:
Each part: 2 Marks
(a) Adding up the sides by taking two at a time, we get:
5 cm + 5 cm = 10 cm < 80 cm
80 cm + 5 cm = 85 cm > 5 cm
So we see that the sum of one pair of sides when added is not greater than the third side.
Therefore, we cannot draw a triangle with these measurements.
(b) Adding up the sides by taking two at a time, we get
13 cm + 6 cm = 19 cm > 7 cm
13 cm + 7 cm = 20 cm > 6 cm
6 cm + 7 cm = 13 cm = 13 cm
So in each case, the sum of two sides is equal to the third side.
Hence, this triangle cannot be constructed.

:
Steps: 1 Mark
Final answer: 1 Mark
Adding up the sides, taking two at a time, we get
5 cm + 19 cm = 24 cm > 20 cm
5 cm + 20 cm = 25 cm > 19 cm
19 cm + 20 cm = 39 cm > 5 cm
So we see that the sum of the pair of sides when added is greater than the third side.
Therefore, we can construct the triangle with these measurements.

:
Each part: 1 Mark
For SAS, the criterion is as follows:
a) Lengths of any two sides.
b) The measure of the angle between these sides.
For ASA,the criterion are as follows
a)Measures of any two angles.
b) Lengths of the side between these angles.

:
Each part: 1 Mark
(a)

For the lines to be parallel, $\mathrm{âˆ}NOB$ has to be equal to$\mathrm{âˆ}MPC$ (Alternate Interior Angles)
(b) If two parallel lines are cut by a transversal then the corresponding angles are equal.
So, $x={35}^{∘}$.

:
Constructions: 2 Marks
Steps: 1 Mark
Step 1: Draw a line segment $\stackrel{Â}{BC}=48cm$

Step 2: Draw $\mathrm{âˆ}CBD={90}^{∘}$ at B using protractor or compass.
Step 3: With C as centre and radius equal to 50 cm, draw an arc to cut the ray $\stackrel{Â}{BD}$ at A and join $\stackrel{Â}{AC}$
ABC is the required triangle.

Answer: Option C. -> Statement $2$ is right and Statement $1$ is wrong
:
C

It is possible to construct a triangle when the measurements of two sides and included angle are given. (SAS rule of construction)

 

Answer: Option C. -> Statement $2$ is right and Statement $1$ is wrong
:
Given, $\mathrm{âˆ}ACD$ = ${90}^{∘}$ and $CF$ is the angular bisector of $\mathrm{âˆ}ACD$, so $\mathrm{âˆ}DCF={45}^{∘}$. So, $BCF={135}^{∘}$.

Answer: Option C. -> Statement $2$ is right and Statement $1$ is wrong
:

As stated in question AB||CD and if $\mathrm{âˆ}CFE$ = ${90}^{∘}$, so $\mathrm{âˆ}FEB$ is an alternate angle to $\mathrm{âˆ}CFE$ which should be equal.
So, $\mathrm{âˆ}CFE={90}^{∘}$.

Answer: Option C. -> Statement $2$ is right and Statement $1$ is wrong
:

The minimum number of triangles that can be drawn if two sides and one angle has been given is $0$ since the angle might not be an included one.

Answer: Option C. -> $3and1$
:
C

Steps for drawing a line parallel to a given line:

Step 1: Mark a point A, not on the line 'l'.
Step 2: Mark point B on line 'l'.
Step 3: Draw line segment joining points A and B.
Step 4: Draw an arc with B as the centre, such that it intersects line 'l' at D and AB at E.
Step 5: Draw another arc with the same radius and A as the centre, such that it intersects AB at F.  
Step 6: Draw another arc with F as the centre and distance DE as the radius.
Step 7: Mark the point of intersections of this arc and the previous arc as G.
Step 8: Draw line 'm' passing through points A and G.

No. of arcs drawn = $3$
No. of times the distance between compass tip and pencip tip is changed = $1$