NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Constructions 11.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2.

• Constructions Class 9 Ex 11.1
• Constructions Class 9 Ex 11.2

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 11
Chapter Name	Constructions
Exercise	Ex 11.2
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

Ex 11.2 Class 9 Maths Question 1.
Construct a ΔABC, in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
Solution:
Given in Δ ABC, BC = 7 cm, ∠B = 75° and AB + AC = 13 cm

Steps of construction

1. First, draw the base BC = 7 cm
2. At the point B, make an ∠XBC = 75°
3. Cut a line segment BD equal to AB + AC = 13 cm from the ray
4. Now, join
5. Make a ∠DCY at C is equal to
6. Let CY intersects BX at Then join AC.
7. Thus, ABC is the required triangle.

Ex 11.2 Class 9 Maths Question 2.
Construct a Δ ABC, in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Solution:
Given, in Δ ABC, BC = 8 cm, ∠B = 45°
and AB – AC = 3.5 cm
Here, AB > AC, i.e., the side containing base angle is greater than third side.

Steps of construction

1. First, draw the base BC = 8 cm.
2. At the point B, make an ∠XBC = 45°.
3. Cut the line segment BD equal to AB – AC =5 cm from the ray BX.
4. Join DC
5. Draw the perpendicular bisector, say PQ of DC
6. Let PQ intersects BX at a point
7. Join AC
   Thus, ABC is the required triangle.

Ex 11.2 Class 9 Maths Question 3.
Construct a Δ PQR, in which QR = 6 cm, ∠Q = 60°  and PR = PQ 2 cm.
Solution:
Given, in Δ PQR, QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm
Here, PR > PQ, i.e., the side containing base angle is less than third side.

Steps of Construction

(i) Draw the base QR = 6 cm.
(ii) At the point Q, make an ∠XQR = 60°
(iii) Cutline segment QS = PR -PQ = 2 cm from the
line QX extended on opposite side of line
segment QR.
(iv) Join SR.
(v) Draw the perpendicular bisector LM of SR.
(vii) Let LM intersects QX at P.
(vii) Join PR. Thus, PQR is the required triangle.

Ex 11.2 Class 9 Maths Question 4.
Construct a ΔXYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
Given, in Δ XYZ, ∠Y = 30 °, ∠Z = 90 ° and XY + YZ + ZX = 11 cm

Steps of Construction 

1. Draw a line segment BC = XY + YZ +ZX = 11
2. Draw a ray LB making an ∠LBC = 30° at B and a ray MC making an ∠MCB = 90° atC.
3. Draw the angle bisector of ∠LBC and Let these bisectors meet at a point X.
4. Draw perpendicular bisectors DE of XB and FG of
5. Let DE intersects BC at Y and FG intersects BC at Z.
6. Join XY and XZ. Thus, XYZ is the required triangle.

Ex 11.2 Class 9 Maths Question 5.
Construct aright angled triangle whose base is 12 cm and the sum of its hypotenuse and other side is 18 cm.
Solution:
Given, in Δ ABC, base BC = 12 cm, ∠B = 90° and AB + BC = 18 cm

Steps of construction

1. Draw the base BC = 12 cm.
2. At the point B, draw a ray BX making ∠XBC = 90°.
3. Cut a line segment BD = AB + AC = 18 cm from the ray
4. Join DC
5. Now, draw the perpendicular bisector PQ of CD, which intersects BD at a point
6. Join AC. Thus, ABC is the required right angled triangle.

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