KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2

In this chapter, we provide KSEEB SSLC Class 9 Maths Chapter 6 Constructions Ex 6.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 9 Maths Chapter 6 Constructions Ex 6.2 pdf, free KSEEB SSLC Class 9 Maths Chapter 6 Constructions Ex 6.2 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 9 Maths Chapter 5 Triangles Ex 5.2

Karnataka Board Class 9 Maths Chapter 6 Constructions Ex 6.2

Question 1.
Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm,
Solution:
Construction of a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 1
Steps of Construction :

  1. Draw BC = 7 cm line segment.
  2. Construct ∠B = 75° and produce it upto Bx.
  3. With B as centre draw an arc BD = 13 cm such that it intersects Bx at D.
  4. Construct PQ which is the perpendicular bisector of BD.
  5. PQ intersects BD at A.
  6. Join AC.
  7. Now ∆ABC, with measurement AB + AC = 13 cm, and ∠B = 75°, BC = 7 cm is constructed.

Question 2.
Construct a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
Solution:
Constructing a triangle ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 2
Steps of Construction :

  1. Draw a line segment BC = 8 cm.
  2. Draw ∠B = 45° with the help of a protractor. Bx is joined.
  3. Draw an arc BD = 3.5 cm.
  4. When perpendicular bisector DC is drawn, that intersects Bx at A. AC is joined.
  5. Now, DABC with BC = 8 cm, ∠B = 45° and AB – AC = 3.5 cm. is constructed.

Question 3.
Construct a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
Solution:
Constructing a triangle PQR in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 3
Steps of Construction :

  1. Draw a line segment QR = 6 cm.
  2. With the help of a protractor draw a line segment Qx and ∠Q = 60°.
  3. Produce line segment xQ upto y. mark T such that QT = 2 cm. Join TR.
  4. If the perpendicular bisector of TR is drawn, that meets line segment Qx at P. PR is joined.
    Now, ∆PQR is constructed.

Question 4.
Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
Solution:
Constructing a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 4
Steps of Construction :

  1. Draw AB = 11 cm line segment with measurement XY + YZ + ZX.
  2. With the help of protractor construct ∠A = 30°, ∠B = 90°.
  3. Construct angular bisectors of ∠CAB and ∠DBA which meet at x. Ax, Bx are joined.
  4. If perpendicular bisector AX is drawn, it intersects AB at Y and AB at Z.
  5. By joining XY and XZ, ∆XYZ is constructed.

Question 5.
Construct a right triangle whose base is 12 cm and the sum of its hypotenuse and the other side is 18 cm.
Solution:
Constructing a right-angled triangle ABC with base, AB =12 cm and the sum of hypotenuse BC and another side AC, AC + BC = 18 cm.
KSEEB Solutions for Class 9 Maths Chapter 6 Constructions Ex 6.2 5
Steps of Construction :

  1. Draw a line segment AB = 12 cm.
  2. With the help of a protractor construct ∠A = 90° and produce upto AX.
  3. Mark ‘D’ on AX such that AD = 18 cm.
  4. Join BD.
  5. Let the perpendicular bisector BD intersects AD at C. Join BC.
  6. Now required triangle ABC is constructed with Base AB = 12cm, Side AC = – 5 cm, Hypotenuse BC = 13 cm and ∠A = 90°.

All Chapter KSEEB Solutions For Class 9 Maths

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All Subject KSEEB Solutions For Class 9

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