NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2

NCERT Solutions for Class 9 Maths Chapter 11 Constructions Ex 11.2 are the part of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for Chapter 11 Constructions Ex 11.2.

Ex 11.2 Class 9 Maths Question 1.
Construct a ∆ABC in which BC = 7 cm, B = 75° and AB + AC = 13 cm.

Solution:
Steps of Construction:
Step I : Draw a ray BX.
Step II : Along BX, cut off a line segment BC = 7 cm.
Step III : At B, construct
CBY = 75°.
Step IV : From the ray BY, cut off BD = 13 cm (= AB + AC).
Step V : Join DC.
Step VI : Draw a perpendicular bisector of CD which meets BD at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle.


Ex 11.2 Class 9 Maths Question 2.
Construct a ∆ABC in which BC = 8 cm, B = 45° and AB – AC = 3.5 cm.

Solution:
Steps of Construction:
Step I : Draw a ray BX.
Step II : Along BX, cut off a line segment BC = 8 cm.
Step III : At B, construct
CBY = 45°.
Step IV : From BY, cut off BD = 3.5 cm (= AB – AC).
Step V : Join DC.
Step VI : Draw PQ, perpendicular bisector of DC, which intersects BY at A.
Step VII: Join AC.

Thus, ∆ABC is the required triangle.

 

Ex 11.2 Class 9 Maths Question 3.
Construct a ∆ABC in which QR = 6 cm, Q = 60° and PR – PQ = 2 cm.

Solution:
Steps of Construction:
Step I : Draw a ray QX.
Step II : Along QX, cut off a line segment QR = 6 cm.
Step III : Construct a line YQY’ such that
RQY = 60°.
Step IV : Cut off QS = 2 cm (= PR – PQ) on QY’.
Step V : Join SR.
Step VI : Draw MN, perpendicular bisector of SR, which intersects QY at P.
Step VII: Join PR.

Thus, ∆PQR is the required triangle.

 

Ex 11.2 Class 9 Maths Question 4.
Construct a ∆XYZ in which Y = 30°, Z = 90° and XY + YZ + ZX = 11 cm.

Solution:
Steps of Construction:
Step I : Draw a line segment AB = 11 cm = (XY + YZ + ZX)
Step II : Construct
BAP = 30°
Step III : Construct
ABQ = 90°
Step IV : Draw AR, the bisector of
BAP.
Step V : Draw BS, the bisector of
ABQ. Let AR and BS intersect at X.
Step VI : Draw perpendicular bisector of AX, which intersects AB at Y.
Step VII: Draw perpendicular bisector of XB, which intersects AB at Z.
Step VIII: Join XY and XZ.

Thus, ∆XYZ is the required triangle.

 

Ex 11.2 Class 9 Maths 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 BC = 12 cm.
Step II : At B, construct
CBY = 90°.
Step III : Along BY, cut off a line segment BX = 18 cm.
Step IV : Join CX.
Step V : Draw PQ, the perpendicular bisector of CX, which meets BX at A.
Step VI : Join AC.

Thus, ∆ABC is the required triangle.


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NCERT Solutions for Maths Class 10

NCERT Solutions for Maths Class 12

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