Yes, the quadrilateral MIST can be constructed with \(\angle\)M = 100° instead of 75°.

Steps of construction:

I. Draw AB = 6.3 em.
II. At B, draw \(\overrightarrow { BX } \) , such that \(\angle \)ABX = 90°
III. From \(\overrightarrow { BX } \), cut off BC = 6.3 cm.
IV. With centre C and radius = 6.3 cm, draw an arc.
V. With centre A and radius = 6.3 cm, draw another arc to intersect the previous arc at D.
VI. Join DA and CD.
Thus, ABCD is the required square.

steps of construction:

I. Draw a line segment AB = 4.5 em.
II. At B, construct \(\angle \)ABX = 90°.
III. From \(\overrightarrow { BX } \) , cut off BC = 5.1 em.
IV. AT C, draw\(​​\overrightarrow { CY } \) such that \(\angle \)BCY= 120°.
V. From\(​​\overrightarrow { CY } \), cut off CD = 3.9 em.
VI. Join AD.
Thus, ABCD is the required quadrilateral.

Following properties are used in constructing the kite EASY:
(i) Two Diagonals intersect at right angles.
(ii) One of the diagonal bisects the other.
(iii) Pairs of consecutive sides are equal.
Let us draw a rough sketch to visualise the kite, which is given below.

Steps of construction
Step I Draw a line segment AY = 8 cm.
Step II Draw a perpendicular bisector of AY.
Let it be MN.
Step III With centre as Y draw the arc of YE = 4 cm and with centre A, draw the arc of AE = YE = 4 cm on MN.
Step IV Join EY and EA.
Step V Draw the arc of 6 cm with centre Y and draw another arc of 6 cm with centre A on MN. The intersection point of arcs give a point S.
Step VI Join YS and AS.

Thus, Easy is the required kite

We know that, a rectangle is a parallelogram whose opposite sides are equal and each of the angle is 90°.
If PQ and QR are given, then
PQ =RS and QR =PS
Steps of construction
Step I Draw PQ and make \(\angle\)POX = 90°.

Step II Cut QR from QX.
Step III Draw \(\angle\)QRY = 90° and cut RS = PQ from RY
Step IV Join SP.

Thus, we get the quadrilateral PQRS, which is a rectangle.

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.

Since, opposite sides of a parallelogram are equal.
\(\therefore\)  AB = DC = 4 cm \(\Rightarrow\) BC = AD = 5 cm
Steps of construction
Step I Draw AB = 4 cm.
Step II Draw ray BX such that \(\angle\) ABX = 60°.
Step III Mark a point C such that, BC = 5 cm.
Step IV With C and A as centre, draw arcs of length 4 cm and 5 cm respectively.
Step V These axes intersecting at D. Join AD and CD.

Hence, ABCD is the required parallelogram.

\(\because\)  \(\angle \)A+\(\angle \)D = 180°
\(\therefore\) 105°+\(\angle \)D=180°  \(\Rightarrow\) \(\angle \)D = 75°
Steps of construction
Step I Draw AB = 4 cm.
Step II Draw  \(\overline { AX } \) such that, \(\angle \)BAX = 105°.
Step III Mark a point D on AX such that AD = 3 cm.
Step IV Draw \(\overline { DY } \)  such that \(\angle \)ADY = 75°.
Step V Mark a point C such that CD = 8 cm.
Step VI Join BC.

Hence, ABCD is the required trapezium.

Let us draw a rough sketch of the required quadrilateral RAIL.

Steps of construction
Step I Draw AI = 4.5 cm and draw angle \(\angle \)lAX =105° and \(\angle \)AIY =110°.
Step II Draw an arc of 6 cm with centre A on AX and get the point R. So, AR = 6 cm.
Step III Now, make an angle of 60° on R.
Step IV Let it intersect IY at L.

Thus, quadrilateral RAIL is the required quadrilateral.

Steps of construction
StepI Draw MT = 6 cm and draw \(\angle\)TMX = 45°.
StepII Cut MA from MX, so that MA = 10 cm.
StepIII Draw an arc of 5 cm with centre T and draw an arc of 3 cm with centre A.
StepIV Mark the intersection point of both the arcs as E. Join TE and AE.

Thus, quadrilateral TEAM is constructed.

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.

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.

TRUE is the required quadrilateral.

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.