Haryana State Board HBSE 9th Class Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

## Haryana Board 9th Class Maths Solutions Chapter 11 Constructions Exercise 11.2

Question 1.

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

Solution:

Steps of construction:

Step – I: Draw a line segment BC of length 7 cm.

Step – II: At B construct ∠CBX = 75° with BC.

Step – III: Cut BD = 13 cm from BX and join CD.

Step – IV: Draw the perpendicular bisector of CD intersecting BD at A.

Step – V: Join AC, then ABC is the required triangle.

Question 2.

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

Solution:

Steps of construction:

Step – I: Draw a line segment BC = 8 cm.

Step – II: At point B construct ∠CBX = 45° with BC.

Step – III: Cut BD = 3.5 cm from BX and join CD.

Step – IV: Draw the perpendicular bisector of CD intersecting BX at point A.

Step – V: Join AC, then ABC is the required triangle.

Question 3.

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

Solution:

Steps of construction:

Step – I: Draw a line segment QR = 6 cm.

Step – II: At the point Q construct ∠RQX = 60° with QR.

Step – III: Extend XQ to S to opposite side of QR such that QS = 2 cm.

Step – IV: Join SR and draw the perpendicular bisector of SR intersecting QX at P.

Step – V: Join PR, then PQR is the required triangle.

Question 4.

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

Solution:

Steps of construction:

Step – I: Draw a line segment PQ = 11 cm (= XY + YZ + ZX).

Step – II: At the point P construct ∠QPR = 30° and at the point Q construct ∠PQS = 90°.

Step – III: Draw the bisectors of ∠QPR and ∠PQS intersecting at point X.

Step – IV: Draw the perpendicular bisectors KL of PX and MN of XQ intersecting PQ at Y and Z respectively.

Step – V: 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:

Step – I: Draw a line segment BC = 12 cm.

Step – II: At the point B, construct ∠CBP = 90° with BC.

Step – III: Cut BD = 18 cm from BP and join CD.

Step – IV: Draw the perpendicular bisector of CD intersecting BD at point A.

Step – V: Join AC, then ABC is the required triangle.