NCERT Solutions For Class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.3 - 2025-26
NCERT Class 7 Maths Chapter 6 Exercise 6.3 Solutions of, The Triangle and its Properties are designed to help students understand the fundamental concepts of triangles. This exercise focuses on important properties such as the angle sum property, types of triangles based on angles, and the criteria for congruence of triangles. These concepts are crucial for building a strong foundation in geometry.
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The detailed NCERT Solutions for Class 7 Maths solutions provided by Vedantu’s expert teachers simplify complex problems, making it easier for students to follow and learn. Focus on understanding each property and how to apply them to different problems. Students’ understanding and problem-solving talents in this CBSE Class 7 Maths Syllabus can be improved by practicing these solutions.
Glance on NCERT Solutions Maths Chapter 6 Exercise 6.3 Class 7 | Vedantu
Class 7 Exercise 6.3, The Angle Sum Property of a Triangle simply means that when you add up all the angles inside any triangle, the total always equals 180 degrees.
This rule applies to all triangles, whether they are big or small, and helps solve problems involving angles in geometry.
Calculating unknown angles within a triangle based on given angles.
Understanding the relationship between the angles in different types of triangles.
Applying geometric principles to prove theorems and solve problems involving triangles.
Establishing a foundational understanding of geometric concepts in mathematics.
Providing a basis for further exploration into more complex geometrical figures and properties.
This article contains exercise notes, important questions, exemplar solutions, exercises, and video links for Exercise 6.3 - The Triangle and its Properties, which you can download as PDFs.
There are 2 fully solved questions in Class 7 Maths Chapter 6 Exercise 6.3 Solutions.
Exercise 6.3
1. Find the value of \[x\] in the following diagrams.
i.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle ABC$, $\angle BAC + \angle ACB + \angle ABC = 180^\circ $.
$ \Rightarrow x + 60^\circ + 50^\circ = 180^\circ $
$ \Rightarrow x + 110^\circ = 180^\circ \\ $
Subtract $110^\circ $ from both sides and simplify.
$\Rightarrow x + 110^\circ - 110^\circ = 180^\circ - 110^\circ $
$\Rightarrow x = 70^\circ $
The value of \[x\] is $70^\circ $.
ii.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle PQR$, $\angle RPQ + \angle PQR + \angle QRP = 180^\circ $.
$ \Rightarrow 90^\circ + 30^\circ + x = 180^\circ $
$ \Rightarrow x + 120^\circ = 180^\circ $
Subtract $120^\circ $ from both sides and simplify.
\[ \Rightarrow x + 120^\circ - 120^\circ = 180^\circ - 120^\circ \]
\[ \Rightarrow x = 60^\circ \]
The value of \[x\] is \[60^\circ \].
iii.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In $\vartriangle XYZ$, $\angle ZXY + \angle XYZ + \angle YZX = 180^\circ $.
\[ \Rightarrow 30^\circ + 110^\circ + x = 180^\circ \]
\[ \Rightarrow x + 140^\circ = 180^\circ \]
Subtract \[140^\circ \] from both sides and simplify.
\[\Rightarrow x + 140^\circ - 140^\circ = 180^\circ - 140^\circ \]
\[ \Rightarrow x = 40^\circ \]
The value of \[x\] is \[40^\circ \].
iv.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ + x + x = 180^\circ \].
Hence,
\[50^\circ + 2x = 180^\circ \]
Subtract \[50^\circ \] from both sides and simplify.
\[\Rightarrow 50^\circ - 50^\circ + 2x = 180^\circ - 50^\circ \]
\[\Rightarrow 2x = 130^\circ \]
Divide both sides by 2 and simplify.
\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{130^\circ }}{2} \]
\[\Rightarrow x = 65^\circ \]
The value of \[x\] is \[65^\circ \].
v.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[x + x + x = 180^\circ \].
Hence,
\[3x = 180^\circ \]
Divide both sides by 3 and simplify.
\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3} \]
\[ \Rightarrow x = 60^\circ \]
The value of \[x\] is \[60^\circ \].
vi.
Ans: The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[90^\circ + x + 2x = 180^\circ \].
\[ \Rightarrow 90^\circ + 3x = 180^\circ \]
Subtract \[90^\circ \] from both sides and simplify.
\[\Rightarrow 90^\circ - 90^\circ + 3x = 180^\circ - 90^\circ \]
\[\Rightarrow 3x = 90^\circ \]
Divide both sides by 3 and simplify.
\[\Rightarrow \dfrac{{3x}}{3} = \dfrac{{90^\circ }}{3} \]
\[\Rightarrow x = 30^\circ \]
The value of \[x\] is \[30^\circ \].
2. Find the value of \[x\] and $y$ in the following diagrams.
i.
Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.
$x + 50^\circ = 120^\circ $
Subtract $50^\circ $ from both sides of the equation.
