NCERT Solutions Class 7 Maths Chapter 7 - Congruence of Triangles Exercise 7.4

The NCERT Solutions Class 7 Maths Chapter 7: Congruence of Triangles concludes the congruence rules in Exercise 7.4 by introducing the critical criterion for congruence in right-angled triangles, the RHS (Right angle-Hypotenuse-Side) Congruence Criterion. 

This exercise is entirely focused on pairs of right-angled triangles, asking students to demonstrate that the corresponding Hypotenuses and corresponding Sides are equal. The RHS rule is an essential rule to master because it allows for a more direct proof showing congruence in 90-degree triangles than SSS or SAS do. The exercise also has real-world problems that utilise congruence to prove that a larger triangle is isosceles.

1.0Download Class 7 Maths Chapter 7 Ex 7.4 NCERT Solutions PDF

Enhance your geometric proof skills, especially about right-angled triangles! Download the complete, step-by-step NCERT Solutions for Class 7 Maths Chapter 7 Exercise 7.4. This is an excellent resource for studying for your geometry exams.

2.0Key Concepts of Exercise 7.4 Class 7 Chapter 7 Solutions

• RHS Congruence Criterion: Two right triangles are congruent if the hypotenuse of one triangle is equal to the hypotenuse of the other triangle and one side of the first triangle is equal to the corresponding side of the second triangle. (Right angle, Hypotenuse, Side). 
• Identifying Hypotenuse: The first step in applying RHS correctly is identifying the hypotenuse (which is opposite the 90-degree angle). 
• Proving Isosceles Triangles: To prove the two non-right-angle sides of a triangle are equal, use a congruence proof (typically RHS or ASA) on a pair of smaller right triangles inside a larger triangle, and then C.P.C.T. to prove the two non-right angle sides of the large triangle are equal, thus proving the larger triangle is isosceles.
• Direct RHS Application: Questions present illustrations of two right triangles with instructions to state the three necessary equal parts (RHS) to prove congruence. 
• Missing the Information: Problems ask you to name the other pair of corresponding parts required to apply the RHS criterion. 
• Theorem-Based Proofs: Lengthier questions again require you to draw an altitude (forming two right triangles) and prove that the two triangles are congruent using RHS or SSS/SAS to establish further properties, such as showing that the altitude is a bisector.

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

4.0Detailed CBSE Class 7 Chapter 7 Exercise 7.4 Solutions

1. Check if a triangle exists for each of the following set of lengths.

(i) 1,100,100

(ii) 3,6,9

(iii) 1,1,5

(iv) 5,10,12

Sol. Triangle inequality states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

(i) Here, 1+100>100 and 100+100>1

So, triangle is possible.

(ii) Here, 3+6=9

So, triangle cannot be formed.

(iii) Here, 1+1<5

So, triangle cannot be formed.

(iv) Here, 5+10>12,

10+12>5 and 12+5>10

So, triangle is possible.

2. Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any side length? Justify your answer.

Sol. Yes, an equilateral triangle is possible with side lengths 50,50 and 50 .

Since, an equilateral triangle is a triangle in which all three sides are equal.

Hence, there exist an equilateral triangle of any side length.

3. For each of the following, give atleast 5 possible values for the third length, so there exists a triangle having these as side lengths (decimal values could also be chosen).

(i) 1,100

(ii) 5,5

(iii) 3,7

Sol. By the triangle inequality, the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

Also, the difference between the lengths of any two sides of a triangle is always smaller than the length of the third side.

Let x be the length of third side of the triangle.

(i) Given, two sides of a triangle are 1 and 100.

Then, x satisfies

⇒​100−1<x<100+199<x<101​

∴ The five possible values for x are

99.5, 100, 100.2, 100.5, 100.9.

(ii) Given, two sides of a triangle are 5 and 5.

Then, x satisfies

⇒​5−5<x<5+50<x<10​

∴ The five possible values of x are 1,3,5,7 and 9 .

(iii) Given, two sides of a triangle are 3 and 7.

Then, x satisfies

⇒​7−3<x<7+34<x<10.​

∴ The five possible values for x are 4.5,5,6,6.8 and 6.9.

5.0Key Features and Benefits: Class 7 Maths Chapter 7 Exercise 7.4

• Specialized Rule: Introduces the RHS rule, simplifying proofs for right triangles when you only know two sides and 90 degrees.
• Advanced Proof Structure: Helps transition students from filling in the blanks to fully writing proofs in a formal geometry structure with justification.
• Geometric Properties: Provides a proven method of establishing the basis for properties of triangles including that isosceles triangles have equal base angles.