Answer: Option C. -> Some kites are parallelograms
:
C
The kites that have all the sides equal and twopairs of sides parallel are parallelograms. These kites are rhombuses.
Hence, some kites are parallelograms.

Answer: Option B. -> False
:
B
The quadrilateral formed between two pairs of parallel lines is a parallelogram. In a trapezium, the only one pair of opposite sides is parallel.

Answer: Option C. -> IV  
:
C

A line segment connecting any two non-consecutive vertices of a polygon is known as diagonal.A polygon with an exterior diagonal is known as concave polygon.

Only in these two options, there exists an exterior diagonal. So, these are concave polygons. But II is a concave pentagon. Only IV is a concave quadrilateral. Moreover, a concave polygon hasone or more interior angle greater than 180°.

:
We know that, in a parallelogram, opposite sides are parallel and equal.
The diagonals bisect each other but not always at right angles.
The diagonals are not perpendicular to each other in allparallelograms.
All angles are equalin allparallelograms.
Hence, option (a) is correct.

:
Since,

We have ,
PO = RU = 5.5 cm, OU = RP = 7.2 cm
(Since the opposite sides of a parallelogram are equal)
Steps of Construction
Step I Draw PO = 5.5 cm.
Step II Construct $\angle POX={70}^{\circ }$.
Step III With O as centre and radius OU = 7.2 cm, draw an arc.
Step IV With U as centre and radius UR = 5.5 cm, draw an arc.
Step V With P as centre and radius PR = 7.2 cm, draw an arc to cut the arc drawn in Step IV.
Step VI Join PR and UR.
Hence, POUR is the required parallelogram.

:
We know that.
$\angle A+\angle D=180$ [$\because$ sum of adjacent angle of a trapezium is ${180}^{\circ }$]
${105}^{\circ }+\angle D={180}^{\circ }$
$\angle D={75}^{\circ }$

Steps of Construction
Step I Draw AB = 4 cm.
Step II Draw $\overline{AX}$ such that $\angle BAX={105}^{\circ }$.
Step III Mark a point D on AX such that AD = 3 cm.
Step IV Draw $\overline{DY}$ such that $\angle ADY={75}^{\circ }$.
Step V Mark a point C such that CD = 8 cm.
Step VI Join BC.
Hence, ABCD is the required trapezium.

:
Let us assume that a convex polygon has four or more acute angles. If an interior angle is acute, then the corresponding exterior angle will be obtuse ( greater than ${90}^{\circ }$). So, the sum of the exterior angles of such a polygon will begreater than $4\times {90}^{\circ }={360}^{\circ }$.
However, this is impossible, since the sum of the exterior angles of a polygon must always be equal to ${360}^{\circ }$. Hence, a polygon can have, at most, 3 obtuse exterior angles.

:
We know that, if in a parallelogram one angle is of ${90}^{\circ }$, then all angles will be of ${90}^{\circ }$ and a parallelogram with all angles equal to ${90}^{\circ }$ is called a rectangle.

:
Given, $\angle RAI={35}^{\circ }$.
$\therefore \angle PRA={35}^{\circ }$ . [$PR\parallel AI$ and AR is transversal]
$\Rightarrow \angle ARI={90}^{\circ }-\angle PRA={90}^{\circ }-{35}^{\circ }={55}^{\circ }$.
$\because RM=IM,\angle MRI=\angle MIR={55}^{\circ }$.
In $\Delta RMI,$
$\angle RMI+\angle MRI+\angle MIR={180}^{\circ }$ . [Angle sum property]
$\Rightarrow \angle RMI={180}^{\circ }-{55}^{\circ }-{55}^{\circ }={180}^{\circ }-{110}^{\circ }={70}^{\circ }$
Also, $\angle PMA=\angle RMI$[vertically opposite angles]
$\Rightarrow \angle PMA={70}^{\circ }$

:
28 cm
Perimeter of a parallelogram = 2(Sum of lengths of adjacent sides)
$=2(5+9)=2\times 14=28cm$.