$ \Rightarrow x = 120^\circ - 50^\circ $
$ \Rightarrow x = 70^\circ $
The value of \[x\] is $70^\circ $.
The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ + 70^\circ + y = 180^\circ \].
\[ \Rightarrow 120^\circ + y = 180^\circ \]
Subtract $120^\circ$ from both sides and simplify.
\[ \Rightarrow 120^\circ + y - 120^\circ = 180^\circ - 120^\circ \]
\[\Rightarrow y = 60^\circ \]
The value of \[y\] is \[60^\circ \].
ii.
Ans: Since the vertical opposite angles are equal, $y = 80^\circ $.
The value of \[y\] is $80^\circ $.
The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ + 80^\circ + x = 180^\circ \].
\[ \Rightarrow 130^\circ + x = 180^\circ \]
Subtract \[130^\circ \] from both sides and simplify.
\[ \Rightarrow 130^\circ - 130^\circ + x = 180^\circ - 130^\circ \]
\[ \Rightarrow x = 50^\circ \]
The value of \[y\] is \[50^\circ \].
iii.
Ans: The exterior angle property of a triangle states that the exterior angle is equal to the sum of the opposite non-adjacent interior angles.
$50^\circ + 60^\circ = x$.
Hence,
$x = 110^\circ $
The value of \[x\] is $110^\circ $.
The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[50^\circ + 60^\circ + y = 180^\circ \].
Hence,
\[110^\circ + y = 180^\circ \]
Subtract $110^\circ $ from both sides and simplify.
\[\Rightarrow 110^\circ + y - 110^\circ = 180^\circ - 110^\circ \]
\[\Rightarrow y = 70^\circ \]
The value of \[y\] is \[70^\circ \].
iv.
Ans: Since the vertical opposite angles are equal, $x = 60^\circ $.
The value of \[x\] is $60^\circ $.
The sum of the internal angles of a triangle is $180^\circ $. In the given triangle,
\[\Rightarrow 60^\circ + 30^\circ + y = 180^\circ \]
\[ \Rightarrow 90^\circ + y = 180^\circ \].
Subtract \[{90^\circ }\] from both sides and simplify.
\[\Rightarrow 90^\circ - 90^\circ + y = 180^\circ - 90^\circ \]
\[\Rightarrow y = 90^\circ \]
The value of \[y\] is \[90^\circ \].
v.
Ans: Since the vertical opposite angles are equal, $y = 90^\circ $.
The value of \[y\] is \[90^\circ \].
The sum of the internal angles of a triangle is $180^\circ $. In the given triangle, \[x + x + 90^\circ = 180^\circ \].
Hence,
\[90^\circ + 2x = 180^\circ \]
Subtract \[90^\circ \] from both sides and simplify.
\[ \Rightarrow 90^\circ + 2x - 90^\circ = 180^\circ - 90^\circ \]
\[ \Rightarrow 2x = 90^\circ \]
Divide both sides by 2 and simplify.
\[ \Rightarrow \dfrac{{2x}}{2} = \dfrac{{90^\circ }}{2} \]
\[ \Rightarrow x = 45^\circ \]
The value of \[x\] is \[45^\circ \].
vi.
Ans: Since the vertical opposite angles are equal, $y = x$.
The sum of the internal angles of a triangle is $180^\circ $.
In the given triangle, \[x + x + x = 180^\circ \].
Hence,
\[3x = 180^\circ \]
Divide both sides by 3 and simplify.
\[ \Rightarrow \dfrac{{3x}}{3} = \dfrac{{180^\circ }}{3} \]
\[ \Rightarrow x = 60^\circ \]
The value of \[x\] is \[60^\circ \] and the value of \[y\] is \[60^\circ \].
Conclusion
NCERT Chapter 6 Class 7 Exercise 6.3 - The Triangle and Its Properties is necessary for understanding triangle properties. Learning the Angle Sum Property, which maintains that the interior angles of any triangle add up to 180 degrees, is important. This activity helps to improve geometry skills by solving triangle-related questions. Understanding these concepts is important for handling more advanced maths topics and performing well on tests. Vedantu's solutions provide detailed explanations, helping people understand and apply geometric ideas better. Download a FREE PDF for easy step-by-step solutions for Class 7 Exercise 6.3.
Class 7 Maths Chapter 6: Exercises Breakdown
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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.
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FAQs on NCERT Solutions For Class 7 Maths Chapter 6 The Triangle And Its Properties Exercise 6.3 - 2025-26
1. How do you find the measure of a third angle in a triangle if two angles are given, as per the NCERT solutions for Class 7 Maths Chapter 6?
To find the third angle of a triangle when two are known, you must use the Angle Sum Property of a triangle. The correct method is:
Step 1: Add the measures of the two given angles.
Step 2: Subtract the resulting sum from 180°.
