NCERT Solutions Class 7 Maths Congruence of Triangles Chapter 7 Exercise 7.9

This imaginary Exercise 7.9 would be the last stage of geometric mastery in NCERT Solutions Class 7 Maths Chapter 7: Congruence of Triangles, which was created to expose students to Higher Order Thinking Skills (HOTS) problems and concepts commonly found in mathematics Olympiads.

The exercise involves a complete synthesis of all congruence rules (SSS,SAS,ASA, and RHS), angle theorems (e.g., the Exterior Angle Property), and algebraic substitution within complex geometric figures. The objective here is to develop the intuitive and deductive reasoning to solve non-routine, multi-layered proofs.

1.0Download Class 7 Maths Chapter 7 Ex 7.9 NCERT Solutions PDF

Take your skills to the next level! You can download the full NCERT Solutions for Class 7 Maths for this final HOTS review set, which includes fully detailed advanced proofs, and advanced methods for solving complex problems.

2.0Key Concepts of Exercise 7.9 (HOTS Focus) Class 7 Chapter 7 - Congruence of Triangles

• Indirect proof/proof by contradiction: Proving inequalities involving relationships between sides and angles can be obtained by reasoning based on the fact that the angle opposite the longer side is larger, and vice-versa.
• Algebraic substitution in proofs: Establishing expressions for sides or angles in which you may have proved a relationship using CPCT correspondence, and using value substitutions to prove a related value.
• Auxiliary construction: The ability to add an essential line (altitude, median, or line parallel to a side) to a figure in order to set up congruent triangles, which is a proof style.
• Low Threshold: Problems that impart little information to the student that require the student to prove congruence and derive more complicated properties.
• Use of Exterior Angle Theorem: Problems that use the theorem that an exterior angle of a triangle is equal to the sum of the two opposite interior angles.
• Proof of Collinearity or Concurrency: Complex proofs that demonstrate that three points lie on the same line or three lines intersect at a single point.

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

4.0Detailed CBSE Class 7 Chapter 7 Exercise 7.9 Solutions

1. Construct a △ABC with BC=5 cm,AB=6 cm,CA=5 cm. Construct an altitude from A to BC .

Sol. The steps of construction are given as

(i) Draw a line segment AB of length 6 cm using a ruler and a pencil.

(ii) Place the compass pointer at point A and open it to 5 cm . Draw a circle.

(iii) Place the compass pointer at point B and open it to 5 cm . Draw another circle.

(iv) Let point C be one of the points of intersection of these two circles.

Now, join AC and BC . Then, △ABC is the required triangle.

(v) Construction of an altitude

Keep the ruler aligned to the base of △ABC. Place the set square on the ruler such that one of the edges of the right angle touches the ruler and slide it along ruler till the vertical edge touches C. Then, draw an altitude to AB through C.

2. Construct a triangle TRY with RY=4 cm,TR=7 cm∠R=140∘.

Construct an altitude from T to RY.

Sol. The steps of construction are given as

(i) Draw a line segment TR of length 7 cm using a ruler and a pencil. 

(ii) At point R , draw an angle 140∘ using protector and mark X .

(iii) At point R , mark a point Y along RX such that RY=4 cm

(iv) Join the point Y to T . Then, ΔTRY is the required triangle.

(v) Construction of an altitude

Keep the ruler aligned to the base RY. Place the set square on the ruler such that one of the edges of the right angle touches the ruler and slide it along ruler till the vertical edge touches T. Then, draw the altitude TM on extended RY.

Hence, an altitude from T to RY is outside of a △TRY.

3. Through construction, explore if it is possible to construct an equilateral triangle that is

(i) right angled

(ii) obtuse angled.

Also, construct an isosceles triangle that is

(i) right angled

(ii) obtuse angled

Sol. (i) Consider the length of sides of an equilateral triangle to be 3 cm .

The steps of the construction are given as

1.Draw a line segment AB of length 3 cm using pencil and ruler.

2.At point A , draw an angle of 90∘ using protector and mark X .

3.At point A , mark a point C along AX such that AC=3 cm.

4.Join BC.

Here, we observe that the length of BC is no longer 3 cm .

Hence, it is not possible to construct an equilateral triangle that is right angled but can construct an isosceles triangle.

(ii) Consider the length of sides of an equilateral triangle to be 3 cm and an obtuse angle to be 110∘.

The steps of the construction are given as

1.Draw a line segment AB of length 3 cm using ruler and pencil.

A−3 cmB

2.At point A , draw an angle of 110∘ using protector and mark X .

3.At point A mark a point C along AX such that AC=3 cm.

4.Join BC.

Here, we observe that the length of BC is no longer 3 cm .

Hence, it is not possible to construct an equilateral triangle that is obtuse angled but can construct an isosceles triangle.

5.0Key Features and Benefits: Class 7 Maths Chapter 7 Exercise 7.9

• HOTS Skill Development: Introduces the very important skill of Auxiliary Construction - a significant skill to have when writing high-level proofs in geometry. 
• Competitive Edge: Problems will have the same style and difficulty level of the types of questions asked on all levels of competitions/Olympiads. 
• Comprehensive Review: This will take full conceptual understanding of all the congruence criteria and fundamental triangle theorems.