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NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.4

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Last updated date: 14th Jul 2024
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NCERT Solutions for Class 7 Maths The Triangle and its Properties Exercise 6.4 - FREE PDF Download

NCERT Maths Class 7 Exercise 6.4 Chapter 6, "The Triangle and its Properties." explores various properties and characteristics of triangles, focusing specifically on the concept of the exterior angle and its relationship with the interior opposite angles. You will also focus on the Angle Sum Property of a triangle, which states that the sum of the angles in a triangle is always 180 degrees in ex 6.4 class 7. These foundational concepts are crucial for understanding the geometry of triangles and solving related problems. Keep practising NCERT Class 7 Maths Exercise 6.4 Solutions to do well on exams. Vedantu’s Class 7 Maths NCERT Solutions provide step-by-step solution to help you excel in your exams.

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Table of Content
1. NCERT Solutions for Class 7 Maths The Triangle and its Properties Exercise 6.4 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 6 Exercise 6.4 Class 7 | Vedantu
3. Access NCERT Solutions for Maths Class 7 Chapter 6 - The Triangle and its Properties
4. Class 7 Maths Chapter 6: Exercises Breakdown
5. CBSE Class 7 Maths Chapter 6 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 6 Exercise 6.4 Class 7 | Vedantu

  • Exercise 6.4 focuses on the concepts of the median and altitude of a triangle. Learn how to identify and draw medians and altitudes, which are crucial for understanding the properties and characteristics of triangles.

  • Exercise 6.4 also focuses on the angle sum property of triangles, which states that the sum of the interior angles of a triangle is always 180 degrees.

  • Learn about the exterior angle property, where the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

  • In ex 6.4 class 7 for Unknown angle problems, apply the angle sum and exterior angle properties to find unknown angles in various triangles.

  • There are 6 questions in Class 7 Math 6.4 Chapter 6 which experts at Vedantu fully solved.

Access NCERT Solutions for Maths Class 7 Chapter 6 - The Triangle and its Properties

Exercise 6.4

1. Is it possible to have a triangle with the following sides?

i) Is it possible to have a triangle with the following sides 2 cm, 3 cm, 5 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and the third side is also 5 cm, it is not possible to have a triangle with the sides 2 cm, 3 cm and 5 cm.

ii) Is it possible to have a triangle with the following sides 3 cm, 6 cm, 7 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. First, take the two sides as 3 cm and 6 cm. Since $3 + 6 = 9$ and $9 > 7$, the property of the triangle is satisfied. Now take the two sides as 6 cm and 7 cm. Since $6 + 7 = 13$ and $13 > 3$, the property of the triangle is satisfied.  Now, take the two sides as 7 cm and 3 cm. Since $7 + 3 = 10$ and $10 > 6$, the property of the triangle is satisfied. Hence, it is possible to have a triangle with the sides 3 cm, 6 cm and 7 cm.

iii) Is it possible to have a triangle with the following sides 6 cm, 3 cm, 2 cm?

Ans: A triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. Take the two sides as 2 cm and 3 cm. Since $2 + 3 = 5$ and 5 is not greater than 6, it is not possible to have a triangle with the sides 6 cm, 3 cm and 2 cm.

2. Take any point $O$ in the interior of a triangle $PQR$. Is the following inequality satisfied?

Triangle PQR with point O in interior

i) Is $OP + OQ > PQ$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.

Triangle PQR , joining OR,OQ and OP


From the diagram, it can be seen that $OPQ$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OP + OQ > PQ$. Yes, $OP + OQ > PQ$.


ii) Is $OQ + OR > QR$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.


Triangle PQR with joining OR,OQ and OP


From the diagram, it can be seen that $OQR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OQ + OR > QR$. Yes, $OQ + OR > QR$.

iii) Is $OR + OP > RP$?

Ans: In the given triangle, join $OR$, $OQ$ and $OP$.


Triangle PQR with joining OR,OQ and OP


From the diagram, it can be seen that $OPR$ is a triangle and a triangle can be possible only when the sum of the lengths of any two sides would be greater than the length of the third side. So, $OR + OP > RP$. Yes, $OR + OP > RP$.

3. AM is the median of a triangle $ABC$. Is $AB + BC + CA > 2AM$?


Triangle ABC, with AM median


Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\Delta ABM$, $AB + BM > AM$ and in $\Delta AMC$, $AC + MC > AM$.

Add both the inequalities and simplify.

$AB + BM + AC + MC > AM + AM$

$ \Rightarrow AB + AC + \left( {BM + MC} \right) > 2AM$

Substitute $BC$ for $BM + MC$.

$ \Rightarrow AB + AC + BC > 2AM$

Hence, $AB + BC + CA > 2AM$ is true.

4. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA > AC + BD$?


Quadrilateral ABCD


Ans: The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle ABC$, $AB + BC > AC$.

In $\vartriangle ADC$, $AD + DC > AC$.

In $\vartriangle DCB$, $DC + CB > DB$.

In $\vartriangle ADB$, $AD + AB > DB$.

Add all the four inequalities and simplify.

