NCERT Solutions Class 7 Maths Chapter 7: Congruence of Triangles Exercise 7.8

Chapter 7 of Class 7 Maths NCERT Solutions: Congruence of Triangles would finish with this last and superior review in a hypothetical Exercise 7.8. This exercise serves as the last proof testing the overall integration of all congruence rules (SSS,SAS, ASA and RHS), angle properties (such as the Angle Sum Property) and the CPCT rule to prove theorems involving special triangles and quadrilaterals.

The problems contained here essentially reverse the structure of more formal proofs in geometry. They require rigorous, linear logic between steps to establish relationships, such as proving that an exterior angle is greater than either of the interior opposite angles, or demonstrating the property of the perpendicular bisector using a two-step proof.

1.0Download Class 7 Maths Chapter 7 Ex 7.8 NCERT Solutions PDF

Reach mastery in solid geometry! With this last, high-order application task so close to being complete, download the all-inclusive, step-by-step NCERT Solutions for Class 7 Maths , which provide comprehensive, justified proofs to validate your learning!

2.0Key Concepts of Exercise 7.8 (Chapter Synthesis) Class 7 Chapter 7

• Contradiction/Inequality Proof: Utilize the theorem of Exterior Angle Property or the Triangle Inequality where one of the sides/angles is known to be unequal (greater than/less than) etc., rather than simple congruency. 
• Combined Properties: Use the properties from previous chapters (i.e., Linear Pairs, Vertically Opposite Angles, Angle Sum Property, etc.) in a congruence proof.
• Proving Geometric Properties: Solve problems that are essentially minor geometric theorems (i.e., proving that the angle opposite the longer side is larger), or proving that the sum of any two sides of a triangle is greater than the third side (proper proof of the Triangle Inequality), etc.
• Maximal Application: Questions that involve multiple steps and one or more congruence relations to provide a solution to a single question.
• Inequality Proving: Problems focused on proving relationships such as ∠B >∠C given certain side requirements.
• Theorem Justification: Showing the basic logic behind a main theory of geometry stated in the chapter.

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

4.0Detailed CBSE Class 7 Chapter 7 Exercise 7.8 Solutions

1. Find the third angle of a triangle (using a parallel line) when two of the angles are

(i) 36∘,72∘

(ii) 150∘,15∘

(iii) 90∘,30∘

(iv) 75∘,45∘

Sol. (i) Given, the two angles of a triangle are 36∘ and 72∘.

Let the third angle be a.

Let △ABC be any triangle with angles

∠B=36∘∠C=72∘ and ∠A=a.

Now, draw a line XY parallel to BC , passing through A .

Then, ∠XAB=36∘ and angle YAC=72∘[∵ alternate interior angles ]

We know that ∠XAB,∠YAC and ∠BAC together form a straight line.

So, ∠XAB+∠YAC+∠BAC=180∘

⇒36∘+72∘+a=180∘

⇒108∘+a=180∘

⇒a=72∘

Hence, the third angle of a triangle is 72∘.

(ii) Given, the two angles of a triangle are 150∘ and 15∘.

Let the third angle be a.

Let △ABC be any triangle with angles

∠B=150∘∠C=15∘ and ∠A=a.

Now, draw a line XY parallel to BC , passing through A .

Then, ∠XAB=150∘ and angle YAC=15∘[∵ alternate angles ]

We know that ∠XAB,∠YAC and ∠BAC together form a straight line.

So, ∠XAB+∠YAC+∠BAC=180∘

⇒150∘+15∘+a=180∘

⇒165∘+a=180∘

⇒a=15∘

Hence, the third angle of a triangle is 15∘.

(iii) Given, the two angles of a triangle are 90∘ and 30∘.

Let the third angle be a.

Let △ABC be any triangle with angles

∠B=90∘∠C=30∘ and ∠A=a.

Now, draw a line XY parallel to BC , passing through A .

Then, ∠XAB=90∘ and angle YAC=30∘[∵ alternate angles ]

We know that ∠XAB,∠YAC and ∠BAC together form a straight line.

So, ∠XAB+∠YAC+∠BAC=180∘

⇒90∘+30∘+a=180∘

⇒120∘+a=180∘

⇒x=60∘

Hence, the third angle of a triangle is 60∘.

(iv) Given, the two angles of a triangle are 75∘ and 45∘.

Let the third angle be a.

Let △ABC be any triangle with angles

∠B=75∘∠C=45∘ and ∠A=a.

Now, draw a line XY parallel to BC , passing through A .

Then, ∠XAB=75∘ and angle YAC=45∘[∵ alternate angles ]

We know that ∠XAB,∠YAC and ∠BAC together form a straight line.

So, ∠XAB+∠YAC+∠BAC=180∘

⇒75∘+45∘+a=180∘

⇒120∘+a=180∘

⇒a=60∘

Hence, the third angle of a triangle is 60∘.

2. Can you construct a triangle all of whose angles are equal to 70∘ ? If two of the angles are 70∘, what would the third angle be? If all the angles in a triangle have to be equal, then what must its measure be? Explore and find out.

Sol. We know that the sum of the angles of a triangle is 180∘.

Here, 70∘+70∘+70∘=210∘>180∘.

So, we cannot construct a triangle whose all of angles are equal to 70∘.

If two angles are 70∘ each, then the sum of these two angles is

70+70∘=140∘

Therefore, the third angle =180∘−140∘=40∘.

Let all the angles of a triangle to be x .

Then, by angle sum property,

x+x+x=180∘

⇒3x=180∘

⇒x=60∘

Hence, if all the angles of a triangle are equal, then each of them measures to 60∘.

3. Here is a triangle in which we know ∠B=∠C and ∠A=50∘. Can you find ∠B=∠C ?

Sol. We know that the sum of all angles of a triangle is 180∘.

∴⇒⇒⇒⇒⇒​∠A+∠B+∠C=180∘50∘+∠B+∠C=180∘∠B+∠C=180∘−50∘=130∘∠B+∠B=130∘2∠ B=130∘∠B=2130​=65∘​

Therefore, the values of ∠B and ∠C are 65∘.

5.0Key Features and Benefits: Class 7 Maths Chapter 7 Exercise 7.8

• Involves proof situations that require students to combine different ideas and steps, while using geometric concepts, in a single, extended argument. 
• Provides the basis necessary for writing formal geometric proofs (like the Perpendicular Bisector Theorem) interestingly with an emphasis on several key observations. 
• Focuses on breaking multi-step, or even multi-stage, complex problems, down to simpler problems, focusing on two congruent triangles at a time.