NCERT Solutions for Maths Class 7 Chapter 6 Exercise 6.1 - 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.
- 4.1Exercise 6.1
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.
FAQs on NCERT Solutions for Class 7 Maths Chapter 6 - The Triangle and its Properties Exercise 6.1
1. Where can I find the NCERT Class 7 Maths Chapter 6 Exercise 6.1 Solutions?
Vedantu, India's top online learning portal, offers NCERT Class 7 Maths Chapter 6 Exercise 6.1 Solutions The Triangle and its Properties were meticulously developed by highly qualified and experienced teachers following the most recent CBSE criteria. These solutions include exact and thorough solutions to every sum in the class 7 NCERT Maths textbook. On Vedantu's official website (Vedantu.com), you may quickly and gratis download PDF versions of these study guides. You can also get the Vedantu mobile app.
2. How many Questions are there in Maths Class 7 Chapter 6 Exercise 6.1?
Maths Class 7 Chapter 6 Exercise 6.1 consists of a total of 3 questions. In nearly all of the questions in Chapter 6.1, a special emphasis is placed on Pythagoras' Theorem, one of the most significant topics in geometry. You can consult Vedantu, India's most popular online platform if you're looking for NCERT solutions for Class 7 Math. At Vedantu, all of the chapter exercises are collected in one location and solved step-by-step by a qualified teacher following the NCERT book's instructions.
3. What do you mean by Pythagoras' theorem in Maths Class 7 Chapter 6 Exercise 6.1?
In a right-angled triangle, the square of the hypotenuse side is equal to the sum of the squares of the other two sides, according to Pythagoras's Theorem. These triangles' three sides are known as the perpendicular, base, and hypotenuse. Due to its position opposite the 90° angle, the hypotenuse in this case is the longest side. When the positive integer sides of a right triangle (let's say sides a, b, and c) are squared, the result is an equation known as a Pythagorean triple. It is a crucial area of mathematics that describes how a right-angled triangle's sides relate to one another. Pythagorean triples are another name for the sides of the right triangle.
4. What are the topics discussed in Maths Class 7 Chapter 6 Exercise 6.1?
The topics discussed in Class 7 Chapter 6 The Triangle and Its Properties, are:
Introduction
Medians of Triangle
Altitudes of a Triangle
Exterior Angle of a Triangle and its Property
Angle Sum Property of a Triangle
Two Special Triangles - Equilateral and Isosceles
Sum of the Lengths of Two Sides of Triangle
Right Angles Triangle and Pythagoras Property
5. Is Class 7 Chapter 6 Maths Exercise 6.1 The Triangle and its Properties important?
"Triangle and its Properties" is the first topic that comes to mind when thinking about the CBSE Class 7 Maths curriculum. Unquestionably, this is one of the most important chapters on the course outline. Your foundation, which will be crucial at higher levels, will be built in this chapter. Triangle and its Properties will test your fundamental comprehension skills. To make the ideas clear and encourage intelligent learning, one of the key goals of this chapter was to carefully break down each essential section.
6. What are the altitudes of a triangle in Class 7 Chapter 6 Maths Exercise 6.1?
Altitudes of a triangle are perpendicular lines drawn from each vertex to the opposite side. They help determine the height of the triangle from each vertex, crucial in various geometric calculations.
7. How do you calculate the length of a median in a triangle in Class 7 Chapter 6 Maths Exercise 6.1?
The length of a median can be calculated using the formula Length of Median= $\frac{1}{2}\times \sqrt{2a^{2}+2b^{2}-2c^{2}}$, where a, b, c are the sides of the triangle. Class 7 Maths Chapter 6 Exercise 6.1 Question 1 formula helps find the segment connecting a vertex to the midpoint of the opposite side.
8. Why are altitudes and medians important in geometry in Class 7 Chapter 6 Maths Exercise 6.1?
Altitudes help in calculating the area of triangles, while medians divide triangles into two equal areas. Understanding these concepts enhances problem-solving skills and provides insights into geometric properties.
9. How can altitudes help in finding the area of a triangle in Maths Class 7 Chapter 6 Exercise 6.1?
Altitudes are essential in calculating triangle area using the formula Area = $\frac{1}{2}\times base\times altitude$. They provide a direct method to determine the perpendicular height of the triangle from each vertex.
10. In Maths Class 7 Chapter 6 Exercise 6.1 what is the centroid of a triangle?
The centroid is the point where all three medians intersect. It divides each median in a 2:1 ratio, meaning it is closer to the midpoint of the longer segment.
11. In Maths Class 7 Chapter 6 Exercise 6.1 how do altitudes and medians relate to each other in a triangle?
Altitudes and medians are both important geometric properties. Altitudes help in calculating heights and area, while medians divide the triangle into symmetrical parts and locate the centroid.