PQ = 5 cm, \(\angle PQR\) =105°, \(\angle QRP\) = 40°
\(\therefore\) Sum of angles of a triangle is 180°
\(\therefore\)  \(\angle PQR+\angle QRP+\angle QPR\) = 180° 
\(\therefore\) 105° + 40° + \(\angle QPR\) = 180°
\(\therefore\) 145° + \(\angle QPR\)  = 180°
\(\angle QPR\) = 180° - 145°
\(\angle QPR\) = 35°
Steps of Construction:
(a) Draw a line segment PQ of 5 em.

(b) From point P draw an angle of 35°, such that \(\angle QPX\) = 35°
(c) From point Q draw an angle of 105°, such that \(\angle PQY\) = 150°
(d) These lines PX and QY cut at R.
PQR is required triangle.

Steps of construction
(i) Draw a line AC= 4 cm.
(ii) Construct \(\angle XAC=75^0\) at A.
(iii) Also, construct \(\angle YCA=60^0\)at C.
(iv) Ray X and ray Y cut each other at point B.

Hence, \(\triangle ABC\) is the required triangle, having \(\angle A=75^0,\angle C=60^0\) and AC = 4 cm.

Steps of construction
(i) Draw a line I and mark a point C outside it.
(ii) Take a point B on line I and join BC.
(iii) Draw line parallel to line I passing through C.
(iv) Mark a point D on line m, at a distance of 4 cm from C.
(v) Through D draw AD II BC.
∴ Line I is parallel to line m

Also, AD II BC, AB=DC= 4 cm

In ΔDEF, we have DE = 4.5 cm, EF = 5.5 cm and DF = 4 cm
Here, DE + DF = 4.5 + 4 = 8.5 cm and EF = 5.5 cm
∴ DE + DF > EF
Therefore, ΔDEF can be constructed.
To construct a triangle, we use the following steps.
Steps of construction
Step I Firstly, draw a rough sketch of triangle with given measures marked on it.

Step II Draw a line segment, EF=5.5 cm.

Step III With E as centre, and radius 4.5 cm, draw an arc.

Step IV With F as centre and radius 4 cm, draw an arc to intersect the previous arc at D.

Step V Join DE and DF.
Thus, ΔDEF is the required triangle.

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.

Given, an isoscelesright angled ΔABC in which
m ㄥACB =90°
and AC = BC =6 cm
∵ In an isoscelesright angled triangle, two sides, i.e. base and height will be equal.
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 BC = 6 cm.

Step III At point C, draw CX 丄 BC.

Step IV With C as centre and radius 6 cm, draw an arc to intersect ray CX at A.

Step V Join AB .

Thus, ΔABC is the required isosceles right angled triangle.

Given, PQ = 5 cm, mㄥPQR = 105° and m ㄥQRP = 40°
In ΔPQR, by angle sum property, we have
ㄥPQR + ㄥQRP + ㄥRPQ = 180°
⇒ 105° + 40° + ㄥRPQ = 180°
⇒ ㄥRPQ = 180° -145° = 35°
Thus, we have PQ = 5 cm, ⇒ ⇒P = 35° and ⇒Q = 105°
Now, to draw ΔPQR, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch of traingle with measures marked on it.

Step II Draw a line segment PQ = 5 cm.

Step III At point P, draw a ray PX making an angle of 350 with PQ i.e., ㄥQPX = 35⁰.

Step IV At point Q, draw a ray QY making an angle of 105° with PQ i.e. ㄥPQY = 105°.

Step V Extend the ray PX and QY. Let rays PX and QY intersect at R.

Thus, ΔPQR is the required triangle.

To draw a line parallel to AB using ruler and compasses, we use the following steps:
Steps of construction
Step I Draw a line segment AB and then take any point P on it.

Step II Take a point C outside AB and join PC.

Step III With P as centre, draw an arc cutting AB and PC at X and Y respectively.

Step IV With C as as centre and the same radius as in step III, draw an arc on the opposite side of PX to cut PC at Q.

Step V Place the pointed tip of the compasses at X and adjust the opening, so that the pencil tip is at Y with this opening and with Q as centre, draw an arc cutting the arc drawn in step IV at R.

Step VI Join CR and produce it in both sides to obtain the required line l.

Thus,  l II AB.

Given, two sides and an angle of ΔABC are BC = 7.5 cm, AC = 5 cm and m ㄥC = 60⁰
To construct a triangle with these two sides and included angle, we use the following steps:
Step I Firstly, we draw a rough sketch of triangle with given measures marked on it.

Step II Draw a line segment BC = 7.5 cm.

Step III At point C, draw a ray CX making an angle of 60° with BC.

Step VI With C as centre and radius 5 ern, draw an arc which intersects CX at A.

Step V Join BA

Thus, MBC is the required triangle