The Triangle And Its Properties Class 7 Questions and Answers - 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.
Table of Content
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.
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?
i) Is $OP + OQ > PQ$?
Ans: In the given triangle, join $OR$, $OQ$ 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 $OP$.
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 $OP$.
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 $AB + BC + CA > 2AM$?
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 $AB + BC + CD + DA > AC + BD$?
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 $BD$.
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
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.
Important Related Links for NCERT Class 7 Maths
Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.
FAQs on NCERT Solutions For Class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.4 - 2025-26
1. What are the major concepts for which solutions are provided in NCERT Class 7 Maths Chapter 6, 'The Triangle and its Properties'?
The NCERT solutions for this chapter provide step-by-step methods for problems related to the following key concepts as per the 2025-26 syllabus:
- Medians and Altitudes of a Triangle
- The Exterior Angle Property
- The Angle Sum Property of a Triangle
- Properties of Special Triangles (Isosceles and Equilateral)
- The Triangle Inequality Property (Sum of the lengths of two sides)
- The Pythagoras Property and its applications in Right-Angled Triangles
2. What is the correct method to solve problems using the Pythagoras property in the Chapter 6 exercises?
The correct method demonstrated in the NCERT solutions involves three main steps:
- First, accurately identify the right-angled triangle in the problem and label its sides: the hypotenuse (the side opposite the 90° angle), the base, and the perpendicular.
- Clearly state the property being used: By the Pythagoras Property, (hypotenuse)² = (base)² + (perpendicular)².
- Finally, substitute the given side lengths into the formula and solve the resulting equation to find the value of the unknown side.
3. Why is it crucial to check the Triangle Inequality Theorem when solving problems with given side lengths?
It is crucial because the Triangle Inequality Theorem is the fundamental condition that determines if a triangle can exist. The theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. The NCERT solutions apply this property to verify if a set of given lengths can form a valid triangle, which is a common type of question in the exercises. Without this check, you might attempt to solve for a geometrically impossible figure.
4. What is a common mistake students make when applying the exterior angle property in the Chapter 6 exercises?
A very common mistake is confusing the interior adjacent angle with the interior opposite angles. The correct application of the property states that an exterior angle of a triangle is equal to the sum of its two interior opposite angles. Many students incorrectly equate it with the adjacent interior angle. The step-by-step solutions for Chapter 6 emphasize correctly identifying the two opposite angles first to prevent this error.
5. How do the NCERT Solutions for Chapter 6 help differentiate between a 'median' and an 'altitude' in practical problems?
The NCERT solutions clarify this critical difference by focusing on their unique definitions within the problem's context:
- A median is a line segment that connects a vertex to the midpoint of the opposite side. The solutions highlight its function of dividing a side into two equal lengths.
- An altitude is a line segment from a vertex that is perpendicular to the opposite side, meaning it forms a 90° angle. The solutions emphasize the right angle as its defining feature.
6. Where can I find reliable, step-by-step NCERT Solutions for Class 7 Maths Chapter 6 for the 2025-26 academic session?
For the 2025-26 session, you can refer to Vedantu for detailed and accurate NCERT Solutions for Class 7 Maths Chapter 6. The solutions are created by subject matter experts to align perfectly with the latest CBSE guidelines, providing a clear, step-by-step explanation for every problem in the textbook exercises.
7. In an exam, why is it important to write down the name of the property (e.g., Angle Sum Property) before solving a question from this chapter?
Writing down the property (e.g., 'Using Angle Sum Property' or 'By Pythagoras Theorem') is a crucial step in your answer. According to the CBSE evaluation pattern, this shows the examiner your conceptual understanding and the logical basis for your calculation. It makes your solution clear and structured, which can help you earn full marks for the method, even if a minor calculation error occurs. It proves you know why your steps are correct.
