Firstly, we draw a rough sketch of rectangle OKAY, which helps us in deciding steps of construction.

We know that, in a rectangle, opposite sides are equal and parallel and angles between two adjacent sides is 90°.
In rectangle OKAY,
OK = AY = 7 cm, KA = OY = 5 cm
and \(\angle \)OKA=\(\angle \)KOY = 90°
Steps of construction
Steps IDraw OK = 7 cm.
Step IIAt K, draw a ray KX making an \(\angle \)OKX = 90°
Step III Cut KA = 5 cm from ray KX.
Step IV With A as centre and radius 7 cm, draw an arc.
Step V With O as centre and radius 5 cm, draw another arc which intersects the arc drawn in Step IV at Y.
Step VI Join AY and OY.

We know that, in a rhombus, all sides are equal and diagonals bisect each other (perpendicularly). Here, two diagonals of rhombus are given. So, firstly we draw a rough sketch of rhombus BEND, which helps us In deciding our

Steps of construction.
Step IDraw DE = 6.5 cm
Step II With D as centre and radius more than \(\frac { 1 }{ 2 } \) DE, draw two arcs on both sides of DE.
Step III With E as centre and same radius as taken in Step II, draw two arcs on both sides of DE, which intersect arcs drawn in Step II, at P and Q, respectively.
Step IV Join PQ. Let it intersects DE at M. Then, M is the mid-point of DE.
Step V Now, cut MN=\(\frac { 1 }{ 2 } \times 5.6\quad cm\) = 2.8 cm from MP.
Step VI Cut MB=\(\frac { 1 }{ 2 } \times 5.6\quad cm\) = 2.8 cm from MQ.
Step VII Join DN, NE, EB and BD.

Thus, BEND is the required rhombus.

We know that, the length PQ and QR of a rectangle PQRS. Also, we know that in a rectangle opposite sides have equal length and each of the angle is 90°. 
Thus, we have PQ=RS and QR=PS and \(\angle \)PQR = 90°.
Steps of construction
Step I Draw PQ.
Step II Makes \(\angle \)PQX=90°.
Step III Cut QR from QX
Step IV From P cut-off an arc equal to QR.
Step V From R, cut-off an arc equal to PQ.
Step VI Mark S as the intersection point of both the arcs. Join PS and RS to get the required rectangle PQRS.
 

Let us draw a rough sketch to visualise the quadrilateral. See the following figure.

Steps of construction
Step I From the rough sketch, it is easy to see that a \(\Delta \)BCD can be constructed using SSS construction condition.
Draw the \(\Delta \)BCD.

Step II Now, we will locate a point A, which would be on the side opposite to C with reference to B. 
A is 5.5 cm away from D. So, with D as centre, draw an arc of radius 5.5 cm.

Step III A is 5 cm away from B. So, with B as centre, draw an arc of radius 5 cm.

Step IV A should be the intersection point of both the arcs drawn. Mark A and join BA and DA.

Hence, ABCD is the required quadrilateral.

Yes, to construct a unique parallelogram whose two adjacent sides are given, we need the angle included between them, because if angle is not given, then we cannot find a unique parallelogram. So, we cannot construct the parallelogram with the given information.

No, as we know that the sum of measures of angles of a quadrilateral is 360°.
Here, \(\angle\)A + \(\angle\)B + \(\angle\)C =120° + 105° + 160°
= 385° > 360°
So, it is not possible to have such a quadrilateral.

In a parallelogram, the lengths of adjacent sides are known. So, we do not still need measures of the angles to construct a parallelogram, because opposite sides of a parallelogram are equal and parallel.

No, we cannot construct a quadrilateral PQRS, because we cannot draw the \(\Delta\)QSP as SQ + PQ \(\ngtr \)SP.

In case of a quadrilateral, it is necessary to have atleast the knowledge of five parts to be able to construct it uniquely. We shall need data about specified parts of the quadrilateral as follows:
(a) When four sides and one diagonal are given.
(b) When two diagonals and three sides are given.
(c) When two adjacent sides and three angles are given.
(d) When three sides and two included angles are given.
(e) When some special properties are given.

Steps of construction
Step I Let AB = 4 cm and draw \(\angle\)ABX = 90°.
Step II Cut the length BC = 4 cm from BX.
Step III Draw \(\angle\)BAY = 90°.
Step IV Cut AD = 4 cm from AY.
Step V Join DC.

Thus, we get a square ABCD, whose one of the side is 4 cm.

We know that, in a rhombus, all four sides are equal in length and diagonals are perpendicular bisector of each other.
In rhombus ABCD,
Diagonals AC = 5.2 cm and BD =6.4 cm 
Firstly,we draw a rough sketch of rhombus say ABCD, which helps us in deciding steps of construction.
Steps of construction
Step IDraw AC = 5.2 cm.
Step II With A as centre and radius more than \(\frac { 1 }{ 2 } \) AC, draw two arcs on both sides of AC,
Step III With C as centre and same radius as taken in Step II, draw two arcs on both sides of AC, which intersect arcs drawn in Step II, at X and Y respectively.
Step IV Join XY. Let XY meet AC at point O. Then O is the mid-point of AC.
Step V Cut-off OB=\(\frac { 6.4 }{ 2 } =3.2\)cm from OX and \(OD=\frac { 6.4 }{ 2 } =3.2\) cm from OY.
Step VI Join AB, BC, CD and DA

Thus, ABCD is the required rhombus.

Firstly, draw a rough sketch of quadrilateral DEAR, which helps us in deciding the steps of construction.

Steps of construction
Step I Draw EA = 5 cm.
Step II At A, draw a ray AX making and \(\angle \)EAX = 90 °
Step III Cut AR = 4.5 cm from ray AX.
Step IV At E, draw a ray EY making \(\angle \)AEY = 60°
Step V Join DR.

Thus, DEAR is the required quadrilateral.

TRUE is the required quadrilateral.