**Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.**

Solution:

- Draw a line segment BC = 7 cm
- Make an ∠B = 75° i.e., ∠CBX = 75°
- Cut a line segment BX at D such that BD = AB + AC = 13 cm
- Join DC
- Make an ∠DCY = ∠BDC
- Let CY intersect BD at A .
- Then the required triangle is ABC

**Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.**

Solution:

- Draw a line segment BC = 8c
- Make an angle ∠B = 45° i.e., ∠CBX = 45°
- Cut the line segment BD = 3.5 cm from the ray BX
- Join DC
- Draw perpendicular bisector MN of DC and intersect it to BX at A
- Join AC
- Then ABC is the required triangle.

**Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.**

Solution:

- Draw a line segment QR = 6cm
- Make an angle ∠Q = 60° i.e., ∠RQX = 60°
- Cut the line segment QX such that, QS = 2cm = PR – PQ which is extended in opposite side
- Join SR . Draw perpendicular bisector of MN of SR which intersects QZ at P
- Join RP then PQR is the required triangle.

**Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.**

Solution:

- Let AB = XY + ZX + XY = 11 cm
- Then ∠BAP = 30° and ∠ABR = 90°
- Bisect ∠LBC and ∠MCB . Let these bisector meet at X
- Draw perpendicular bisectors DE and FG to XB and XC respectively.
- Let DE intersect BC and FC intersect BC at Y and Z respectively.
- Join XY and XZ . Then XYZ is the required triangle.

**Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.**

Solution:

- Let the base of an right angled triangle is BC = 12 cm
- At B make an angle XBC = 90°
- Cut the line segment BX at D i.e., at 18cm = AB+AC
- Draw a perpendicular bisector MN to DC which cut the line segment BX at A
- Join AC
- Therefore triangle ABC is the required right angled triangle.

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