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

Steps of Construction :

- Draw BC = 7 cm line segment.
- Construct ∠B = 75° and produce it upto Bx.
- With B as centre draw an arc BD = 13 cm such that it intersects Bx at D.
- Construct PQ which is the perpendicular bisector of BD.
- PQ intersects BD at A.
- Join AC.
- 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.

Steps of Construction :

- Draw a line segment BC = 8 cm.
- Draw ∠B = 45° with the help of a protractor. Bx is joined.
- Draw an arc BD = 3.5 cm.
- When perpendicular bisector DC is drawn, that intersects Bx at A. AC is joined.
- 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.

Steps of Construction :

- Draw a line segment QR = 6 cm.
- With the help of a protractor draw a line segment Qx and ∠Q = 60°.
- Produce line segment xQ upto y. mark T such that QT = 2 cm. Join TR.
- 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.

Steps of Construction :

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

Steps of Construction :

- Draw a line segment AB = 12 cm.
- With the help of a protractor construct ∠A = 90° and produce upto AX.
- Mark ‘D’ on AX such that AD = 18 cm.
- Join BD.
- Let the perpendicular bisector BD intersects AD at C. Join BC.
- 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**

—————————————————————————–**All Subject KSEEB Solutions For Class 9**

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