We know that, a quadrilateral has 8 elements, i.e. 4 sides and 4 angles. Firstly, draw a rough sketch of quadrilateral ABCD, which helps us in deciding of construction.

Steps of construction
Step IDraw AB = 4.5 cm.
Step II With A as centre and radius 7 em, draw an arc.
Step III With B as centre and radius 5.5 cm, draw another arc which intersects the arc drawn in Step II at C.
Step IV With A as centre and radius 6 cm, draw an arc on the side opposite to B with reference to AC.
Step V With C as centre and radius 4 cm, draw another arc which intersects the.arc drawn in Step IV at D.
Step VI Join BC, CD, DA and AC.

JUMP is the required quadrilateral.

We know that, a quadrilateral has 8 elements, i.e. 4 sides and 4 angles. Firstly, draw a rough sketch of quadrilateral ABCD, which helps us in deciding of construction.

Steps of construction
Step I Draw AB = 4.5 cm.
Step II With A as centre and radius 7 em, draw an arc.
Step III With B as centre and radius 5.5 cm, draw another arc which intersects the arc drawn in Step II at C.
Step IV With A as centre and radius 6 cm, draw an arc on the side opposite to B with reference to AC.
Step V With C as centre and radius 4 cm, draw another arc which intersects the.arc drawn in Step IV at D.
Step VI Join BC, CD, DA and AC.

Thus, ABCD is the required quadrilateral.

Steps of construction

Step I Draw ER = 6 cm.
Step II Draw an arc of 2 cm with centre R and draw an arc of 7 cm with centre E.
StepIII Mark the intersection of both the arcs as O. Join OR and OE.
StepIV Draw an arc of 4 cm with centre R and draw an arc of 3 cm with centre O.
StepV Mark the intersection of both the arcs as M. Join OM and EM.
Thus, we get the quadrilateral MORE.

Steps of construction
Step I Draw AC = 7.6 cm.
Step II Draw an arc of 4.5 cm with centre as A draw an arc of 6.4 cm with centre as C.
Step III Mark the intersection point as B. Join AB and BC.
Step IV Now, opposite to the point B with reference to AC. Draw an arc of 5.6 cm with centre as A and draw an arc of 4.8 cm with centre as C.
Step V Mark the intersection point as D. Join AD and DC.

Thus, we get the required quadrilateral ABCD.

We know that, in a parallelogram opposite sides are parallel and equal to each other.

In parallelogram OKAY,
OK = AY = 55 cm
and KA = OY = 4.2 cm
Firstly, draw a rough sketch of parallelogram OKAY which helps us in deciding steps of construction.
Steps of construction
Step I Draw OK = 55 cm.
Step II At K, draw a ray KX making any obtuse angle at K.
Step III Cut KA = 4.2 cm from ray KX.
Step IV Now, take O as centre and radius 4.2 cm, draw an arc above to OK.
Step V Take A as centre and radius 5.5 cm, draw another arc which intersect the arc drawn in Step IV at Y.
Step VI Join OY and AY.

Thus, parallelogram OKAY is the required parallelogram, which is unique.

We know that in a ·rectangle, opposite sides are parallel and D equal and each angle is of 90°. Firstly, we draw a rough sketch of rectangle PQRS, which helps us in deciding steps of construction.

Steps of construction
Step I Draw PQ = 5 cm.
Step II At Q, draw a ray QX making \(\angle \)PQX = 90°.
Step III Cut QR = 4 cm from ray QX.
Step IV At P, draw a ray PY making \(\angle \)QPY = 90°.
Step V Cut PS = 4 cm from PY.
Step VI Join SR.

Thus, PQRS is the required rectangle.

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.

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, since, using angle sum property of a quadrilateral PLAN, we see that
\(\angle\) P+ \(\angle\)L + \(\angle\)A + \(\angle\)N=75° + 150°+ 140° + \(\angle\)N
= 365°+ \(\angle\)N, which is greater than 360°.
But sum of all the interior angles of a quadrilateral must be equal to 360°. So, construction of quadrilateral PLAN is not possible.

Here, PL = 6 cm, \(\angle \)A = 9.5 cm,\(\angle \)P = 75°, \(\angle \)L = 150° and \(\angle \)A = 140°.
By angle sum property of a quadrilateral,
\(\angle \)N+\(\angle \)P+\(\angle \)L+140° = 360°
\(\Rightarrow\) \(\angle \)N + 75° + 150° + 140° = 365° + \(\angle \)N \(\neq \)360°.
But sum of all the angles of a quadrilateral must be 360°.
So, we cannot construct the quadrilateral PLAN.

Yes,the quadrilateral MIST can be constructed with \(\angle \)M = 100° instead of 75°.
e.g. Construct a quadrilateral MIST, where MI = 35 cm,IS = 6.5 cm, \(\angle \)M = 75°, \(\angle \)I = 105° and \(\angle \)S = 120°.
Obviously, we can construct MIST by taking \(\angle \)M = 100°.

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