NCERT Solutions For Class 10 Mathematics Chapter 6 Exercise 6.3

The history of Mathematics is frequently confused with the history of Philosophy, since it is fundamental to culture. Non-Euclidean geometry has sparked new theories about the nature of the cosmos, much like how cosmology and evolution have had a significant influence on how people view themselves. The boundaries of deductive reasoning are also demonstrated by mathematical logic theorems. To comprehend Mathematics and use it in daily life, it’s vital to read the NCERT Mathematics book. The answers to the NCERT questions can be found in the NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.3.In art, Mathematics is also present. The relationship between Mathematics and art has persisted since the most famous mathematician, Pythagoras, discovered the numerical reasons for musical harmony. These characteristics make Mathematics a useful bridge between the humanities and sciences. One’s mind benefits much from learning math in a variety of fascinating ways. It promotes pragmatism, sharpens the mind, encourages rationality, and may be applied to daily life.

Everyday living involves math. Many children find mathematics to be dull, abstract, creative, sophisticated, and extremely challenging to comprehend. But because it’s a part of their studies, youngsters must practise it constantly. However, the answers to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 were created with the needs of students in mind. A lesson from a mathematics tutor is a suitable choice if parents believe they are unable to assist their child in math. These experts not only have the knowledge, but they also understand the methodical techniques for teaching it. This has major significance.

Analytical thinking is aided by mathematics. Arguments can be characterised as ideas aimed at breaking them down into their individual premises or expressions, determining the connections between them and their conclusions, and assessing their accuracy or dependability. Everyone should approach mathematical problems in this manner. Gather information, dissect presumptions, note relationships, save the pieces, and approach the problems in a methodical, logical manner. One will be more equipped to handle real-world issues if they have a solid understanding of mathematics and can develop rational answers. They connect the data they need to make judgments, find the best rationale, and consider potential solutions. The capacity for analytical thought helps us investigate the world around us and discover its realities. People need to look for certain truths based on facts rather than feelings. It is a way of thinking that enables one to recognise flaws, lies, and manipulations in both oneself and others. This is possible because, given verifiable data from the real world, one may reason rationally and clearly about mathematics.

Because solving a problem necessitates seeing a whole, cohesive process, mathematics helps people become more analytical. It might be argued that mathematics is essential to educating children since it teaches them how to think.

It is possible to explain how things work using mathematics. In other words, one was able to accurately, clearly, and coherently convey their thoughts and views. Everyone else understands us and is aware that humans are rational beings with clear minds, which is fundamental and incredibly wonderful. A large component of the image is how they effectively convey themselves.

Everyday life involves mathematics, which is used in both traditional sciences and cutting-edge technologies. In actuality, a large number of everyday phenomena are guided by precise science. Because math programmes stress that a problem’s solution must lead to the truth, they aid and facilitate students’ development of their views. since math is undoubtedly rational and objective.

When presented with hard situations, mathematics speeds up thinking and generally encourages deeper thought. Most people’s lives are made up of decisions, methods, justifications, and circumstances where they encounter issues that need to be resolved. In this way, mathematics promotes mental flexibility and the realisation that there is only one right answer to every problem.

One of the main issues in education nowadays is that high school pupils don’t take mathematics seriously. Even in fields like engineering, statistics, education, and technology, where big salaries are possible, people simply aren’t interested in the subject. Teenagers are sceptical of all the advantages that mathematics may present in the future, including expanded college options and, as previously mentioned, higher earnings. They regard mathematics as uninteresting, difficult, and unimportant to their lives. work, etc.) inside the industry.

Here are seven more reasons why a, individual should study Mathematics and why it’s important to their future.

Strength and endurance in athletics are equivalent to mathematics. It creates the groundwork for a child to excel in his career. Without strength and the absence of health issues, a young athlete cannot become a major sports star. Similar to other subjects, learning mathematics requires practise, hence, referring to the Extramarks NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 is advised.

