RS Aggarwal Solutions Class 9 Maths Chapter 9: Congruent triangles are those with corresponding sides and angles that are equal. Understanding congruence helps solve geometry problems and prove theorems.
This chapter also covers triangle inequalities, which are rules about the relationships between the sides and angles of triangles. By studying this chapter, students can improve their understanding of geometry concepts and become better at solving math problems.RS Aggarwal Solutions Class 9 Maths Chapter 9 PDF
Question 1.
Solution:
In ∆ABC, ∠B = 76° and ∠C = 48° But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle) => ∠A + 76° + 48° = 180° => ∠ A + 124° = 180° => ∠A= 180° – 124° = 56°Question 2.
Solution:
Angles of a triangle are in the ratio = 2:3:4 Let first angle = 2x then second angle = 3x and third angle = 4x 2x + 3x + 4x = 180° (Sum of angles of a triangle) = 9x = 180° = x = 180/9= 20° First angle = 2x = 2 x 20° = 40° Second angle = 3x = 3 x 20° = 60° and third angle = 4x = 4 x 20° = 80° Ans.Question 3.
Solution:
In ∆ABC, 3∠A = 4∠B = 6∠C = x (Suppose)Question 5.
Solution:
In ∆ABC, ∠A+∠B = 125° ...(i) ∠A + ∠C = 113° ...(ii) But ∠A + ∠B + ∠C = 180° ...(iii) (sum of angles of a triangles) Subtracting, (i), from (iii), ∠C = 180°- 125° = 55° Subtracting (ii) from (iii), ∠B = 180°- 113° – 67° ∠A + ∠B = 125° ∠ A + 67° = 125° => ∠ A = 125° – 67° ∠A = 58° Hence ∠A = 58°, ∠B = 67° and ∠ C = 55° Ans.Question 6.
Solution:
In ∆ PQR, ∠ P – ∠ Q = 42° = ∠P = 42°+∠Q ...(i) ∠Q – ∠R = 21° ∠Q – 21°=∠R ...(ii) But ∠P + ∠Q + ∠R = 180° (Sum of angles of a triangles) 42° + ∠Q + ∠Q + ∠Q – 21°= 180° = 21° + 3∠Q = 180° = 3∠Q = 180°- 21° = 159° from ∠Q = 53° (i)∠P = 42° + ∠Q = 42° + 53° = 95° and from (ii) ∠R = ∠Q – 21° = 53° – 25° = 32° Hence ∠P = 95°, ∠Q = 53° and ∠R = 32° Ans.Question 8.
Solution:
Let ∠ A, ∠ B and ∠ C are the three angles of the ∆ ABC Let ∠ A = ∠ B = x then ∠C = x + 48° But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle) x + x + x + 18° = 180° = 3x + 18° = 180° = 3x = 180° – 18° = 162° x = 54° ∠A = 54°, ∠B = 54° and ∠C = 54° + 18° = 72° Hence angles are 54°, 54 and 72° Ans.Question 9.
Solution:
Let the smallest angle of a triangle = x° their second angle = 2x° and third angle = 3x° But sum of angle of a triangle = 180° x + 2x + 3x = 180° = 6x = 180° = x = 30° Hence smallest angle = 30° Second angle = 2 x 30° = 60° and third angle = 3 x 30° = 90° Ans.Question 10.
Solution:
In a right angled triangle. one angle is = 90° Sum of other two acute angles = 90° But one acute angle = 53° Second acute angle = 90° – 53° = 37° Hence angle of the triangle with be 90°, 53°, 37° Ans.Question 13:
Solution:
Given. In ∆ABC, Each angle is less than the sum of the other two angles ∠A< ∠B + ∠C ∠B < ∠C + ∠A and ∠C< ∠A + ∠C Proof : ∠ A < ∠B + ∠C Adding ∠ A both sides, ∠A + ∠A < ∠A + ∠B + ∠C => 2 ∠ A < 180° (∴ ∠A+∠B+∠C=180°) ∠A = ∠A< 90 Similarly, we can prove that, ∠B < 90° and ∠C < 90° ∴ each angle is less than 90° Hence, triangle is an acute angled triangle. Hence proved.