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NCERT Solutions
Class 7
Maths
Chapter 7 A Tale of Three Intersecting Lines
Exercise 7.7

NCERT Solutions Class 7 Maths Chapter 7: Congruence of Triangles Exercise 7.7

With the conclusion of NCERT Solutions Class 7 Maths Chapter 7: Congruence of Triangles, we have an advanced review, in a potential Exercise 7.7, which puts students to work solving an advanced, multi-step problem that requires them to use SSS, SAS, ASA, AAS, and RHS to prove triangles are congruent (or use congruence to prove a derived property). Experimentation with foreign geometric figures (not just triangles) such as parallelograms or figures with intersecting lines would occur in these problems. 

The intention for this class is not to just prove triangles congruent, but to use triangles to prove a derived property—the specific one being to prove that the diagonals bisect each other or that a specific line segment is the perpendicular bisector of another segmented line. These final components of triangulation will be essential to prepare students for high school geometry.

1.0Download Class 7 Maths Chapter 7 Ex 7.7 NCERT Solutions PDF

Develop your problem-solving skills! Download the complete NCERT Solutions for Class 7 Maths Exercise 7.7 application in a precise, step-by-step process using the final application set. This resource will provide proofs and logical rationales to your complicated geometric problems.

NCERT Solutions Class 7 Maths Chapter 7 Ex 7.7

2.0Key Concepts of Exercise 7.7 (Advanced Application) Class 7 Chapter 7

  • Sequential Congruence: Using the CPCT result resulting from the first proof from a congruence proof (△1≅△2) as a needed "Given" condition when proving a second pair of triangles congruent (△3≅△4). 
  • Proving Properties Of Figures: Using triangle congruence to show properties/relationships of larger figures. 

Show that sides of a quadrilateral are parallel.

Show that a point is a midpoint or that an intersection is 90 degrees.  

  • Logical Deductive Reasoning: Moving beyond simply applying rules to build a complete, logical argument for a theorem.
  • Proofs with Two Parts: Problems that need two subsequent congruency proofs to make the final conclusion.
  • Proofs with Quadrilaterals: Problems where you have to prove that two triangles made with a diagonal are congruent to show a quadrilateral property.
  • Complex Use of CPCT: Problems where the objective is the equality of a specific angle or side based on a correspondence that isn't as obvious.

3.0NCERT Solutions Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines : Other Exercises

NCERT Solutions for Class 7 Chapter 7: Exercise 7.1

NCERT Solutions for Class 7 Chapter 7: Exercise 7.2

NCERT Solutions for Class 7 Chapter 7: Exercise 7.3

NCERT Solutions for Class 7 Chapter 7: Exercise 7.4

NCERT Solutions for Class 7 Chapter 7: Exercise 7.5

NCERT Solutions for Class 7 Chapter 7: Exercise 7.6

NCERT Solutions for Class 7 Chapter 7: Exercise 7.7

NCERT Solutions for Class 7 Chapter 7: Exercise 7.8

NCERT Solutions for Class 7 Chapter 7: Exercise 7.9

4.0Detailed CBSE Class 7 Chapter 7 Exercise 7.7 Solutions

1. For each of the following angles, find another angle for which a triangle is (a) possible, (b) not possible. Find atleast two different angles for each category.

(i) 30∘

(ii) 70∘

(iii) 54∘

(iv) 144∘

Sol. We know that when the sum of two angles is less than 180∘ then the triangle exists and when the sum of two angles is greater than or equal to 180∘ then there exists no triangle.

(i) (a) Two possible angles are 70∘ and 60∘

(b) Two not possible angles are 150∘ and 155∘

(ii) (a) Two possible angles are 105∘ and 50∘

(b) Two not possible angles are 110∘ and 120∘

(iii) (a) Two possible angles are 120∘ and 110∘

(b) Two not possible angles are 126∘ and 130∘

(iv) (a) Two possible angles are 20∘ and 30∘

(b) Two not possible angles are 36∘ and 40∘

2. Determine which of the following pairs can be the angles of a triangle and which cannot.

(i) 35∘,150∘

(ii) 70∘,30∘

(iii) 90∘,85∘

(iv) 50∘,150∘

Sol. We know that if the sum of two angles of a triangle is greater than or equal to 180∘ then triangle is not possible with these angles.

(i) Given, the two angles of a triangle are 35∘ and 150∘.

Here, 35∘+150∘=185∘>180∘.

Therefore, triangle is not possible.

(ii) Given, the two angles of a triangle are 70∘ and 30∘.

Here, 70∘+30∘=100∘<180∘.

Therefore, triangle is possible.

(iii) Given, the two angles of a triangle are 90∘ and 85∘.

Here, 90∘+85∘=175∘<180∘.

Therefore, triangle is possible.

(iv) Given, the two angles of a triangle are 50∘ and 150∘.

Here, 50∘+150∘=200∘>180∘.

Therefore, triangle is not possible.

5.0Key Features and Benefits: Class 7 Maths Chapter 7 Exercise 7.7

  • Knowledge Synthesis: Needs students to identify the right congruence rule from all four criteria learned in the chapter. 
  • Advanced Proof-Writing: Concentrates on the aspect of formal, multi-step geometric proofs which is an important skill needed for success in higher classes. 
  • Real-World Geometry: Illustrates that the simple idea of triangle congruence serves as the basis for proving properties of complicated shapes.

NCERT Solutions Class 7 Maths Chapter 7 : Other Exercises

Exercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

Exercise 7.5

Exercise 7.6

Exercise 7.7

Exercise 7.8

Exercise 7.9

NCERT Solutions for Class 7 Maths: All Chapters

Chapter 1 : Large Numbers Around Us

Chapter 2 : Arithmetic Expressions

Chapter 3 : A Peek Beyond The Point

Chapter 4 : Expressions Using Letter-Numbers

Chapter 5 : Parallel and Intersecting Lines

Chapter 6 : Number Play

Chapter 7 : A Tale of Three Intersecting Lines

Chapter 8 : Working with Fractions

Frequently Asked Questions

Vertically Opposite Angles (VOA) are frequently used as the Angle component in SAS or ASA proofs, especially when two triangles overlap or are formed by intersecting lines. They are automatically equal.

In Ex 7.7, the questions usually ask you to prove a final property (AC=BD). You must use CPCT as the final, absolute justification once the triangles themselves have been proven congruent.

A Common Side is a side that is shared by both triangles in question. It is often the easiest part to establish as equal (Side=Side) and is vital for SSS or SAS proofs involving overlapping figures.

Always look for RHS when you see an altitude (90∘ mark), or any perpendicular line segment. The right angle is the R, and you must then check if the hypotenuse (H) and one side (S) are equal.

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