Step 3: The final value is the measure of the unknown third angle. According to this property, the sum of all three interior angles in any triangle is always 180°.
2. What is the step-by-step method to solve for an unknown angle using the Exterior Angle Property of a triangle?
The NCERT solutions for Chapter 6 explain that an exterior angle of a triangle is equal to the sum of its two opposite interior angles. The steps to solve problems using this property are:
Step 1: Identify the exterior angle and the two interior angles that are opposite to it.
Step 2: Set up an equation where the exterior angle's measure equals the sum of the two interior opposite angles.
Step 3: Solve the equation to find the value of the unknown angle (which could be the exterior angle or one of the interior angles).
3. How does knowing if a triangle is equilateral or isosceles help solve for angles in NCERT Class 7 Maths Chapter 6?
Knowing the type of triangle provides crucial information for solving problems:
For an equilateral triangle, all three angles are equal. Since the total is 180°, each angle must be 60°. If a problem states a triangle is equilateral, you immediately know all its angles.
For an isosceles triangle, the angles opposite the equal sides are also equal. If one of these base angles is known, the other is known too. This reduces the number of unknown angles, making it easier to solve using the Angle Sum Property.
4. What is the correct procedure to verify if three given lengths can form a triangle, according to the Triangle Inequality Property?
To correctly verify if three line segments can form a triangle, you must apply the Triangle Inequality Property. The method is to check if the sum of the lengths of any two sides is greater than the length of the third side. You must check all three combinations:
Is Side 1 + Side 2 > Side 3?
Is Side 1 + Side 3 > Side 2?
Is Side 2 + Side 3 > Side 1?
If all three conditions are true, a triangle can be formed. If even one condition fails, it is not possible to form a triangle with those lengths.
5. How do you apply the Pythagoras theorem to find the length of an unknown side in a right-angled triangle as shown in NCERT solutions?
The Pythagoras theorem applies only to right-angled triangles. The steps to find an unknown side are:
Step 1: Identify the hypotenuse (the side opposite the right angle) and the other two sides (legs).
Step 2: Use the formula a² + b² = c², where 'a' and 'b' are the lengths of the legs and 'c' is the length of the hypotenuse.
Step 3: Substitute the known values into the formula and solve for the unknown side's length.
6. How does an altitude differ from a median in a triangle, and why is this distinction important when solving problems in Chapter 6?
The distinction between an altitude and a median is critical for applying the correct properties:
An altitude is a line segment from a vertex that is perpendicular to the opposite side, forming a 90° angle. It is related to the height of the triangle.
A median is a line segment from a vertex to the midpoint of the opposite side, dividing that side into two equal lengths.
This difference is crucial because properties like the Pythagoras theorem require a right angle, which involves an altitude. In contrast, problems about side lengths being bisected involve a median.
7. Why is it often necessary to use the Angle Sum Property as the main method for finding angles in Chapter 6 problems?
The Angle Sum Property is a fundamental rule that holds true for every type of triangle, whether it is scalene, isosceles, or right-angled. While other properties like the Exterior Angle Property or properties of isosceles triangles are useful, the Angle Sum Property is the universal tool that can be applied in almost any situation where at least one or two other angle-related facts are known. It often serves as the final step to find a missing angle after other properties have been used.
8. Can the Pythagoras theorem be applied to an obtuse or acute triangle? Explain why not based on the concepts in Chapter 6.
No, the Pythagoras theorem cannot be applied to obtuse or acute triangles. The theorem (a² + b² = c²) is derived specifically from the geometric properties of a right-angled triangle, where the square of the hypotenuse is exactly equal to the sum of the squares of the other two sides. In acute and obtuse triangles, this specific relationship does not exist because there is no 90° angle. Attempting to use it would lead to incorrect calculations for side lengths.
9. What steps should be followed to solve a problem where one exterior angle and one interior opposite angle of a triangle are given?
To solve this type of problem, you should use the Exterior Angle Property. The steps are:
Step 1: Recall the property: Exterior Angle = Sum of two interior opposite angles.
Step 2: Let the unknown interior opposite angle be 'x'. Set up the equation: (Given Exterior Angle) = (Given Interior Angle) + x.
Step 3: Solve for 'x' by subtracting the measure of the given interior angle from the measure of the exterior angle. The result will be the other interior opposite angle.
10. How does the step-by-step problem-solving for triangle properties in NCERT solutions build a foundation for real-world applications?
Mastering the step-by-step methods in NCERT solutions for triangle properties builds crucial logical reasoning skills. These methods teach you to:
Identify known and unknown variables.
Select the appropriate theorem or property for a given situation.
Follow a structured process to reach a conclusion.
This systematic approach is the foundation for solving complex problems in fields like architecture, engineering, and navigation, where precise calculations based on geometric principles are essential for designing stable structures or plotting accurate courses.
