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NCERT Solutions for Class 10 Maths Chapter 6 is Triangle, an important study material for CBSE Class 10 students. This chapter aligns perfectly with the CBSE Syllabus for 2025-26, covering many rules and theorems. Sometimes, students get perplexed about which theorem to use. Here, students will get the NCERT solutions for Class 10 in chapter-wise. Practice these solutions for the triangles class 10, and sail the boat of the triangle.
NCERT solutions created at Careers360 are made using the concept in the easiest way that a student can grasp. Subject experts created these solutions, keeping the ease and the board exam. NCERT class 10 triangle solution not just for exams but also for tackling homework, assignments, and to understand the relation of triangles with other concepts too.
CBSE Class 10 exams usually have questions from NCERT textbooks. So, Chapter 6's NCERT Solutions for Class 10 Maths are the best bet for getting ready and tackling any kind of question from this chapter. We strongly recommend practicing these solutions regularly to ace the Class 10 board exams. Math triangle class 10 Theorems of triangles are available free for the students and can be downloaded using the below link.
Similar Triangles - A pair of triangles that have equal corresponding angles and proportional corresponding sides.
Equiangular Triangles:
A pair of triangles that have corresponding angles equal.
The ratio of any two corresponding sides in two equiangular triangles is always the same.
Criteria for Triangle Similarity:
Angle-Angle-Angle (AAA) Similarity
Side-Angle-Side (SAS) Similarity
Side-Side-Side (SSS) Similarity
As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
Basic Proportionality Theorem:
When a line is drawn parallel to one side of a triangle to intersect the other sides in distinct points, the other two sides are divided in the same ratio.
Converse of Basic Proportionality Theorem:
In a pair of triangles when the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar.
These concepts are used to solve Important questions of triangles class 10 with solutions.
Free download NCERT Solutions for Class 10 Maths Chapter 6 Triangles PDF for CBSE Exam.
Maths Chapter 6 Triangles class 10 Excercise: 6.1
Answer:
All circles are similar.
Since all the circles have a similar shape. They may have different radius but the shape of all circles is the same.
Therefore, all circles are similar.
Answer:
All squares are similar.
Since all the squares have a similar shape. They may have a different side but the shape of all square is the same.
Therefore, all squares are similar.
All equilateral triangles are similar.
Since all the equilateral triangles have a similar shape. They may have different sides but the shape of all equilateral triangles is the same.
Therefore, all equilateral triangles are similar.
Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.
Thus, (a) equal
(b) proportional
Q2 (1) Give two different examples of a pair of similar figures.
The two different examples of a pair of similar figures are :
1. Two circles with different radii.
2. Two rectangles with different breadth and length.
Q2 (2) Give two different examples of a pair of non-similar figures.
The two different examples of a pair of non-similar figures are :
1.Rectangle and circle
2. A circle and a triangle.
Q3 State whether the following quadrilaterals are similar or not:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional i.e.
Class 10 Maths Chapter 6 Triangles Excercise: 6.2
Q1 In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i)
Let EC be x
Given: DE || BC
By using the proportionality theorem, we get
(ii)
Let AD be x
Given: DE || BC
By using the proportionality theorem, we get
(i)
Given :
PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
We have
Hence, EF is not parallel to QR.
(ii)
Given :
PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
We have
Hence, EF is parallel to QR.
Q2 (3) E and F are points on the sides PQ and PR respectively of a triangle PQR. For each of the following cases, state whether EF || QR : PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
(iii)
Given :
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
We have
Hence, EF is parallel to QR.
Q3 In Fig. 6.18, if LM || CB and LN || CD, prove that
Answer:
Given : LM || CB and LN || CD
To prove :
Since , LM || CB so we have
Also, LN || CD
From equation 1 and 2, we have
Hence proved.
Q4 In Fig. 6.19, DE || AC and DF || AE. Prove that BF / FE = BE / EC
Given : DE || AC and DF || AE.
To prove :
Since , DE || AC so we have
Also,DF || AE
From equation 1 and 2, we have
Hence proved.
Q5 In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Given : DE || OQ and DF || OR.
To prove EF || QR.
Since DE || OQ so we have
Also, DF || OR
From equation 1 and 2, we have
Thus, EF || QR. (converse of basic proportionality theorem)
Hence proved.
Given : AB || PQ and AC || PR
To prove: BC || QR
Since, AB || PQ so we have
Also, AC || PR
From equation 1 and 2, we have
Therefore, BC || QR. (converse basic proportionality theorem)
Hence proved.
Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.
i.e.
Using basic proportionality theorem, we have
Since
Let P is the midpoint of line AB and Q is the midpoint of line AC.
