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
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 = 56oQuestion 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/6Question 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 ) = 90oQuestion 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:
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