Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

Mathematics (Maths) Class 9

Class 9 : Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

The document Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

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 Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 7 cm.
  • At B, construct ∠CBY = 75º.
  • From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off BD = 13 cm (= AB + BC)Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
  • 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 Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
  • From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , cut off BC = 8 cm.
  • Construct ∠CBY = 45º.
  • From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev cut off Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 3.5 cm.
    Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
  • Join D and C.
  • Draw PQ, the perpendicular bisector of DC, which intersects Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev 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 Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev .
  • From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev, cut off QR = 6 cm.
  • Construct a line YQY' such that ∠RQY = 60º.
    Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
  • 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:

Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev

  • Draw a line segment AB = 11 cm = (XY + YZ + ZX)
  • Construct ∠BAP = 30º = ∠Y
  • Construct ∠ABQ = 90º = ∠Z
  • Draw Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , the bisector of ∠BAP.
  • Draw Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev , the bisector of ∠ABQ, such that Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev and Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev 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 Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev = 12 cm.
  • Construct ∠CBY = 90º.Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev
  • From Ex 11.2 NCERT Solutions- Constructions Class 9 Notes | EduRev 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.

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