Steps of construction:
I. Draw a line segment BC = 4.5 cm. 
II. Construct \(\angle\)CBX = 60° at B.
III. From BX, cut off line segment BA = 5.5 cm,
IV. Join AC
Thus, ABC is the required triangle.

Since, the angle made by an arc A is equal to 60°.
Also, perpendicular made in triangle figure at B. We know that, the sum of all three angles in a triangle is equal to 180°.
\(\therefore 90^0+60^0+\angle m=180^0\)
\(\Rightarrow \angle m=180^0-90^0-60^0\)
\(\angle m=180^0-150^0\)
\(\Rightarrow \angle m=30^0\)

In the given figure, angle made by an arc is equal to 60°.
∴ The sum of all three angles in a triangle is equal to 180°.
So, 60° + 60° + x = 180° \(\Rightarrow\) 120° + x = 180°
\(\Rightarrow\) x = 180° - 120° \(\Rightarrow\) x = 60°

No, we cannot construct a ΔDEF such that EF = 7.2 cm, mㄥE = 110° and mㄥF = 80°
Justification
We know that, the sum of all the three angles of a triangle is 180°. But in given question, sum of two angles.
mㄥE + mㄥF
= 110° + 80° = 190° > 180°
The sum of these two angles should be less than 180°.
So, the triangle with given measures cannot be constructed.

Yes,we can slightly modify the construction of a line parallel to a given line through a point not on the line to use the idea of equal corresponding angles instead of equal alternate angles. For modifying this construction, we change only Step IV, VI and VII and new steps are given below:
Step IV With A as centre and radius same as in Step III draw an arc EF to cut BA (extended) at P.
Step V Place the pointed tip of the compasses at C and adjust the. opening so that the pencil tip is at D.
Step VI With the same opening as in Step V and with Pas centre, draw an arc which cut the arc EF at Q.
Step VII Join AQ to draw a line m, which is parallel to the given line I.
Here, \(\angle ABC,\angle PAQ\) are corresponding angles,

Therefore, I II m.

Given a right angled triangle in which hypotenuse is 6 cm and one leg is 4 cm.
To construct a triangle with these two sides and one right angle, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch of triangle with given measures marked on it.

Step II Draw a line segment MN = 4 cm.

Step III At point M, draw MX 丄 MN.

Step IV With N as centre and radius 6 cm, draw an arc to intersect ray MX at L.

Step V Joint LN.

Thus, ΔLMN is the required triangle.

Given, two sides and an angle of ΔPQR are QR = 8 cm, PR = 10 cm and mㄥQ = 90°.
To construct a triangle with these two sides and one right angle, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with measures marked on it.

Step II Draw a line segment QR = 8 cm.

Step III At point Q, draw QX 丄 QR.

Step IV With R as centre and radius 10 cm, draw an arc which intersects ray QX at P.

Step V Join PR.

Thus, ΔPQR is the required triangle.

In the given figure, a perpendicular is drawn at both Band B'.
\(\angle XBB'=\angle YB'B=90^0\)
∴ The angle between made by rays with BB' is equal to 90°. Hence, both rays BX II B'Y.

Yes, RS II PQ II TU. Since, two lines parallel to a given line are also parallel to each other.

Given, line segment AB = 6 cm. If perpendicular bisector constructeg on AB. Then, the length of equal parts will be \(\frac{6}{2}=3\) cm.
∴ Perpendicular bisector divide the line in two equal parts.