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NCERT Solutions For Class 9 Maths Chapter 12 Constructions Ex 12.2

Here, Below you all know about NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2 Question Answer. I know many of you confuse about finding these Chapter 12 Constructions Ex 12.2 Of Class 9 NCERT Solutions. So, Read the full post below and get your solutions.

TextbookNCERT
BoardCBSE
CategoryNCERT Solutions
ClassClass 9
SubjectMaths
ChapterChapter 12
ExerciseClass 9 Chapter 12 Constructions Exercise 12.2
Number of Questions Solved5
NCERT Solutions For Class 9 Maths Chapter 12 Constructions Ex 12.2

NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2

NCERT TEXTBOOK EXERCISES

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

Solution:

Given that, in ∆ ABC, BC = 7 cm, ∠B = 75° and AS + AC = 13 cm
Steps of construction

  1. Draw the base BC = 7 cm
  2. At the point 6 make an ∠XBC = 75°.
  3. Cut a line segment BD equal to AB + AC = 13 cm from the ray BX.
  4. Join DC.
  5. Make an ∠DCY = ∠BDC.
  6. Let CY intersect BX at A.
    Then, ABC is the required triangle.

Question 2. Construct a ∆ ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.

Solution:

Given that, in ∆ ABC,
BC = 8 cm, ∠B = 45°and AB – AC = 3.5 cm

Steps of construction

  1. Draw the base BC = 8 cm
  2. At the point B make an ∠XBC = 45°.
  3. Cut the line segment BD equal to AB – AC = 3.5 cm from the ray BX.
  4. Join DC.
  5. Draw the perpendicular bisector, say PQ of DC.
  6. Let it intersect BX at a point A
  7. Join AC.

Question 3. Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.

Solution:

Given that, in ∆ ABC, QR = 6 crn ∠Q = 60° and PR – PQ = 2 cm
Steps of construction

  1. Draw the base QR = 6 cm
  2. At the point Q make an ∠XQR = 60°.
  3. Cut line segment QS = PR- PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.
  4. Join SR.
  5. Draw the perpendicular bisector LM of SR.
  6. Let LM intersect QX at P.
  7. Join PR.

Question 4. Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.

Solution:

Given that, in ∆XYZ ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11cm

Steps of construction

  1. Draw a line segment BC = XY + YZ + ZX = 11 cm
  2. Make ∠LBC = ∠Y = 30° and ∠MCB = ∠Z = 90°.
  3. Bisect ∠LBC and ∠MCB. Let these bisectors meet at a point X.
  4. Draw perpendicular bisectors DE of XB and FG of XC.
  5. Let DE intersect BC at Y and FC intersect BC at Z.
  6. Join XY and XZ.
    Then, XYZ is the required triangle.

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

Solution:

Given that, in A ABC, base BC = 12 cm, ∠B = 90° and AB + BC= 18 cm.
Steps of construction

  1. Draw the base BC = 12 cm
  2. At the point 6, make an ∠XBC = 90°.
  3. Cut a line segment BD = AB+ AC = 18 cm from the ray BX.
  4. Join DC.
  5. Draw the perpendicular bisector PQ of CD to intersect SD at a point A

Join AC.
Then, ABC is the required right triangle.

NCERT Solutions for Class 9 Maths Chapter 12 Constructions Exercise 12.2 PDF

For NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2, you may click on the link below and get your NCERT Solutions for Class 9 Maths Chapter 12 Circles Exercise pdf file.

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