Construction – Exercise 11.2 – Class IX

For previous exercise – Construction – Exercise 11.1 – Class IX


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

Solution:

  • Construction - Exercise 11.2 – Class IXDraw a line segment BC = 7 cm
  • Make an ∠B = 75° i.e., ∠CBX = 75°
  • Cut a line  segment BX at D such that BD = AB + AC = 13 cm
  • Join DC
  • Make an ∠DCY = ∠BDC
  • Let CY intersect BD at A .
  • Then the required triangle is ABC

 


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

Solution:

  • Construction - Exercise 11.2 – Class IXDraw a line segment BC = 8c
  • Make an angle ∠B = 45° i.e., ∠CBX = 45°
  • Cut the line segment BD = 3.5 cm from the ray BX
  • Join DC
  • Draw perpendicular bisector MN of DC and intersect it to BX at A
  • Join AC
  • Then ABC is the required triangle.

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

Construction - Exercise 11.2 – Class IX

Solution:

  • Draw a line segment QR = 6cm
  • Make an angle ∠Q = 60° i.e., ∠RQX = 60°
  • Cut the line segment QX such that, QS = 2cm = PR – PQ which is extended in opposite side
  • Join SR . Draw perpendicular bisector of MN of SR which intersects QZ at P
  • Join RP then PQR is the required triangle.

 

 

 


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

Solution:

  • Let AB = XY + ZX + XY = 11 cm
  • Then ∠BAP = 30° and ∠ABR = 90°
  • Bisect ∠LBC and ∠MCB . Let these bisector meet at X
  • Draw perpendicular bisectors DE and FG to XB and XC respectively.
  • Let DE intersect BC and FC intersect BC at Y and Z respectively.
  • Join XY and XZ . Then XYZ is the required triangle.

Construction - Exercise 11.2 – Class IX


  1. Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Solution:

  • Construction - Exercise 11.2 – Class IXLet the base of an right angled triangle is BC = 12 cm
  • At B make an angle XBC = 90°
  • Cut the line segment BX at D i.e., at 18cm = AB+AC
  • Draw a perpendicular bisector MN to DC which cut the line segment BX at A
  • Join AC
  • Therefore triangle ABC is the required right angled triangle.

 

 

 


 

 

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