Class 9 NCERT Solutions- Chapter 11 Constructions - Exercise 11.2

In Class 9 Mathematics, Chapter 11 focuses on constructions a fundamental topic in geometry that deals with creating geometric figures using only a compass and a straightedge. This chapter emphasizes the practical application of geometric principles to construct various shapes and figures with precision. Understanding constructions is crucial for developing spatial reasoning and problem-solving skills in geometry.

Constructions

Constructions in geometry involve creating geometric figures accurately based on the given conditions. These constructions are achieved using simple tools like a compass and a straightedge without the need for measurements. The process includes drawing shapes, lines, and angles with specific properties helping the students visualize and apply geometric concepts effectively. This exercise helps in practicing basic construction techniques and reinforces the understanding of geometric properties.

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

Solution:

Steps of construction:

1. Draw a line segment BC base of cm is drawn.
2. At point B draw an angle of 75°.
3. Cut BD =13cm from BY.
4. Join ∠D which intersect BD at A.
5. 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:

1. Draw  a line segment BC=8cm.
2. At point B, draw angle 45°.
3. Cut BD=3.5 from BY.
4. Join CD.
5. Draw perpendicular bisector of CD, which construct BY at A.
6. 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:

1. Draw a line segment QR=6cm.
2. At point Q draw angle 60°.
3. Extend PQ to Y’.
4. Cut QS =2cm from QY’.
5. Join RS.
6. Draw perpendicular bisector of RS which intersect QY at P.
7. 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:

1. Draw a line segment AB=11cm.
2. At point A draw ∠BAP=30°.
3. At point B draw angle 90°.
4. Draw the bisector of ∠BAP and ∠ABR which intersect each other at X.
5. Join AX and BX.
6. Draw perpendicular bisector of AX and BX which intersect AB on Y and Z respectively.
7. 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:

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

Read More:

• Maths
• Construction of Triangles