RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles Ex 8.6

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:

1. Draw a line segment BC = 7 cm a base line.
   
2. Construct ∠BCY = 50°.
3. From CY cut off a line segment of length 8 cm and mark the point D.
4. Join CD.
5. Draw the perpendicular bisector of BD, intersecting CD at A.
6. 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:

1. Draw a line segment PQ = 6 cm as base line.
   
2. At Q, construct an angle of 60° and produce it.
3. From QY, cut off a line segment of length 8 cm and mark the point M.
4. Join PM.
5. Draw perpendicular bisector of PM, intersecting QM at R.
6. 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:

1. Draw a line segment QR = 5 cm.
   
2. At R, construct at angle of 40° and produce it.
3. By taking R as centre, draw an arc of radius 1 cm cutting RT at S.
4. Join QS.
5. Also, draw ⊥ bisector of QS which meets RT at R
6. 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:

1. Draw a ray PX and cut off a line segment PQ = 12 cm from it.
2. At P, construct ∠YPB = 25° with the help of protractor i.e. [\(\frac { 1 }{ 2 }\) x 50° ]
3. At Q, construct ∠ZQP = 35°
   
4. Draw perpendicular bisectors of AP intersecting PQ at B.
5. Draw perpendicular bisectors of AQ intersecting PQ at C.
6. 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:

1. Draw BC = 8 cm and get mid-point of BC as D.
   
2. 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.
3. 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.
4. Draw CGF = 7.5 cm.
5. Draw BF and extend.
6. From centre D, draw an arc of radius AD = 6 cm (RBSESolutions.com) which intersects BF (produced) at A.
7. 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.