RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles Ex 10.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Exercise 10.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Construction of Triangles |

Exercise |
Ex 10.1 |

Number of Questions |
6 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Ex 10.1

Question 1

Construct a ∆PQR when PQ = 4 cm, QR = 3 cm and RP = 5.5 cm.

Solution:

Steps of construction :

1. Draw a line (RBSESolutions.com) segment QR = 3 cm

2. Make an arc of radius 4 cm from point Q.

3. Make another arc of radius 5.5 cm from point R, which intersects prior arc at point P.

4. Join PQ and PR Thus ∆PQR is required triangle.

Question 2

Construct ∆XYZ when XZ = 6 cm, XY = 4.5 cm and ∠X = 50°,

Solution:

Steps of construction :

1. Draw a line (RBSESolutions.com) segment XY = 4.5 cm M

2. Make ∠YXM = 50° at point X.

3. Make an arc of radius 6 cm from point X which cuts line XM at point Z.

4. Joing Y and Z .

Thus ∆XYZ is required (RBSESolutions.com) triangle.

Question 3

Construct a ∆ABC when AB = 5 cm, ∠A = 45° and ∠B = 60°.

Solution:

Steps of construction :

1. Draw a line segment AB = 5 cm.

2. Make ∠45° at point A and also ∠60° at point B with the (RBSESolutions.com) help of compass, which intersect each other at point C.

Thus ∆ABC is required triangle.

Question 4

Construct ∆DEF when hypotenuse DE = 5 cm, Base DF = 3 cm and ∠D = 90°.

Solution:

Steps of construction :

1. Draw (RBSESolutions.com) a line segment DF = 3 cm.

2. Make ∠90° at point D with the help of

3. Make an arc of radius 5 cm from point F which cuts (RBSESolutions.com) the line making ∠90° at point E.

4. Join E and F.

Thus ∆DEF is required triangle.

Question 5

Construct an equilateral triangle of side 4 cm.

Solution:

Steps of construction :

1. Draw a line segment BC = 4 cm.

2. Make an arc of radius 4 cm form point B and also (RBSESolutions.com) the arc of same radius from point C so that they intersect each other at point A.

Thus ∆ABC is required triangle.

Question 6

Construct a ∆PQR where PQ = 5 cm, ∠P = 75° and ∠R = 55°.

Solution:

Steps of construction :

1. Draw a line segment PQ = 5 cm.

2. Make ∠75° at point P with (RBSESolutions.com) the help of compass.

3. Sum of all angles of a triangle is 180°.

∠P + ∠Q + ∠R = 180°

∠5° + ∠Q + 55° = 180°

130° + ∠Q = 180°

∠Q = 180°- 130° = 50°

4. Make ∠50° at point Q which (RBSESolutions.com) intersect the line making ∠75° at point R.

5. Join RQ.

Thus ∆PQR is required triangle.

We hope the RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles Ex 10.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Exercise 10.1, drop a comment below and we will get back to you at the earliest.

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