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.