RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Exercise 8.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 8 |

Chapter Name |
Triangle and its Properties |

Exercise |
Ex 8.1 |

Number of Questions |
8 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1

Question 1

Find the value of unknown (RBSESolutions.com) angle x in the following figures :

Solution:

Sum of all angles of (RBSESolutions.com) a triangle is 180°.

(i) x + 60° + 85° = 180°

⇒ x + 145° = 180°

⇒ x = 180° – 145° = 35°

(ii) x + 40° + 20° = 180°

⇒ x + 60° = 180°

⇒ x = 180° – 60° = 120°

(ii) x + 40° + 90° = 180°

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°

(iv) x + x + x = 180°

⇒ 3x = 180°

⇒ x = [latex]\frac { { 180 }^{ 0 } }{ 3 }[/latex] = 180°

Question 2

Find the unknown angle x for (RBSESolutions.com) the following figures

Solution:

In any triangle exterior angle is (RBSESolutions.com) always equal to the sum of two interior angles.

(i) x = 80° + 50° = 130°

(ii) x + 35° = 110°

⇒ x = 110° – 35° = 75°

(iii) x = 30° + 35° = 65°

(iv) x + 70° = 120°

⇒ x = 120° – 70° = 50°

(v) x + 35° = 65°

⇒ 65° – 35° = 30°

(vi) x = 55° + 40° = 95°

Question 3

Find the values of the unknown (RBSESolutions.com) angles x and y in the following figures :

Solution:

(i) In a triangle, exterior (RBSESolutions.com) angle and adjecent internal angle make a linear pair.

x + 110° = 180°

⇒ x = 180° – 110° = 70°

In a triangle, sum of all angles is 180°.

∴ x + y + 60° = 180°

⇒ 70° + y + 60° = 180°

⇒ y + 130° = 180°

⇒ y = 180° – 130° = 50°

∴ x = 70°, y = 50°

(ii) In a triangle, exterior angle and (RBSESolutions.com) adjecent internal angle make a linear pair.

x = 45° + 30° = 75°

∵ In a triangle, sum of all angles is 180°.

∴ y + 45° + 30° = 180°

⇒ y + 75° = 180°

∴ y = 180° – 75° = 105°

∴ x = 75°,y= 105°

(iii) In a triangle, exterior angle and adjecent internal (RBSESolutions.com) angle make a linear pair.

∴ y + 80° = 180°

⇒ y = 180° – 80° = 100°

In a triangle, sum of all angles is 180°

∴ x + x + y = 180°

⇒ 2x + 100° = 180°

⇒ 2x = 180° – 100°

⇒ 2x = 80°

∴ x = [latex]\frac { { 80 }^{ 0 } }{ 2 }[/latex] = 40°

∴ x = 40°, y = 100°

(iv) Here x = 70° (vertically opposite angle)

∵ Sum of (RBSESolutions.com) three angles of a triangle = 180°

∴ x + y + 40° = 180°

⇒ 70° + y + 40° = 180°

⇒ 110° + y = 180°

∴ y = 180° – 110° = 70°

∴ x = 70°,y = 70°

Question 4

If one acute angle of a right angled triangle is 45° then find (RBSESolutions.com) another acute angle.

Solution:

Sum of all angles of a triangle is 180°.

∴ 90° + one acute angle + other acute angle = 180°

⇒ 90° + 45° + other acute angle = 180°

⇒ other acute angle = 180 – 135°

⇒ = 45°

Question 5

If two angles of (RBSESolutions.com) a triangle are 50° each then find the third angle.

Solution:

∵ Sum of three angles of a triangle = 180°

∴ 50° + 50° + third angle = 180°

⇒ 100° + third angle = 180°

⇒ third angle = 180° – 100°

⇒ = 80°

Question 6

If the angles of a triangle are in ratio 1 : 2 : 3, then find each angle (RBSESolutions.com) of the triangle.

Solution:

Let the angles of the triangle are x°, 2x° and 3x°

∴ x° + 2x° + 3x° = 180°

⇒ 6x° = 180°

∴ x° = [latex]\frac { { 180 }^{ 0 } }{ 6 }[/latex] = 30°

∴ I angle = x° = 30°

II angle = 2x° = 2 x 30° = 60°

III angle = 3x° = 3 x 30° = 90°

Question 7

Is it possible to construct a right angled triangle (RBSESolutions.com) whose other two angles are 70° and 21° ? If not, then why ? Justify.

Solution:

No, ∴ Sum of three angles = 180

⇒ 70° + 21 + x° = 180°

⇒ x = 180 – 90°

⇒ x = 89°

≠ 9

∵ Third angle is not a right angle.

Hence. Such right angle triangle is not possible.

Question 8

Given below are the triads. Which of the (RBSESolutions.com) following triads represents the angles of a triangle ?

(i) 100°, 30°, 40°

(ii) 30°, 59°,91°

(iii) 45°, 45°, 90°

(iv) 120°, 30°, 50°

Solution:

(i) (100°, 30°, 40°)

∵ Sum of all angles of a triangle = 180°

(i) 100° + 30 + 40° ≠ 180° ⇒ 170 ≠ 180

∴ Triangle is not possible.

(ii) 30° + 59° + 91°= 180° ⇒ 180° = 180°

∴ Triangle is possible.

(iii) 45° + 45° + 90° = 180° ⇒ 180° = 180°

∴ Triangle is (RBSESolutions.com) possible.

(iv) 120° + 30° + 50° = 180° ⇒ 200°= 180°

∴ Triangle is not possible.

We hope the RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Ex 8.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Exercise 8.1, drop a comment below and we will get back to you at the earliest.

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