Math is necessary to prevent loss. The main reason why foolish people invest money in numerous pyramid schemes with the hope of generating a lot of money is that mathematics is not their strong suit. If they are informed, they can recognise financial and smoking scams quite quickly. With the aid of mathematics, one can save money on initiatives and suggestions they believe will be beneficial.

A student who excels in mathematics can travel the world. Mathematics and other “hard” sciences are seen as a way out of poverty and social decay by intelligent children in Eastern Europe, India, and China. In a world that is constantly evolving, mathematics is crucial. New technologies are transforming how people live and work. Using mathematics, they will learn how and why things work the way they do.In the future, mathematics will be expressed more forcefully. Whether you like it or not, mathematics is playing a bigger role in many sectors. Future journalists and politicians will speak less and think more critically. The military and police of the future will employ scientifically created technology. Numbers and technology will also be used by teachers and nurses. In addition to using hammers and wrenches, future carpenters and machinists will also utilise optimization electronics and analytics.

Daily living involves a lot of mathematics. It is their duty as parents to explain to the youngster all the advantages that this course offers. It goes without saying that not everyone needs to be a mathematician or an engineer, but this subject gives students a good chance at the future by teaching them to think critically, analyse situations, and come to the best decisions they can in a variety of real-world situations. This allows them to experience all the possibilities that mathematics has to offer. So, on its website, Extramarks offers the answer to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

NCERT Class 10 Chapter 6 Exercise 6.3 Triangle Solutions

Along with publishing high-quality textbooks, NCERT also produces teaching aids, practise tests, and other materials. Students at the primary, secondary, and collegiate levels can easily pass tests since they have a very firm foundation in each topic (English, Social Studies, etc.). With the aid of these Solutions, students may learn about society, Hindi, science, physics, chemistry, mathematics (like NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3), and biology.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 deal with problems in which various theorems about the similarity of triangles must be applied to solve them. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 covers all aspects of this subject, giving students plenty of opportunity to practise preparing for the exam. Extramarks experts have prepared NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 online using the latest CBSE curriculum with clear step-by-step solutions. This will make students proficient in solving similar problems independently. Subjects like science, Mathematics, and English will become easy to learn if students have access to the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  problem-solving, and other subject solutions.

Access NCERT Solutions For Mathematics Chapter 6 – Triangle Solutions

The main aim of the NCERT program is to make it student-friendly and useful for both students and candidates who are preparing for competitive exams. The book is prepared in detail, all based on the manuals of different boards. The answers to the questions given in the textbook are equally important, and students can find the answers.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are fully compatible with most state boards and CBSE (Central Board of Secondary Education). NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 with detailed explanation, will help students  prepare for and pass 12th Class exams or competitive exams with excellent scores. The NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  play an indispensable role in preparing for the exam. Here are the benefits of settling for NCERT solutions:

Students should learn and practise NCERT solutions as they provide detailed chapter explanations. These solutions also help students complete homework and prepare for exams.

With the help of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students can take brief notes that they can use at the time of their review. Thanks to that, they will be able to remember all the important formulas, keywords, theorems, author names, etc. By practising NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students can easily score good marks in their exams. Students will find that most exam questions are derived from these solutions. This is why they are considered good course material.

These NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are written in an easy-to-understand dialect, so they can easily understand the theory behind each chapter. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are designed under the CBSE program. In addition, by solving these problems, students will know their weaknesses so that they can be overcome.

CBSE Class 10 Mathematics Chapter 6 Exercise 6.3 Solutions

If students want to have a quick look at the various tips and tricks explained in the NCERT Solution for Class 10 Mathematics Exercise 6.3, they can download its PDF format from the website of Extramarks. The NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 PDF is available as a PDF and can be downloaded or printed for ease of use. Students no longer need to worry about internet connection while preparing for tests once they save the PDF file of NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.3 in their computer system.

