RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles Ex 8.6 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 8 Construction of Triangles Ex 8.6.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Construction of Triangles |

Exercise |
Ex 8.6 |

Number of Questions Solved |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 8 Construction of Triangles Ex 8.6

Question 1.

Construct ∆ABC in which BC = 7 cm, ∠C = 50° and AC + AB = 8 cm.

Solution.

We are given base BC = 7 cm, the sum of (RBSESolutions.com) other two sides AB + AC = 8 cm and ∠ACB = 50°.

We are required to construct of ∆ABC.

Steps of construction:

- Draw a line segment BC = 7 cm a base line.

- Construct ∠BCY = 50°.
- From CY cut off a line segment of length 8 cm and mark the point D.
- Join CD.
- Draw the perpendicular bisector of BD, intersecting CD at A.
- Join BA.

Thus, ∆ABC is the required triangle.

Question 2.

Construct a triangle PQR in (RBSESolutions.com) which PQ = 6 cm, ∠Q = 60° and QR + PR = 8 cm.

Solution.

Steps of construction:

- Draw a line segment PQ = 6 cm as base line.

- At Q, construct an angle of 60° and produce it.
- From QY, cut off a line segment of length 8 cm and mark the point M.
- Join PM.
- Draw perpendicular bisector of PM, intersecting QM at R.
- Join P to R.

Thus ∆PQR is the required triangle.

Question 3.

Construct a triangle PQR in which QR = 5 cm, ∠R = 40° and PR – PQ = 1 cm.

Solution.

It is given that QR = 5 cm, the difference (RBSESolutions.com) between two sides PR – PQ = 1 cm and base angle ∠R = 40°.

Steps of construction:

- Draw a line segment QR = 5 cm.

- At R, construct at angle of 40° and produce it.
- By taking R as centre, draw an arc of radius 1 cm cutting RT at S.
- Join QS.
- Also, draw ⊥ bisector of QS which meets RT at R
- Join P to Q.

Hence, ∆PQR is the required triangle.

Question 4.

Construct a triangle ABC having its perimeter 12 cm and (RBSESolutions.com) base angles 50° and 70°.

Solution.

Perimeter of ∆ABC is given

i. e., AB + BC + CA = 12 cm, ∠B = 50° and ∠C = 70°.

We are required to construct the ∆ABC.

Steps of construction:

- Draw a ray PX and cut off a line segment PQ = 12 cm from it.
- At P, construct ∠YPB = 25° with the help of protractor i.e. [\(\frac { 1 }{ 2 }\) x 50° ]
- At Q, construct ∠ZQP = 35°

- Draw perpendicular bisectors of AP intersecting PQ at B.
- Draw perpendicular bisectors of AQ intersecting PQ at C.
- Join AB and AC.

Then, ∆ABC is the required triangle.

Question 5.

Constructs ∆ABC in which BC = 8 cm, and (RBSESolutions.com) medians AD and CF are 6 cm and 7.5 cm respectively.

Solution.

Steps of construction:

- Draw BC = 8 cm and get mid-point of BC as D.

- AD = 6 cm, let G be centroid then GD = \(\frac { 1 }{ 3 }\) x 6 =2 cm. Taking D as centre, draw an arc of radius 2 cm.
- CG = \(\frac { 2 }{ 3 }\) x CF = \(\frac { 2 }{ 3 }\) x 7.5 = 5 cm. From centre C, cut an arc of radius 5 cm to intersect previous arc at G.
- Draw CGF = 7.5 cm.
- Draw BF and extend.
- From centre D, draw an arc of radius AD = 6 cm (RBSESolutions.com) which intersects BF (produced) at A.
- Join AC.

Hence, ∆ABC is the required triangle.

We hope the given RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles Ex 8.6 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 8 Construction of Triangles Ex 8.6, drop a comment below and we will get back to you at the earliest.