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:
(a) Draw a line XY. Take a point A on the line.
(b) Taking radius 6 cm draw an arc which cuts the line at B.

(c) From B draw an angle of 110° i.e.,  \(\angle ABZ\) = 110°
(d) From point B draw an arc of radius 6 cm which cuts the line BZ at C.
(e) Join AC to get the required triangle.

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 AB of length 3 cm.
Step III Draw an angle of 45° on point B and produce it to ray Y.
Step IV With B as centre, draw an arc of 3 cm which intersects ray BYat C.
Step V Join AC.

Thus, \(\triangle ABC\) is the required isosceles triangle.

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.

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 BC = 4.5 cm.
(ii) With centre B and radius 2 cm draw an arc.
(iii) With centre C and radius 4 cm draw an arc which cuts the previous arc at A.
(iv) Join AB and AC.

Hence, \(\triangle ABC\) is the required triangle in which AB= 2 cm, BC= 4.5 cm and AC= 4 cm.

Given, three sides of \(\triangle PQR\) are PQ = 4 cm, QR = 35 cm and PR = 4cm.
To construct a triangle with given sides, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with given measures marked on it.

Step II Draw a line segment QR = 35 cm.

Step III With Q as centre and radius 4 cm, draw an arc.

Step IV Now, with R as centre and radius 4 cm, draw another arc intersecting the previous arc at P.

Step V Join PQ and PR.

Thus, \(\triangle PQR\) is the required triangle.
Here, PQ = PR = 4 cm, i.e. two sides of \(\triangle PQR\) are equal.
Therefore, \(\triangle PQR\) is an isosceles triangle.

Given, length of each side of an equilateral triangle is 5.5 cm.
To construct an equilateral triangle, we use the following steps:
Steps of construction
Step I Firstly, we draw a rough sketch with gIven measures marked on it.

Step II Draw a line segment QR = 55 cm.

Step III With Q as centre and radius 5.5 cm, draw an arc.

Step IV Now, with R as centre and radius 5.5 cm, draw another arc to cut the previous arc at P.

Step V Join PQ and PR.

Thus, \(\triangle PQR\) is the required equilateral triangle.

Given, three sides of \(\triangle XYZ\) are XY = 4.5 cm, yz = 5 cm and ZX=6 cm.
To construct this triangle, 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 yz of length 5 cm, i.e. yz = 5 cm.

Step III With Y as centre and radius 45 cm, draw an arc.

Step IV With Z as centre and radius 6 cm, draw another arc to cut the previous arc at X.

step V Join XY and ZX.

Hence, \(\triangle XYZ\) 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.