Here are some of the benefits of using NCERT Solutions:

Provides authentic information – The content of these books was created and designed after a lot of research. These books provide accurate and comprehensive information in all chapters. It is recommended to solve all the in text questions of the NCERT book which are available on NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3

CBSE Recommended  – According to CBSE, these NCERT books are the best study materials for students. It itself prescribes these books because they help students to understand the concepts better. NCERT books follow CBSE guidelines and cover all chapters of Class 1-12 syllabus. Students must visit NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

Clearing Student Questions – Subject matter experts are responsible for preparing the content for NCERT books. They use simple words so that students can easily understand each topic. Students can also solve NCERT questions to clarify their own doubts. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 can be considered beneficial in solving these problems.

The NCERT textbook is useful. Students are urged to concentrate on their NCERT book and avoid wasting time on other study materials. Most exam questions are comparable to those in the NCERT book, as most students will discover. As the best resources for studying for exams, these books should be takenseriously.

It saves a tonne of time – When examinations are coming up, students must plan their time wisely. They should study the NCERT texts in addition to other reference materials. These books include a lot of questions and are meant to help students clarify ideas. All of these NCERT books and their Solutions are very time-efficient during exams.

Topic Covered In Class 10 Mathematics Chapter 6 Exercise 6.3

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are based on the topic of Similarity between different Triangles. Criteria for similarity of triangles: Two triangles are said to be similar if their corresponding angles are congruent and their corresponding sides are also proportional.

Different Theorems Covered in Exercise 6.3

Theorem 1: If in two triangles the corresponding angles are congruent, then their corresponding sides have the same ratio. So it can be said that the two triangles are similar. Students should refer to the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

Theorem 2: If in two triangles, the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are congruent. Hence, it can be said that the two triangles are similar. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 provides all the answers related to this theorem.

Theorem 3: If an angle of a triangle is equal to an angle of another triangle and the sides including these angles are proportional to each other, then the two triangles are similar. For solutions, students can look at the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

Class 10 Mathematics Exercise 6.3 Solutions

Extramarks provides the answers to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 to assist students in their time of need.They can visit the website of Extramarks to get access to all the answers in the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

Exercise 6.3 Lesson 10 Question 1

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are the answers to questions in the Mathematics Class 10 Chapter 6 Exercise 6.3 is based on the similarity of pairs of triangles. A set of pairs of triangles is given, and students need to determine which pairs are similar and also mention the criteria by which one defines them as congruent triangles. Students must also give the correct notation to describe similar triangles.

Exercise 6.3 Class 10 Question 2

In the second question of the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students have to use the relationship between adjacent angles and the similarity of triangles to find the angle of two similar triangles formed by the intersection of  two intersecting lines and two parallel lines.

Exercise 6.3 Class 10 Question 3

In NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, students are given a trapezium ABCD, and they have to prove that for the intersection O the ratio of the lengths of the diagonals  is AO/OC = OB/OD. They need to apply the theorem to the alternate inscribed angle and the vertical opposite angle to solve this problem.

Exercise 6.3 Class 10 Question 4

In this question, a triangle is given, which has two interior triangles. By applying the theorem that if the sides of a triangle are proportional, then their corresponding angles are congruent, one can prove that the two triangles are similar. The answer is available in the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3

Exercise 6.3 Class 10 Question 5

This is a simple question where two angles of two triangles are congruent (criteria AA), so students can prove that the given triangles are similar. Students are advised to refer to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

Exercise 6.3 Class 10 Question 6

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 use knowledge of similar triangles and CPCT (corresponding parts of similar triangles) to prove that the given triangles are similar.

Exercise 6.3 Class 10 Question 7

In this question, the altitudes of two triangles are given, and students have to prove that the triangles formed by the intersection of the altitudes are congruent. Students will use the concept of vertically opposite angles and common angles to solve this question. The NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 answer this question as well.

