Maths Class 7 Chapter 6 Questions and Answers - Free PDF Download
NCERT Class 7 Maths Chapter 6 Exercise 6.1 Solutions, The Triangle and its Properties Students will find detailed explanations of topics like medians and altitudes of triangles. These solutions are available in PDF format and designed by our expert teachers to help you understand these concepts better. Focus on understanding how medians divide a triangle into equal areas and how altitudes calculate perpendicular distances.
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These concepts are fundamental in geometry and will help in your problem-solving skills. Practice these solutions to learn the principles explained in NCERT Solutions for Class 7 Maths and practicing these solutions might help students improve their understanding of the CBSE Class 7 Maths Syllabus.
Glance on NCERT Solutions Maths Chapter 6 Exercise 6.1 Class 7 | Vedantu
Medians of a Triangle divide it into smaller triangles with equal areas.
The altitude of a Triangle measures the perpendicular distance from a vertex to the opposite side.
The Centroid marks where medians intersect, splitting each into two equal segments.
The incenter is the circle's centre inscribed in the triangle, touching sides at right angles.
Circumcenter is the circle's centre passing through all triangle vertices, affecting geometric properties.
Euler's Line is a straight line through the triangle's centroid, orthocenter, and circumcenter, revealing key geometric relationships.
Understanding these concepts helps in calculating triangle heights and finding centroids, crucial in geometry and problem-solving.
This article contains exercise notes, important questions, exemplar solutions, exercises, and video links for Exercise 6.1 - The Triangle and its Properties, which you can download as PDFs.
There are 3 fully solved questions in NCERT Class 7 Maths Chapter 6 Exercise 6.1 Solutions.
Formulas Used in Class 7 Chapter 6 Exercise 6.1
Altitude Formula: Area = $\frac{1}{2}\times Base\times Altitude$
Median Formula: Length of Median = $\frac{1}{2}\times \sqrt{2a^{2}+2b^{2}-2c^{2}}$
Exercise 6.1
1. In $\vartriangle PQR$, $D$ is the mid-point of $\overline {QR} $.
$\overline {PM} $ is ___________
$PD$ is _______.
Is $QM = MR$?
Ans: It is given a $\vartriangle PQR$, $D$ is the mid-point of $\ overline {QR} $. It means $QD = DR$. Since $\overline {PM} $ is perpendicular to the side $QR$ of triangle $\vartriangle PQR$, $\overline {PM} $ is the altitude of $\vartriangle PQR$. Since $QD = DR$, $PD$ is the median of triangle $\vartriangle PQR$.
No, $QM \ne MR$ because $D$ is the mid-point of $\overline {QR} $. It means $QD = DR$.
2. Draw rough sketches for the following:
a) $\vartriangle ABC$, where $BE$ is a median.
Ans: A median of a triangle is a line segment that is drawn from a vertex to the opposite side of the vertex and it divides the opposite side into two equal parts. The rough sketch of $\vartriangle ABC$, where $BE$ is a median, is drawn below.
Here $BE$ is a median in $\vartriangle ABC$ and $AE = EC$.
b) $\vartriangle PQR$, where $PQ$ and $PR$ are the altitudes of the triangle.
Ans: An altitude of a triangle is defined as a perpendicular drawn from the vertex to the line containing the opposite side of the triangle. The rough sketch of $\vartriangle PQR$, where $PQ$ and $PR$ are the altitudes of the triangle is drawn below.
c) $\vartriangle XYZ$, where $YL$ is an altitude in the exterior of the triangle.
Ans: An altitude of a triangle is defined as a perpendicular drawn from the vertex to the line containing the opposite side of the triangle. The rough sketch of $\vartriangle XYZ$, $YL$ is an altitude in the exterior of the triangle is drawn below.
3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be the same.
Ans: It is given an isosceles triangle. It is required to verify if the median and altitude of the given triangle can be the same. To do this, construct an isosceles triangle. An isosceles triangle has two equal sides.
Construct an isosceles triangle $\vartriangle ABC$ with sides$AB = AC$ and draw a median $AL$ that divides the base of the triangle into two equal parts.
From the triangle, it can be seen that the median makes a ${90^ \circ }$ angle with the base $BC$. So, $AL$ is the altitude of the triangle $\vartriangle ABC$. Hence verified, $AL$ is the median and altitude of the given triangle $\vartriangle ABC$.
Conclusion
NCERT Class 7 Maths Chapter 6 Exercise 6.1 Solutions clearly explain fundamental concepts of The Triangle and its Properties, such as triangle altitudes and medians. It's important to understand how altitudes determine triangle heights and how medians divide triangles into equal parts. Practicing these ideas improves geometric reasoning and problem-solving skills. Learning these concepts prepares students for more difficult geometry topics and provides a solid foundation for future mathematical studies. Using these answers gives clarity and confidence in using geometric principles successfully.
