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RS Aggarwal Solutions Class 9 Maths Chapter 8 Triangles

Here, we have provided RS Aggarwal Solutions Class 9 Maths Chapter 8. Students can view these RS Aggarwal Solutions Class 9 Maths Chapter 8 before exams for better understanding of the chapter.
authorImageAnanya Gupta13 Apr, 2024
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RS Aggarwal Solutions Class 9 Maths Chapter 8

RS Aggarwal Solutions Class 9 Maths Chapter 8: RS Aggarwal Solutions Class 9 Maths Chapter 8 focuses on triangles. It covers different aspects of triangles, like their properties and types. The chapter provides clear explanations and step-by-step solutions to help students understand these concepts easily.

By studying Chapter 8, students can learn important concepts such as the Pythagorean theorem and criteria for congruence and similarity of triangles. Practicing the exercises in this chapter can enhance problem-solving skills and boost confidence in dealing with triangle-related problems.

RS Aggarwal Solutions Class 9 Maths Chapter 8 PDF

You can access the PDF for RS Aggarwal Solutions Class 9 Maths Chapter 8 by clicking on the link provided below. This PDF contains detailed solutions to the exercises and problems covered in the chapter, making it easier for students to understand and practice triangle-related concepts.

RS Aggarwal Solutions Class 9 Maths Chapter 8 PDF

RS Aggarwal Solutions Class 9 Maths Chapter 8

The solutions for RS Aggarwal Class 9 Maths Chapter 8 are provided below. These solutions cover various topics related to triangles and offer step-by-step explanations to help students understand the concepts better. By referring to these solutions, students can clarify their doubts and strengthen their understanding of triangle geometry.

RS Aggarwal Solutions Class 9 Chapter 8 Triangles Exercise- 8.8

Question 1.

Solution:

Since, sum of the angles of a triangle is 180o ∠A + ∠B + ∠C = 180o ⇒ ∠A + 76o + 48o = 180o ⇒ ∠A = 180o – 124o = 56o ∴ ∠A = 56o

Question 2.

Solution:

Let the measures of the angles of a triangle are (2x)o , (3x)o and (4x)o . Then, 2x + 3x + 4x = 180 [sum of the angles of a triangle is 180o ] ⇒ 9x = 180 ⇒ x = 180/9= 20 ∴ The measures of the required angles are: 2x = (2 × 20)o = 40o 3x = (3 × 20)o = 60o 4x = (4 × 20)o = 80o Question 3. Solution: Let 3∠A = 4∠B = 6∠C = x (say) Then, 3∠A = x ⇒ ∠A = x/3 4∠B = x ⇒ ∠B = x/4 and 6∠C = x ⇒ ∠C = x/6

Question 9.

Solution:

Let ∠C be the smallest angle of ABC. Then, ∠A = 2∠C and B = 3∠C Also, ∠A + ∠B + ∠C = 180o ⇒ 2∠C + 3∠C + ∠C = 180o ⇒ 6∠C = 180o ⇒ ∠C = 30o So, ∠A = 2∠C = 2 (30o ) = 60o ∠B = 3∠C = 3 (30o ) = 90o

Question 12.

Solution:

Given : ∆ABC in which ∠A = 90o , AL ⊥ BC To Prove: ∠BAL = ∠ACB Proof : In right triangle ∆ABC, ⇒ ∠ABC + ∠BAC + ∠ACB = 180o ⇒ ∠ABC + 90o + ∠ACB = 180o ⇒ ∠ABC + ∠ACB = 180o – 90o ∴ ∠ABC + ∠ACB = 90o ⇒ ∠ ACB = 90o – ∠ABC ....(1) Similarly since ∆ABL is a right triangle, we find that, ∠BAL = 90o – ∠ABC ...(2) Thus from (1) and (2), we have ∴ ∠BAL = ∠ACB (Proved)

Question 13.

Solution:

Let ABC be a triangle. So, ∠A < ∠B + ∠C Adding A to both sides of the inequality, ⇒ 2∠A < ∠A + ∠B + ∠C ⇒ 2∠A < 180o [Since ∠A + ∠B + ∠C = 180 o ] ⇒ ∠A < 180/2 = 90o Similarly, ∠B < ∠A + ∠C ⇒ ∠B < 90o and ∠C < ∠A + ∠B ⇒ ∠C < 90o ∆ABC is an acute angled triangle.

Question 14.

Solution:

Let ABC be a triangle and ∠B > ∠A + ∠C Since, ∠A + ∠B + ∠C = 180o ⇒ ∠A + ∠C = 180o – ∠B Therefore, we get ∠B > 180o – ∠B Adding ∠B on both sides of the inequality, we get, ⇒ ∠B + ∠B > 180o – ∠B + ∠B ⇒ 2∠B > 180o ⇒ ∠B > 180/2= 90o i.e., ∠B > 90o which means ∠B is an obtuse angle. ∆ABC is an obtuse angled triangle.

Question 18.

Question 25:

Benefits of RS Aggarwal Solutions Class 9 Maths Chapter 8 - Triangles

Here are some key benefits of RS Aggarwal Solutions Class 9 Maths Chapter 8:
  • Clarity of Concepts: The solutions provide clear explanations of the concepts related to triangles, making it easier for students to understand.
  • Step-by-Step Approach: Each solution is presented in a step-by-step manner, allowing students to follow along and grasp the solution method.
  • Practice Material: The chapter provides ample practice problems, allowing students to reinforce their understanding of triangle geometry.
  • Exam Preparation: By solving the problems in this chapter, students can prepare effectively for their exams, including both school exams and competitive exams.
  • Self-Assessment: The solutions enable students to assess their understanding of the concepts by checking their answers against the provided solutions.
CBSE Class 9 Maths Syllabus CBSE Class 9 Science Syllabus
CBSE Class 9 Computer Application Syllabus CBSE Class 9 Social Science Syllabus

RS Aggarwal Solutions Class 9 Maths Chapter 8 FAQs

How are angles classified in triangles?

Angles in triangles are classified as acute, obtuse, or right angles based on their measures.

What is the Pythagorean theorem?

The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

How are congruent triangles defined?

Two triangles are said to be congruent if their corresponding sides and angles are equal in measure.

What are the properties of an equilateral triangle?

An equilateral triangle has all three sides of equal length and all three angles of equal measure (60 degrees).

How are the medians of a triangle defined?

The medians of a triangle are line segments drawn from each vertex to the midpoint of the opposite side. They intersect at a point called the centroid.
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