Exercise 6.3 Class 10 Question 8

This question has a parallelogram ABCD, and a line from point B that intersects DC at F and extends to point E. Point D is extended until it meets E, forming the triangle AEB. Using the opposite angle property of parallelograms and the concept of alternating angles, students can take the help of the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 to prove that △ABE and △CFB are similar.

Exercise 6.3 Class 10 Question 9

This is again a simple question where two right triangles have a common side and students have to prove that both triangles are similar triangles. They can easily prove the common angle and the corresponding sides of the triangle. Similarity and the concept of similarity of angles with the help of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

Exercise 6.3 Class 10 Question 10

Question 10 of this exercise is related to the bisector theorem and the similar angle substitution criterion to prove that two triangles formed by bisectors of an angle are congruent. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 can help students understand how to correctly answer the question.

Exercise 6.3 Class 10 Question 11

In this problem, an isosceles triangle is given, one of its sides is stretched, and a perpendicular is dropped from the point of extension to the opposite side of the triangle. Students must demonstrate that the triangle formed by stretching and the triangle formed by dropping perpendicular to the base of the original triangle are congruent. The properties of isosceles triangles and the alternating angle criterion given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are used to solve this problem.

Exercise 6.3 Class 10 Question 12

There are two triangles ABC and PQR in this problem, and it is known that the medians of the two triangles and their adjacent sides are proportional. Students have to prove that triangles ABC and PQR are similar triangles. The median dividing the opposite side, using this concept with SSS (for congruent triangles) and corresponding angles of similar triangles given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, this problem can be solved.

Exercise 6.3 Class 10 Question 13

In triangle ABC, draw line AD on side BC  and ∠ADC is equal to ∠BAC. Students have to prove that CA² = CB.CD. This is done using the concepts of common angle, alternative angle and corresponding angle from similar triangle rules. Students can refer to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 for answers.

Exercise 6.3 Class 10 Question 14

This question is similar to question 12 in that the sides of two triangles and their medians are proportional, and students have to prove that the two triangles are similar. This question is solved by extending the medians of two triangles by equal lengths, then using the rules for parallelograms, SSS, SAS, and corresponding angles from the Congruent Triangle Standard given on NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  to prove that the triangles are similar.

Exercise 6.3 Class 10 Question 15

It is a question about a vertical pillar and its shadow with the shadow of a tower. Given the height of the column and the length of the shade,  students need to find the height of the tower. Using the AAA (angle-angle-angle) congruence theorem, one can deduce the height of the tower. They can also take the help of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3

Exercise 6.3 Class 10 Question 16

Given two similar triangles, the student must prove that the ratios of the sides of the triangle and their medians are equal. By using the properties of the mean and the corresponding angle criterion given in NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3, they can solve this problem.

Key Features Of NCERT Solutions For Class 10 Mathematics Exercise 6.3

Extramarks provides a lot of courses to students and these solutions are easily accessible to them on the Extramarks website.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are prepared in simple language for students to understand it.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are prepared by the subject experts of Extramarks.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 give accurate answers to all the questions in NCERT textbooks.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 will help the students with a quick revision.

NCERT Solutions For Class 10 Mathematics Chapter 6 All Other Exercises

Exercise 6.1: It comprises Three Questions and Answers, and these are Short Answer Type Questions. Students can access the solutions of this exercise like NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 on Extramarks

Exercise 6.2: It consists of 10 Questions and Answers including 9 Short Answers and 1 long answer which are available in resources like NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 and others provided by Extramarks

Exercise 6.4: It consists of 9 Questions and Answers with 7 Short and 2 long Answers. Extramarks provide the answers to these questions in a similar format as the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3.

Exercise 6.5: It comprises 17 Questions and Answers i.e, 15 short answers and 2 long answers. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 gives a blueprint of the kinds of answers that will be helpful for solving the questions.

Exercise 6.6: It consists of 10 Questions and Answers. 5 Short answers and 5 long answers. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 is an example of the kind of answers that Extramarks provides for this exercise.

