RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles Additional Questions 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 Additional Questions.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Construction of Triangles |

Exercise |
Additional Questions |

Number of Questions Solved |
4 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 8 Construction of Triangles Additional Questions

Question 1.

Construct a ∆ABC in which AB = AC, BC = 3.8 cm and AD the altitude from A to BC = 4.3 cm.

Solution.

Given: ABC is (RBSESolutions.com) a triangle, AB = AC and BC = 3.8, AD ⊥ BC, AD = 4.3

To construct: ∆ABC

Steps of construction:

- Draw a straight line BC = 3.8 cm.

- Draw the perpendicular bisector AX.
- From D, cut off DA = 4.3 cm.
- Join BA and CA. Hence, ∆ABC is the required triangle.

Question 2.

Construct a ∆ABC in (RBSESolutions.com) which BC = 3.8 cm, (AB + AC) = 5 cm and ∠ABC = 60°.

Solution.

Given: ABC is a triangle, BC = 3.8 cm,

(AB + AC) = 5 cm and ∠ACB = 60°.

To construct: ∆ABC

Steps of construction:

- Draw a straight line BC = 3.8 cm.
- At C, make ∠BCX = 60°.

- At CX, cut off CD = 5 cm.
- Join BD.
- At B, make ∠DBA = ∠BDA where A lies in DC.

Hence, ∆ABC is the required triangle.

Question 3.

Construct a ∆ABC in (RBSESolutions.com) which BC = 3.6 cm, (AC – AB) = 1.6 cm and ∠ACB = 30°.

Solution.

Given: ABC is a triangle, ∠ACB = 30°, BC = 3.6 cm, AC – AB =1.6 cm

To construct: ∆ABC

Steps of construction:

- Draw a straight line BC = 3.6 cm.
- At C, make ∠BCX = 30°.

- At CX, cut off CD = 1.6 cm.
- Join BD.
- At B, make ∠DBY = ∠BDX.
- Let BY cuts DX at A.

Hence, ∆ABC is the required triangle.

Question 4.

Construct a triangle ABC in (RBSESolutions.com) which AB = 3.3 cm, AC = 2.8 cm and altitude AD = 2.3 cm.

Solution.

Given: ABC is a triangle, in which AB = 3.3 cm, AC = 2.8 cm and altitude AD = 2.3 cm.

To construct ∆ABC.

Steps of construction:

- Draw a straight line PQ and mark a point D in it.
- Draw DX ⊥ PQ.
- From DX cut off DA = 2.3 cm.
- With A as (RBSESolutions.com) centre and radii 3.3 cm, 2.8 cm, draw arcs to cut PQ at B, and C respectively.
- Join AB and AC.

Hence, ∆ABC is the required triangle.

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