NCERT Solutions for Class 9 Maths Chapter 11 - Exercise 11.2 Constructions

Q1. Construct a triangle ABC in which BC = 7 cm, ∠B = 75º and AB + AC = 13 cm.
Ans: Steps of construction:

• Draw a ray BX.
• From  cut off  = 7 cm.
• At B, construct ∠CBY = 75º.
• From  cut off BD = 13 cm (= AB + BC)
• Join D and C.
• Bisect DC such that the bisector of DC meets BD at A.
• Join AC.

Thus, ΔABC is the required triangle.

Q2. Construct a triangle ABC in which BC = 8 cm, ∠ B = 45º and AB ∠ AC = 3.5 cm.
Ans: Steps of construction: 

• Draw a ray
  
• From  , cut off BC = 8 cm.
• Construct ∠CBY = 45º.
• From  cut off  = 3.5 cm.
  
• Join D and C.
• Draw PQ, the perpendicular bisector of DC, which intersects  at A.
• Join AC.

Thus, ABC is the required triangle.

Q3. Construct a triangle PQR in which QR = 6 cm, ∠Q = 60º and PR ∠ PQ = 2cm.
Ans: Steps of construction:

• Draw a ray  .
• From , cut off QR = 6 cm.
• Construct a line YQY' such that ∠RQY = 60º.
  
• Cut off QS = 2 cm (from QY').
• Join ‘S’ and ‘R’.
• Draw MN, the perpendicular bisector of SR, which intersects QY at P.
• Join P and R.

Thus, PQR is the required triangle.

Q4. Construct a triangle XYZ in which ∠ Y = 30º , ∠ Z = 90º and XY + YZ + ZX = 11 cm.
Ans: Steps of construction:

• Draw a line segment AB = 11 cm = (XY + YZ + ZX)
• Construct ∠BAP = 30º = ∠Y
• Construct ∠ABQ = 90º = ∠Z
• Draw  , the bisector of ∠BAP.
• Draw  , the bisector of ∠ABQ, such that  and  intersect each other at X.
• Draw perpendicular bisector of AX, which intersects AB at Y.
• Draw perpendicular bisector of XB, which intersects AB at Z.
• Join XY and XZ.

Thus, XYZ is the required triangle.

Q5. Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.
Ans: Steps of construction:

• Draw  = 12 cm.
• Construct ∠CBY = 90º.
• From  cut off BX = 18 cm.
• Join CX.
• Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.
• Join AC.

Thus, ABC is the required triangle.