RBSE Solutions for Class 9 Maths Chapter 8 Construction of Triangles Miscellaneous Exercise 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 Miscellaneous Exercise.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Construction of Triangles |

Exercise |
Miscellaneous Exercise |

Number of Questions Solved |
7 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 8 Construction of Triangles Miscellaneous Exercise

Question 1.

Construct a triangle whose perimeter is 12 cm and the (RBSESolutions.com) sides are in the ratio 1 : 2 : 3.

Solution.

Suppose sides of the triangle are x, 2x and 3x

Here it is given that

Perimeter of Δ = 12 cm

x + 2x + 3x = 12 cm

⇒ 6x = 12

⇒ x = 2

Sides of Δ are 2 cm, 4 cm and 6 cm.

Steps of construction:

- Draw BC = 4 cm as base line.
- Take B as centre, draw an arc of radius 2 cm.
- Taking C as centre, draw an arc of radius 6 cm.
- Join A to B and C.
- Hence, ΔABC is the required triangle whose sides are in the ratio 1 : 2 : 3 and having perimeter 12 cm.

Question 2.

Construct a triangle ABC in which ∠B = 90°, ∠C = 60° and c = 5 cm.

Solution.

First, of all we will find (RBSESolutions.com) the third angle of ΔABC

i.e. ∠A = 180° – (90° + 60°) = 30°

Steps of construction:

- First draw AB = 5 cm as base line.
- At B, draw an angle of 90° with the help of compass.
- Also at A, construct an angle of 30°, which meets at C.

Hence, ΔABC is the required triangle.

Question 3.

Construct a right angle triangle ABC in (RBSESolutions.com) which hypotenuse BC = 8.2 cm and one side = 4.2 cm.

Solution.

Steps of construction:

- Draw a line segment AB = 4.2 cm.

- At A, draw an angle of 90°.
- From B, draw an arc of radius 8.2 cm which cuts the line segment at C.

Hence, ΔABC is the required triangle.

Question 4.

Construct a triangle ABC in which ∠B = 45°, ∠C = 60° and the length (RBSESolutions.com) of perpendicular AD from A on BC is 4 cm.

Solution.

Steps of construction:

- Draw a straight XY as base line.
- At any point say D, on XY line segment, draw perpendicular DZ. By taking D as centre cut an arc of radius 4 cm from DZ and marked it as DA.

- Now, draw ∠DAB = 45° and ∠DAC = 30° which meets XY at B and C respectively.

Hence, ΔABC is the required triangle.

Question 5.

Construct (RBSESolutions.com) a triangle ABC in which a = 5.6 cm, b + c = 10.2 cm and ∠B – ∠C = 30°.

[Hint: Construct an angle of 90°+ [latex]\frac { 1 }{ 2 }[/latex] (∠B – ∠C) i.e. 105° at the vertex B]

Solution.

Steps of construction:

- First of all, draw BC = 5.6 cm as base line.
- Construct an angle of 105° at B.

- By taking C as centre, cut an arc of radius 10.2 cm, which intersects arm of the angle ∠CBD at E. Join EC.
- Draw perpendicular bisector of EB, which intersects CE at A. Join AB. Hence, ΔABC is the required triangle.

Question 6.

Construct a triangle such that the lengths (RBSESolutions.com) of its three medians are 4.2 m, 4.8 cm and 5.4 cm respectively.

Solution.

Steps of construction:

- 1. Draw AG = 2.8 cm and produce to a point P such that AG = GP.

- From centre G, cut an arc

GC = [latex]\frac { 2 }{ 3 }[/latex] x CF = [latex]\frac { 2 }{ 3 }[/latex] x 5.4 = 3.6 cm

and from P cut an arc

CP = BG = [latex]\frac { 2 }{ 3 }[/latex] x BE

= [latex]\frac { 2 }{ 3 }[/latex] x 4.8 = 3.2 cm

to intersect each other at C. - From centre G, cut an arc

GB = [latex]\frac { 2 }{ 3 }[/latex] x BE = 3.2 cm and from P cut an arc PB = CG = 3.6 cm to intersect each other at B. Draw BC. - Join A to B and A to C.

Hence, ΔABC is the required triangle.

Question 7.

Construct an isosceles triangle in (RBSESolutions.com) which height is 6 cm and equal sides are 7 cm. Measure the base.

Solution.

Steps of construction:

- Draw a line PQ.
- Draw perpendicular on any point on the line PQ, mark as M.

- By taking M as centre, cut an arc of radius 6 cm and mark that point as A.
- From A, cut an arc of 7 cm either side (RBSESolutions.com) of the perpendicular and mark them as B and C. Join AB and AC.

Hence, ΔABC is the required triangle.

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

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