RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle (Updated For 2024)

RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle: Be it your Class 9 Maths assignments or exam preparation, RS Aggarwal Solutions Class 9 Maths has got your back.The solutions of RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle are reliable and accurate, thanks to the subject matter experts.

You can download the Free PDF of RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle by using the link given in the blog. To know more, read the whole blog.

Download RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle PDF

RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle

Access The RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle

Question 1.
Solution:
In ∆ ABC, ∠A = 70° and AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)

=> 70° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B = 180°- 70° = 110°
∠B = 110o2 = 55° and
Hence ∠B = 55°,∠C = 55° Ans.

Question 2.
Solution:
In ∆ ABC, ∠A= 100°
It is an isosceles triangle

∴AB = AC
∠B = ∠C
(Angles opposite to equal sides)
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> 100° + ∠B + ∠B = 180°
(∴ ∠B = ∠C)
=> 2∠B + 100° = 180°
=> 2∠B = 180°- 100° = 80°
∠B = 80o2 = 40°
and ∠C = 40°
∴ Base angles are 40°, 40° Ans.

Question 3.
Solution:
In ∆ ABC,
AB = AC
∴∠C = ∠B
(Angles opposite to equal sides)

But ∠B = 65°
∴ ∠ C = 65°
But ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠A + 65° + 65° = 180°
=> ∠ A + 130° = 180°
=> ∠ A = 180° – 130°
=> ∠ A = 50°
Hence ∠ A = 50° and ∠ C = 65° Ans.

Question 4.
Solution:
In ∆ ABC
AB = AC
∴ ∠C = ∠B
(Angles opposite to equal sides)

But ∠A = 2(∠B + ∠C)
=> ∠B + ∠C = 12 ∠A 2
But ∠A + ∠B + ∠C = 180°
(sum of angles of a triangle)
=> ∠A+ 12 ∠A = 180°
=> 32 ∠A = 180°
=> ∠A = 180° x 23 = 120°
and ∠B + ∠C = 12 ∠A = 12 x 120°
= 60°
∴ ∠ B = ∠ C
∴ ∠ B = ∠ C = 60o2 = 30°
Hence ∠ A = 120°, ∠B = 30°, ∠C = 30° Ans.

Question 5.
Solution:
In ∆ ABC,
AB = BC and ∠B = 90°
∴ AB = BC
∴ ∠ C = ∠ A
(Angles opposite to equal sides)

Now ∠A + ∠B + ∠C = 180°
(Sum of angles of a triangle)
=> ∠ A + 90° + ∠ A = 180°
(∴ ∠C = ∠A)
=> 2∠A + 90° – 180°
=> 2∠ A = 180° – 90° = 90°
∠ A = 90o2 = 45°
∴ ∠ C = 45° (∴ ∠ C = ∠ A)
Hence, each of the equal angles is 45° Ans.

Question 6.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC

Base BC is produced to both sides upto D and E respectively forming exterior angles
∠ ABD and ∠ ACE
To Prove : ∠ABD = ∠ACE
Proof : In ∆ ABC
∴ AB = AC (given)
∴ ∠C = ∠B
(Angles opposite to equal sides)
=> ∠ ABC = ∠ACB
But ∠ ABC + ∠ABD = 180° (Linear pair)
Similarly ∠ACB + ∠ACE = 180°
∠ ABC + ∠ABD = ∠ACB + ∠ACE
But ∠ ABC = ∠ ACB (proved)
∠ABD = ∠ACE
Hence proved.

Question 7.
Solution:
∆ ABC is an equilateral triangle
∴ AB = BC = CA
and ∠A = ∠B = ∠C = 60°

The sides of the ∆ ABC are produced in order to D,E and F forming exterior angles
∠ACP, ∠BAE and ∠CBF
∆ACD + ∠ACB = 180°
=> ∠ACD + 60° = 180°
=> ∠ACD = 180° – 60°
=> ∠ACD = 120°
Similarly, ∠BAE + ∠BAC = 180°
=> ∠BAE + 60° = 180°
=> ∠BAE = 180° – 60° = 120°
and ∠BAF + ∠ABC = 180°
=> ∠BAF + 60° = 180°
=> ∠BAF = 180° – 60° = 120°
Hence each exterior angle of an equilateral triangle is 120°.

Question 8.
Solution:
Given : In the figure,
O is mid-point of AB and CD.
i.e. OA = OB and OC = OD
To Prove : AC = BD and AC || BD.
Proof : In ∆ OAC and ∆ OBD,
OA = OB {given}
OC = OD {given}
∠AOC = ∠BOD
(Vertically opposite angles)
∴∆ OAC ≅ ∆ OBD (S.A.S. axiom)
∴AC = BD (c.p.c.t.)
and ∠ C = ∠ D
But these are alt. angles
∴AC || BD
Hence proved.

Question 9.
Solution:
Given : In the figure,
PA ⊥ AB, QB ⊥ AB and
PA = QB, PQ intersects AB at O.

To Prove : O is mid-point of AB and PQ.
Proof : In ∆ AOP and ∆ BOQ,
∠ A = ∠ B
AP = BQ (given)
∠ AOP = ∠BOQ
(Vertically opposite angles)
∴ ∆AOP ≅ ∆ BOQ (A.A.S. axiom)
∴OA = OB (c.p.c.t)
and OP = OQ (c.p.c.t)
Hence, O is the mid-point of AB as well as PQ

Question 10.
Solution:
Given : Two line segments AB and CD intersect each other at O and OA = OB, OC = OD
AC and BD are joined.
To Prove : AC = BD
Proof : In ∆ AOC and ∆ BOD,
OA = OB {given}
OC = OD
∠AOC = ∠BOD
(vertically opposite angles)
∴ ∆ AOC ≅ ∆ BOD (S.A.S. axiom)
∴ AC = BD (c.p.c.t)
and ∠A = ∠D (c.p.c.t.)
Hence AC ≠ BD
Hence proved.

RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle – Overview

In the Chapter 9 of RS Aggarwal Solutions Class 9 Maths, you will deal with the study of congruence of triangles. You will also know about various types of triangles as well as the properties of triangles and the different criteria for proving that two triangles are congruent to each other.

• The Chapter begins with a proper explanation of triangles and its types such as Equilateral triangles, Right-angled Triangle, Scalene Triangles and Isosceles Triangles. 
• Then you will learn the properties of triangles and the criteria on which two triangles can be proved congruent.
•  You will learn the various criteria such as Angle Side Angle property, Side Side Side property, Side Angle Side Property and Right-angle Hypotenuse Side property. 
• There are 2 exercises in RS Aggarwal Solutions Class 9 Chapter 9. 
• The first exercise has 27 questions and the second exercise has 16 questions. 

This is the complete blog on RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle. To know more about the CBSE Class 9 Maths exam, ask in the comments.

FAQs on RS Aggarwal Solutions Class 9 Maths Chapter 9 Congruence of Triangles and Inequalities in a Triangle