Class 9 NCERT Solutions- Chapter 11 Constructions – Exercise 11.2

**Question 1. Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.**

**Solution:**

Steps of construction:

- Draw a line segment BC base of cm is drawn.
- At point B draw an angle of 75°.
- Cut BD =13cm from BY.
- Join ∠D which intersect BD at A.
- Join AC. Now triangle ABC is the required triangle

**Question 2. Construct a ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.**

**Solution:**

Steps of construction:

- Draw a line segment BC=8cm.
- At point B, draw angle 45°.
- Cut BD=3.5 from BY.
- Join CD.
- Draw perpendicular bisector of CD, which construct BY at A.
- Join AC. NOW, ABC is the required triangle.

**Question 3. Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.**

**Solution:**

Steps of construction:

- Draw a line segment QR=6cm.
- At point Q draw angle 60°.
- Extend PQ to Y’.
- Cut QS =2cm from QY’.
- Join RS.
- Draw perpendicular bisector of RS which intersect QY at P.
- Join PR. Now, PQR is the required triangle.

**Question 4. Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.**

**Solution:**

Steps of construction:

- Draw a line segment AB=11cm.
- At point A draw ∠BAP=30°.
- At point B draw angle 90°.
- Draw the bisector of ∠BAP and ∠ABR which intersect each other at X.
- Join AX and BX.
- Draw perpendicular bisector of AX and BX which intersect AB on Y and Z respectively.
- Join XY and XZ. Then XYZ is the required triangle.

**Question 5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.**

**Solution:**

Steps of construction:

- Draw a line segment of BC=12cm.
- At point B draw angle b=90°
- Cut BD =18cm.
- Join CD.
- Draw perpendicular bisector of CD which intersect BD at point A.
- Join AC. Now ABC is the required triangle.

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