NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.2 (For CBSE, ICSE, IAS, NET, NRA 2022)

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Construction of Triangle

Q-1 Construct a triangle JAB in which

Solution:

Image Title: Triangle JAB

Image Description: Triangle JAB,

Give,

Step of Construction

  • A line segment AB of 7 cm is drawn.
  • At point A, an angle is constructed such that it is equal to .
  • A line segment cm is cut on AC (which is equal to ) .
  • DB is joined and is made.
  • Let BE intersect AC at J.

Thus, required triangle is

Q-2 Construct a triangle DAB in which .

Solution:

Triangle DAB

Give

Steps of Construction

  • A line segment cm is drawn and at point A, make an angle of i.e..
  • Cut the line segment (equal to ) on ray .
  • Join GB and draw the perpendicular bisector PQ of GB.
  • Let it intersect AC at point D. Join DB.

Thus, is the required triangle.

Q-3 Construct a triangle EAR in which

Solution:

Triangle EAR

Give, Triangle EAR,

Steps of Construction

  • A ray AX is drawn and cut off a line segment AR = 6 cm from it.
  • A ray AY is constructed making an angle of 60º with AR and YA is produced to form a line YAY ′
  • Cut off a line segment AB = 2cm from AY ′ . RB is joined.
  • Draw perpendicular bisector of RB intersecting AY at a point E. ER is joined.

Thus, ΔEAR is the required triangle.

Q-4 Construct a triangle HFI in which

Solution:

Triangle HFI

Give, Triangle HFI

Steps of Construction:

  • A line segment cm is drawn. ()
  • An angle, and an angle
  • And are bisected. The bisectors of these angles intersect each other at point H.
  • Perpendicular bisectors PQ of CH and RV of AH are constructed.
  • Let PQ intersect CA at F and RV intersect CA at I. FH and HI are joined.

Thus, is the required triangle.

Q-5 Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Solution:

Triangle PAJ

Give, right triangle base and othrside is

Steps of Construction:

  • A ray AX is drawn and a cut off a line segment is made on it.
  • is constructed.
  • Cut off a line segment cm is made on AY. JQ is joined.
  • Perpendicular bisector of JQ is constructed intersecting AQ at P. PJ is joined.
  • Thus, is the required triangle.