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

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:

1. Draw a straight line BC = 3.8 cm.
   
2. Draw the perpendicular bisector AX.
3. From D, cut off DA = 4.3 cm.
4. 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:

1. Draw a straight line BC = 3.8 cm.
2. At C, make ∠BCX = 60°.
   
3. At CX, cut off CD = 5 cm.
4. Join BD.
5. 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:

1. Draw a straight line BC = 3.6 cm.
2. At C, make ∠BCX = 30°.
   
3. At CX, cut off CD = 1.6 cm.
4. Join BD.
5. At B, make ∠DBY = ∠BDX.
6. 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:

1. Draw a straight line PQ and mark a point D in it.
2. Draw DX ⊥ PQ.
3. From DX cut off DA = 2.3 cm.
4. With A as (RBSESolutions.com) centre and radii 3.3 cm, 2.8 cm, draw arcs to cut PQ at B, and C respectively.
5. Join AB and AC.
   Hence, ∆ABC is the required triangle.

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