PQ is the line joining midpoints P and Q of line AB and AC, respectively.
i.e.
we have,
From equation 1 and 2, we get
Draw a line EF passing through point O such that
To prove :
In
So, by using basic proportionality theorem,
In
So, by using basic proportionality theorem,
Using equation 1 and 2, we get
Hence proved.
Draw a line EF passing through point O such that
Given :
In
So, by using basic proportionality theorem,
However, its is given that
Using equation 1 and 2 , we get
Therefore, ABCD is a trapezium.
Class 10 Maths Chapter 6 Triangles Excercise: 6.3
(i)
So ,
(ii) As corresponding sides of both triangles are proportional.
(iii) Given triangles are not similar because corresponding sides are not proportional.
(iv)
(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides
(vi) In
In
Given :
In
Since ,
In
Hence proved.
Q5 S and T are points on sides PR and QR of
Given :
To prove RPQ ~
In
Q6 In Fig. 6.37, if
Answer:
Given :
To prove ADE ~
Since
In
and
Therefore,
Q7 (1) In Fig. 6.38, altitudes AD and CE of
To prove :
In
Q7 (2) In Fig. 6.38, altitudes AD and CE of
To prove :
In
Q7 (3) In Fig. 6.38, altitudes AD and CE of
To prove :
In
Q7 (4) In Fig. 6.38, altitudes AD and CE of
To prove :
In
Q8 E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that
To prove :
In
Q9 (1) In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
To prove :
In
Q9 In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that
To prove :
In
Hence proved.
Q10 (1) CD and GH are respectively the bisectors of
To prove :
Given :
In
Hence proved.
Q 10 (2) CD and GH are respectively the bisectors of
To prove :
Given :
In
Q10 (3) CD and GH are respectively the bisectors of
To prove :
Given :
In
To prove :
Given: ABC is an isosceles triangle.
In
AD and PM are medians of triangles. So,
Given :
In
In
Therefore,
Q13 D is a point on the side BC of a triangle ABC such that
In,
Produce AD and PM to E and L such that AD=DE and PM=DE. Now,
join B to E,C to E,Q to L and R to L.
AD and PM are medians of a triangle, therefore
QM=MR and BD=DC
AD = DE (By construction)
PM=ML (By construction)
So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.
Similarly, PQLR is also a parallelogram.
Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR
Similarity,
Adding equation 1 and 2,
In
CD = pole
AB = tower
Shadow of pole = DF
Shadow of tower = BE
In
Hence, the height of the tower is 42 cm.
Q16 If AD and PM are medians of triangles ABC and PQR, respectively where
Answer:
AD and PM are medians of triangle.So,
From equation 1 and 3, we have
In
Similarity of triangles
Theorems based on similar triangles
Areas of similar triangles
Theorems related to Trapezium
Pythagoras theorem
Also get the solutions of individual exercises-
Comprehensive Coverage: 10th class maths triangles solutions cover all the topics and concepts included in the CBSE syllabus for this chapter.
Detailed Explanations: The class 10 triangles solutions provide step-by-step and clear explanations for each problem, making it easy for students to understand the concepts and solutions.
CBSE-Aligned: Triangles questions class 10 are closely aligned with the CBSE curriculum for Class 10, ensuring that students are well-prepared for their board exams.
Illustrative Examples: The triangle solutions class 10 often include illustrative examples to help students grasp the application of mathematical concepts.
Clarity: Concepts are explained in a simple and straightforward manner, ensuring that students can follow along and build a strong foundation in mathematics.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | Triangles |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 |
First of all, go through all the concepts, theorems and examples given in the chapter.
This chapter needs so much use of theorems. So you have to memorize these conditions and theorems to solve the problems.
After this, you can directly jump to practice exercises.
While solving the practising the exercises, if you face any problem in a question then you take the help of NCERT solutions for class 10 maths chapter 6.
Once you have done the practise exercises you can move to previous year questions.
Keep working hard & happy learning!
NCERT Class 10 Maths Chapter 6, "Triangles," focuses on the similarity of triangles, including criteria for similarity (AA, SSS, SAS), and theorems like the Basic Proportionality Theorem and Pythagoras Theorem.
To prove the Pythagorean theorem
Thales's Theorem or Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Similar triangles have various real-life applications, including measuring inaccessible heights or distances, understanding how zoom works on cameras, and in fields like cartography, engineering, and construction.
Congruent triangles are identical in both shape and size, while similar triangles have the same shape but can differ in size, with corresponding sides being proportional. These concepts are discussed in class 10 maths chapter 6 solutions, practice these to command the concepts.
Admit Card Date:03 February,2025 - 18 March,2025
Admit Card Date:03 February,2025 - 04 April,2025
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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