$AB + BC + AD + DC + DC + CB + AD + AB > AC + AC + DB + DB $

$\Rightarrow \left( {AB + AB} \right) + \left( {BC + BC} \right) + \left( {AD + AD} \right) + \left( {DC + DC} \right) > 2AC + 2DB $

$\Rightarrow 2AB + 2BC + 2AD + 2DC > 2\left( {AC + DB} \right) $

$\Rightarrow 2\left( {AB + BC + AD + DC} \right) > 2\left( {AC + DB} \right) $

Divide both sides by 2 and simplify.

$\Rightarrow \dfrac{2}{2}\left( {AB + BC + AD + DC} \right) > \dfrac{2}{2}\left( {AC + DB} \right) $

$\Rightarrow AB + BC + AD + DC > AC + DB$

Hence, $AB + BC + CD + DA > AC + BD$ is true.

5. $ABCD$ is a quadrilateral. Is $AB + BC + CD + DA < 2\left( {AC + BD} \right)$?

Ans: Draw a quadrilateral $ABCD$. Join $AC$ and $BD$.

Quadrilateral ABCD, diagonals AC and BD


The sum of two sides of a triangle is always greater than the third side of the triangle.

In $\vartriangle AOB$, $AB < OA + OB$.

In $\vartriangle BOC$, $BC < OB + OC$.

In $\vartriangle COD$, $DC < OC + OD$.

In $\vartriangle AOD$, $DA < OD + OA$.

Add all the four inequalities and simplify.

\[ AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA \]

\[ \Rightarrow AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD \]

\[ \Rightarrow AB + BC + CD + DA < 2\left[ {\left( {AO + OC} \right) + \left( {DO + OB} \right)} \right] \]

Substitute $AC$ for $AO + OC$ and $BD$ for $DO + OB$.

\[ \Rightarrow AB + BC + CD + DA < 2\left[ {AC + BD} \right]\]

Hence, $AB + BC + CD + DA < 2\left( {AC + BD} \right)$ is true.

6. The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Ans: It is given that two sides of a triangle are 12 cm and 15 cm. The sum of two sides of a triangle is always greater than the third side of the triangle.

Therefore, the third side of the triangle should be less than $12 + 15 = 27$ cm.

Also, the third side cannot be less than the difference of two sides of a triangle. Therefore, The third side should be more than $15 - 12 = 3$ cm. 

Therefore, the length of the third side should be between 3 cm and 27 cm.


Conclusion

In Class 7 Maths Chapter 6.4, we have reinforced our understanding of key geometric principles. By exploring the Exterior Angle Property and the Angle Sum Property, we learned how to determine unknown angles in triangles. These properties, which are covered in Class 7 Maths Chapter 6 Exercise 6.4 Solutions are fundamental in geometry, providing the foundation for more advanced concepts. Mastering these principles allows you to solve various geometric problems with confidence. 


Class 7 Maths Chapter 6: Exercises Breakdown

Exercise

Number of Questions

Exercise 6.1

3 Questions & Answers

Exercise 6.2

2 Questions & Answers

Exercise 6.3

2 Questions & Answers

Exercise 6.5

8 Questions & Answers


CBSE Class 7 Maths Chapter 6 Other Study Materials


Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.


FAQs on NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.4

1. How many questions are there in Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4?

Class 7 Maths, Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4 consists of five questions. Exercise 6.4, "The Triangle and its Properties," contains various sections where students can learn about inequality based on the triangle, and triangle properties. it deals with questions based on the traits of triangle sides. Chapter 6 The Triangle and Its Properties from the NCERT Solutions for Class 7 Math (EX 6.4) Vedantu is mentioned in Exercise 6.4.

2. What do you mean by the median of a triangle in Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4

A line segment joining a vertex to the middle of the other side of a triangle, cutting that side in half, is called the median of a triangle. Each vertex of a triangle has the same number of medians, which all cross at the triangle's centroid, which is the location where a triangle's three medians cross.

3. Is the Vedantu website providing answers fo rNCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4?

Yes, you can download all of the Class 7 Math NCERT Solutions PDFs. These solutions are created using a special method by the knowledgeable faculty of Vedantu. Additionally, they offer free PDF solutions for the NCERT textbooks for classes 1 through 12. For those who want to perform well on exams, the NCERT Textbook is advised. Vedantu's in-house subject matter specialists have carefully, and by all CBSE requirements, solved the problems and questions from the exercise.

4. Mention the key ideas covered in NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4?

The key ideas covered in Chapter 6 of the Class 7 Math NCERT Solutions Triangles and Their Properties are:

  1. Medians of a Triangle

  2. Altitudes of a Triangle

  3. Exterior Angle Property of a Triangle

  4. Angle Sum Property of a Triangle

  5. Two Special Triangles: Equilateral and Isosceles

  6. Sum of the Lengths of a Triangle's Two Sides

  7. Right-Angled Triangles and Pythagoras Property

5. Do I need to solve NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties (EX 6.4) Exercise 6.4?

Studying chapters alone is insufficient for a subject like mathematics; you must first complete the related assignments independently. You will improve as a practitioner and as a result of your practice. Before you start solving difficulties, you won't be able to pinpoint your areas for growth. For a better understanding and to discover several methods for tackling any issue, check Vedantu's NCERT Solutions for Class 7 Math Chapter 6. The Triangle and its Properties (EX 6.4) Exercise 6.4