If two parallel lines are intersected by a transversal:

(i) alternate interior angles are equal i.e., $\angle$2=$\angle$8 and $\angle$3=$\angle$5
(ii) alternate exterior angles are equal i.e., $\angle$1=$\angle$7 and $\angle$4=$\angle$6
(iii) corresponding angles are equal i.e., $\angle$1=$\angle$5,$\angle$4=$\angle$8,$\angle$2=$\angle$6 and $\angle$3=$\angle$7
(iv) co-interior angles are supplementary i.e., $\angle$2 +$\angle$5= ${180}^{\circ }and\angle$3 +$\angle$8 =${180}^{\circ }$.
Hence, if the alternate angles are equal then the lines are parallel.

If he was able to draw a triangle with two sides and an angle which is not included, then the triangle should be a right angled triangle.

We can construct a line parallel to a given line by using alternate angles as well as corresponding angles concept. So, the statement is false.

If two parallel lines are intersected by a transversal:

(i) alternate interior angles are equal i.e., $2=8and3=5$

(ii) alternate exterior angles are equal i.e., $1=7and4=6$

(iii) corresponding angles are equal i.e., $1=5,4=8,2=6and3=7$

(iv) co-interior angles are supplementary i.e., $2+5={180}^{\circ }and3+8={180}^{\circ }$
  A line can be constructed parallel to a given line by making any of the above mentioned pairs of angles equal.

We can only draw one line parallel to a given line passing through a point that is not on the line.

The triangle inequality states that for any triangle, the sum of the lengths of any two of its sides must be greater than the length of the third side. ​We, thus, cannot construct a triangle with any three lengths.

Angle: 0.5 Mark
Type: 0.5 Mark
The angle between the two hands is 90 degrees.
It is an isosceles right-angled triangle.

Correct answer: 2 Marks
With the given lengths, a triangle cannot be constructed because the sum of two sides of a triangle is less than the third side.
3 + 5 < 9
5 + 9 > 3
3 + 9 > 5

Steps: 1 Mark
Construction: 1 Mark
Draw a line segment AB = 5 m
Draw AC making an angle of 60 degrees with AB.
Taking B as centre, draw a circle of radius 9 m.
The point where the circle meets the line will be the other point of the triangle.

Steps: 1 Mark
Construction: 1 Mark

 

Steps of construction:
(i) Draw a line l.
(ii) Take a point M on it.
(iii) Draw an angle of ${90}^{\circ }$ at M with l which is perpendicular to l at M.
(iv) With M as centre and radius 5.2 cm, draw an arc which intersects the above perpendicular at point P. MP is the required perpendicular.
(v) At P, draw an angle of ${90}^{\circ }$ with PM and make a line q.
Line q is the required line parallel to line l.

Steps: 1 Mark
Construction: 2 Marks
Draw a line with given length of 3 m, let it be AB.
Draw a line making 60 degrees with the line AB, let it be AC.
With B as centre, draw a circle of radius 8 m. 
The drawn arc meets the line AC at c.
Join all the points and it will form a triangle with side 3 m and 8 m with an angle of ${60}^{\circ }$.