B. -> False
:
B
Sum of all angles of a quadrilateral is ${360}^{\circ }$.
$Fourthangle={360}^{\circ }-({75}^{\circ }+{90}^{\circ }+{75}^{\circ })={120}^{\circ }$
Since the pairs of opposite angles of a parallelogram are equal, this quadrilateral cannot be a parallelogram since it has three different angles and only one pair of equal angles.

Answer: Option A. -> True
:
A
A parallelogram is a quadrilateral whose both pairs of opposite sides are equal and parallel. Since the pairs of opposite sides equal and parallel in a square, rhombus and rectangle,they are all parallelograms.

Answer: Option B. -> Circle
:
B
A quadrilateral is a polygon with four sides and four vertices.
A quadrilateral has four sides, four angles and four vertices. Square, trapezium and rhombus are quadrilaterals. A circle can be approximated as a polygon with infinite sides but it is not a polygon.
It is because a polygon must have straight edges (sides). A circle doesn't have straight sides.

Answer: Option B. -> 360∘​
:
B
Akite is a quadrilateral.
We know that,the sum of the angles of a quadrilateralis equal to 360°.
$\Rightarrow$ The sum of the angles of a kiteis also equal to 360°.

Answer: Option C. -> Parallelogram
:
C
A parallelogram is a quadrilateral whose both pairs of opposite sides are parallel.

Answer: Option B. -> 36∘, 72∘, 108∘, 144∘​
:
B
Let the smallest angle of quadrilateral be $x^{\circ }$.
Then, the angles of the quadrilateral will be $x^{\circ }$, $2x^{\circ }$, $3x^{\circ }$ and$4x^{\circ }$.
Sum of the angles of a quadrilateral is ${360}^{\circ }$.
$\Rightarrow {360}^{\circ }=x+2x+3x+4x$
$\Rightarrow 10x={360}^{\circ }$
$\therefore x={36}^{\circ },2x={72}^{\circ },3x={108}^{\circ }and4x={144}^{\circ }$