NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3

NCERT Solutions for class 7 maths cover the CBSE syllabus for class 7 Maths. In Exercise 6.3, you will learn about the different types of triangles. You will find out how to classify triangles based on their sides and angles.

Understanding the properties of isosceles, equilateral, and scalene triangles is important. The NCERT Solutions will help you identify the different types of triangles and apply their characteristics to solve problems.

The explanations are clear, and there are many examples and practice questions. By mastering the concepts in this exercise, you will develop strong skills in analyzing and classifying triangles. This will prepare you for more advanced geometry topics.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Exercise 6.3

Ex 6.3 Class 7 Maths Question 1.

Find the value of the unknown x in the following diagrams:

Solution:

(i) By angle sum property of a triangle, we have

∠x + 50° + 60° = 180°

⇒ ∠x + 110° = 180°

∴ ∠x = 180° – 110° = 70°

(ii) By angle sum property of a triangle, we have

∠x + 90° + 30 = 180° [∆ is right angled triangle]

⇒ ∠x + 120° = 180°

∴ ∠x – 180° – 120° = 60°

(iii) By angle sum property of a triangle, we have

∠x + 30° + 110° – 180°

⇒ ∠x + 140° = 180°

∴ ∠x = 180° – 140° = 40°

(iv ) By angle sum property of a triangle, we have

∠x + ∠x + 50° = 180°

⇒ 2x + 50° = 180°

⇒ 2x = 180° – 50°

⇒ 2x = 130°

∴ \(x=\frac{130^{0}}{2}=65^{\circ}\)

(v) By angle sum property of a triangle, we have

∠x + ∠x +∠x =180°

⇒ 3 ∠x = 180°

∴ \(\angle x=\frac{180^{0}}{3}=60^{\circ}\)

(vi) By angle sum property of a triangle, we have

x + 2 x + 90° = 180° (∆ is right angled triangle)

⇒ 3x + 90° = 180°

⇒ 3x = 180° – 90°

⇒ 3x = 90°

∴ \(x=\frac{90^{0}}{3}=30^{\circ}\)

Ex 6.3 Class 7 Maths Question 2.

Find the values of the unknowns x and y in the following diagrams:

Solution:

(i) ∠x + 50° = 120° (Exterior angle of a triangle)

∴ ∠x = 120°- 50° = 70°

∠x + ∠y + 50° = 180° (Angle sum property of a triangle)

70° + ∠y + 50° = 180°

∠y + 120° = 180°

∠y = 180° – 120°

∴ ∠y = 60°

Thus ∠x = 70 and ∠y – 60°

(ii) ∠y = 80° (Vertically opposite angles are same)

∠x + ∠y + 50° = 180° (Angle sum property of a triangle)

⇒ ∠x + 80° + 50° = 180°

⇒ ∠x + 130° = 180°

∴ ∠x = 180° – 130° = 50°

Thus, ∠x = 50° and ∠y = 80°

(iii) ∠y + 50° + 60° = 180° (Angle sum property of a triangle)

∠y + 110° = 180°

∴ ∠y = 180°- 110° = 70°

∠x + ∠y = 180° (Linear pairs)

⇒ ∠x + 70° = 180°

∴ ∠x = 180° – 70° = 110°

Thus, ∠x = 110° and y = 70°

(iv) ∠x = 60° (Vertically opposite angles)

∠x + ∠y + 30° = 180° (Angle sum property of a triangle)

⇒ 60° + ∠y + 30° = 180°

⇒ ∠y + 90° = 180°

⇒ ∠y = 180° – 90° = 90°

Thus, ∠x = 60° and ∠y = 90°

(v) ∠y = 90° (Vertically opposite angles)

∠x + ∠x + ∠y = 180° (Angle sum property of a triangle)

⇒ 2 ∠x + 90° = 180°

⇒ 2∠x = 180° – 90°

⇒ 2∠x = 90°

∴ \(\angle x=\frac{90^{\circ}}{2}=45^{\circ}\)

Thus, ∠x = 45° and ∠y = 90°

(vi) From the given figure, we have

Adding both sides, we have

∠y + ∠1 + ∠2 = 3∠x

⇒ 180° = 3∠x (Angle sum property of a triangle)

∴ \(\angle x=\frac{180^{\circ}}{3}=60^{\circ}\)

∠x = 60°, ∠y = 60°

 

Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3 FAQs

What is the main focus of Exercise 6.3 in NCERT Solutions for Class 7 Maths Chapter 6?

The primary focus of Exercise 6.3 is on the classification of triangles based on their sides and angles. This exercise will help you understand the properties of different types of triangles.

What are the different types of triangles covered in this exercise?

In this exercise, you will learn about isosceles triangles, equilateral triangles, and scalene triangles. You will explore the unique characteristics of each type of triangle.

Why is it important to understand the properties of different triangles?

Knowing the properties of various triangles is essential because it allows you to identify and classify different types of triangles. This knowledge can then be applied to solve a variety of geometry problems.

Why is it important to understand the properties of different triangles?

Knowing the properties of various triangles is essential because it allows you to identify and classify different types of triangles. This knowledge can then be applied to solve a variety of geometry problems.

What kind of practice problems will be included in Exercise 6.3?

NCERT Solutions will provide practice problems that require you to identify the type of triangle, apply the properties of triangles, and solve problems involving triangle classification.

How will the NCERT Solutions help me master the concepts in this exercise?

NCERT Solutions offer clear explanations, step-by-step examples, and ample practice questions to help you thoroughly understand and apply the concepts related to the classification of triangles. The solutions also provide helpful tips and strategies.