RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle

In this chapter, we provide RS Aggarwal Solutions for Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle Maths pdf, free RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle Maths book pdf download. Now you will get step by step solution to each question.

RS Aggarwal Solutions Class 9 Chapter 5 Congruence of Triangles and Inequalities in a Triangle

RS Aggarwal Solutions Class 9 Chapter 5 Exercise 5A

Question 1:

Question 2:

Consider the isosceles triangle ∆ABC.
Since the vertical angle of ABC is 100° , we have, ∠A = 100°.
By angle sum property of a triangle, we have,

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Question 3:

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Question 5:

In a right angled isosceles triangle, the vertex angle is ∠A = 90° and the other two base angles are equal.
Let x° be the base angle and we have, ∠B = ∠C = 90°.
By angle sum property of a triangle, we have

Question 6:
Given: ABC is an isosceles triangle in which AB=AC and BC
Is produced both ways,

Question 7:

Let be an equilateral triangle.
Since it is an equilateral triangle, all the angles are equiangular and the measure of each angle is 60°
The exterior angle of ∠A is ∠BAF
The exterior angle of ∠B is ∠ABD
The exterior angle of ∠C is ∠ACE
We can observe that the angles ∠A and ∠BAF, ∠B and ∠ABD, ∠C and ∠ACE and form linear pairs.
Therefore, we have

Similarly, we have

Also, we have

Thus, we have, ∠BAF = 120°, ∠ABD = 120°, ∠ACE = 120°
So, the measure of each exterior angle of an equilateral triangle is 120°.

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Question 33:
Let AB be the breadth of a river. Now take a point M on that bank of the river where point B is situated. Through M draw a perpendicular and take point N on it such that point, A, O and N lie on a straight line where point O is the mid point of BM.

Question 34

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Question 44:
Given : ABC is a triangle and O is appoint inside it.

To Prove : (i) AB+AC > OB +OC
(ii) AB+BC+CA > OA+OB+OC
(iii) OA+OB+OC > [latex]frac { 1 }{ 2 }  [/latex] (AB+BC+CA)
Proof:
(i) In ∆ABC,
AB+AC > BC ….(i)
And in , ∆OBC,
OB+OC > BC ….(ii)
Subtracting (i) from (i) we get
(AB+AC) – (OB+OC) > (BC-BC)
i.e. AB+AC>OB+OC
(ii) AB+AC > OB+OC [proved in (i)]
Similarly, AB+BC > OA+OC
And AC+BC > OA +OB
Adding both sides of these three inequalities, we get
(AB+AC) + (AC+BC) + (AB+BC) > OB+OC+OA+OB+OA+OC
i.e. 2(AB+BC+AC) > 2(OA+OB+OC)
Therefore, we have
AB+BC+AC > OA+OB+OC
(iii) In ∆OAB
OA+OB > AB ….(i)
In ∆OBC,
OB+OC > BC ….(ii)
And, in ∆OCA,
OC+OA > CA
Adding (i), (ii) and (iii) we get
(OA+OB) + (OB+OC) + (OC+OA) > AB+BC+CA
i.e 2(OA+OB+OC) > AB+BC+CA
⇒ OA+OB+OC > [latex]frac { 1 }{ 2 }  [/latex] (AB+BC+CA)

Question 45:
Since AB=3cm and BC=3.5 cm
∴ AB+BC=(3+3.5) cm =6.5 m
And CA=6.5 cm
So AB+BC=CA
A triangle can be drawn only when the sum of two sides is greater than the third side.
So, with the given lengths a triangle cannot be drawn.

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