∠ABC = 90°
So, ΔABC is a right angled triangle.

From the figure, we get \(\angle ACB=\angle ABC=60^0\)
\(\therefore\triangle ABC\) is an isosceles triangle \(\Rightarrow\) AC = AB = 6 cm

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.

In the given figure, \(\angle YBC=90^0\)
The angle made by first arc is 60° and second arc is 120°.
Each arc = 60°
Bisector of 60° = 30°,
So, 60° + 30° = 90°

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.

Taking point Q as centre, he drew an arc of radius 3 cm and taking point R as centre, he drew an arc of radius 2 cm. But these two arcs do not intersect each other at a point. So, the student could not get P as a point of intersection.

In this connection, we remember the ptoperty of a triangle, "The sum of any two sides of a triangle is always greater than the third side.
No, such a triangle can never exist, because a triangle is possible only when the sum of the length of any two sides is always greater than the length of third side. Here, 2 cm + 3 cm = 5 cm < 6 cm. So, this condition is not satisfied.

x=45°

Steps of construction
(i) Draw a line segment AB = 5 cm.
(ii) Draw an arc of radius 5 cm from point A.
(iii) Now, draw another arc of radius 5 cm from point B to cut previous arc at C.
(iv) Join A to C and B to C.

Hence, \(\triangle ABC\) is the required triangle. The value depicted here is to make the people aware of traffic rules for safety measures.

Given, two sides and an angle of \(\triangle DEF\) are DE = 5 cm, DF = 3 cm and m \(\angle EDF=90^0\).
To construct a triangle with these two sides and included 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 DE = 5 cm.

Step III At point D, construct \(\angle EDX=90^0\).

Step IV With D as centre and radius 3 cm, draw an arc which intersects DX at F.

Step V Join EF.

Thus, \(\triangle DEF\) is the required triangle.