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 6.1 - 2025-26
1. Where can I find reliable, step-by-step NCERT Solutions for Class 7 Maths Chapter 6 for the 2025-26 session?
You can find detailed, step-by-step NCERT Solutions for Class 7 Maths Chapter 6, The Triangle and its Properties, on Vedantu. These solutions are prepared by subject matter experts and are fully aligned with the latest CBSE 2025-26 syllabus. Each answer explains the correct method to solve the problems as given in the NCERT textbook exercises.
2. What is the correct method to solve questions on identifying medians and altitudes in a triangle in Chapter 6?
The correct method to solve these problems involves understanding their precise definitions as per the NCERT textbook. To solve for a median, you must identify the line segment that connects a vertex to the midpoint of the opposite side. For an altitude, you must identify the line segment from a vertex that is perpendicular (forms a 90° angle) to the opposite side. The solutions demonstrate how to draw and label these correctly for different types of triangles.
3. How do I apply the Exterior Angle Property to solve problems in Exercise 6.2?
To correctly apply the Exterior Angle Property, follow these steps:
Identify the exterior angle of the triangle in the given problem.
Identify the two interior opposite angles (the angles that are not adjacent to the exterior angle).
Set up an equation where the measure of the exterior angle is equal to the sum of the measures of the two interior opposite angles.
Solve this equation to find the value of the unknown angle.
4. What is the step-by-step method to solve problems using the Pythagoras property in right-angled triangles from Exercise 6.5?
To solve problems using the Pythagoras property, follow this precise method:
First, confirm that the triangle is a right-angled triangle.
Identify the hypotenuse (the side opposite the right angle) and the other two sides (legs).
Apply the formula: a² + b² = c², where 'a' and 'b' are the lengths of the legs, and 'c' is the length of the hypotenuse.
Substitute the known values into the equation and solve for the unknown side.
5. How should I use the Triangle Inequality Property to check if a triangle can be formed with given side lengths?
The correct procedure is to check if the sum of the lengths of any two sides of the triangle is greater than the length of the third side. For sides a, b, and c, you must verify all three conditions as per the Triangle Inequality Property:
Is (a + b) > c?
Is (a + c) > b?
Is (b + c) > a?
If all three conditions are true, a triangle can be formed. If even one condition is false, a triangle cannot be formed with those side lengths.
6. Why is it important to show every step while solving problems in Chapter 6, 'The Triangle and its Properties'?
Showing every step is crucial because it demonstrates your understanding of the underlying geometric properties. For example, simply writing the answer for an unknown angle does not show whether you used the Angle Sum Property or the Exterior Angle Property. Step-wise solutions, as shown in the NCERT Solutions, help teachers award partial marks and allow you to trace back your logic to find errors in your method.
7. What is a common mistake students make when applying the Pythagoras property, and how can the NCERT Solutions help avoid it?
A very common mistake is incorrectly identifying the hypotenuse. Students sometimes add the squares of the hypotenuse and one side to find the other. The NCERT Solutions help by consistently emphasizing the first step: always identify the side opposite the 90° angle as the hypotenuse ('c' in a² + b² = c²). Following this structured approach prevents errors in setting up the equation.
8. How do the NCERT solutions for this chapter clarify the difference between a median and an altitude in problem-solving?
The NCERT solutions clarify this by showing their different functions in problems. A median is always related to the midpoint of a side and is used in problems about dividing a side into two equal halves. An altitude is always related to the perpendicular height and is used in problems involving right angles. In an isosceles or equilateral triangle, a median and an altitude can be the same line, and the solutions for those specific cases help solidify this advanced concept.
9. How do the solutions for the Angle Sum Property and the Exterior Angle Property relate to each other?
The solutions demonstrate that these two properties are interconnected. The Angle Sum Property states that the three interior angles of a triangle add up to 180°. The Exterior Angle Property can be derived from it. If you have an exterior angle and its adjacent interior angle, they form a linear pair (sum to 180°). This means the exterior angle equals the sum of the other two interior angles, as (180° - adjacent angle) = sum of the other two angles. The solutions show you can often use either property to find missing angles.
10. How do I solve questions related to the properties of isosceles and equilateral triangles in this chapter?
To solve these problems, you must apply their unique properties. For an isosceles triangle, the key is knowing that the angles opposite the equal sides are also equal. For an equilateral triangle, you must know that all three sides are equal and all three angles are equal to 60°. The NCERT Solutions show how to use these facts to set up equations and find unknown angles or sides.