Q.1 State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

Ans.

(i) Corresponding angles are equal in the given triangles. i.e.,A=P=60°,B=Q=80°andC=R=40°.Therefore, by AAA similarity criterion, ΔABC~ΔPQR.(ii) Ratios of the corresponding sides of the given pair oftriangles are equal. i.e., ABQR=BCRP=CAPQ=12Therefore, by SSS similarity criterion, ΔABC~ΔQRP.(iii)The given pair of triangles are not similar as theircorresponding sides are not proportional.(iv)The given pair of triangles are not similar as theircorresponding sides are not proportional.(v)The given pair of triangles are not similar as theircorresponding sides are not proportional.(vi) In ΔDEF,      D+E+F=180°70°+80°+F=180°F=180°150°=30°In ΔPQR,      P+Q+R=180°P+80°+30°=180°P=180°110°=70°We find out that corresponding angles are equal in ΔDEF and ΔPQR. i.e.,D=P=70°,E=Q=80°andF=R=30°.Therefore, by AAA similarity criterion, ΔDEF~ΔPQR.

Q.2 In the following figure,

ΔODCΔOBA, BOC=125°and CDO=70°. Find DOC, DCO and OAB. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8541@

Ans.

DOB is a straight line.BOC+DOC=180°DOC=180°BOC=180°125°=55°In ΔDOC,      DOC+DCO+ODC=180°DCO=180°DOCODCDCO=180°55°70°=55°It is given that ΔODC~ΔOBA and so corresponding anglesin these triangles are equal.OAB=DCO=55°

Q.3

Diagonals AC and BD of a trapezium ABCD with ABDC intersect each other at the point O. Usinga similarity criterion for two triangles, show that OAOC=OBOD.

Ans.

In ΔDOC and ΔBOA,CDO=ABO [Alternate interior angles as ABCD]DCO=BAO [Alternate interior angles as ABCD]DOC=BOA [Vertically opposite angles]ΔDOC~ΔBOA [AAA similarity critarion]DOBO=OCOA [Corresponding sides are proportional]OAOC=OBOD

Q.4 

In the following figure, QR QS = QT PR and 1=2. Show that ΔPQSΔTQR. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@7401@

Ans.

In ΔPQR,        1=2 (Given)PQ=PR ...(1)It is given that QRQS=QTPRQRQS=QTQP [From (1), PR=QP]    ...(2)Q is common in ΔPQS and ΔTQR and the sides includingthis angle in both triangles are proportional as shown inequation (2). Therefore, by SAS critarion, ΔPQS~ΔTQR.

Q.5

S and T are points on sides PR and QR of ΔPQR such that P=RTS. Show that ΔRPQ~ΔRTS.

Ans.

In ΔRPQ and ΔRTS,       P=T (Given)and R=R Therefore, by AA criterion, ΔRPQ~ΔRTS.

Q.6

In the following figure, if ΔABEΔACD, show that ΔADE~ΔABC.

Ans.

It is given that ΔABEΔACD.Therefore, by CPCT,         AB=ACand AD=AESo, ADAB=AEACIn ΔADE and ΔABC,       A=A and ADAB=AEACTherefore, by SAS criterion, ΔADE~ΔABC.

Q.7

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:(i) ΔAEP~ΔCDP(ii) ΔABD~ΔCBE(iii)ΔAEP~ΔADB(iv)ΔPDC~ΔBEC

Ans.

(i)In ΔAEP and ΔCDP,       APE=CPD (Vertically opposite angles)and AEP=CDP=90° Therefore, by AA similarity critarion, ΔAEP~ΔCDP.(ii)In ΔABD and ΔCBE,          ADB=CEB=90°and ABD=CBETherefore, by AA similarity critarion, ΔABD~ΔCBE.(iii)In ΔAEP and ΔADB,          AEP=ADB=90°and PAE=BADTherefore, by AA similarity critarion, ΔAEP~ΔADB.(iv)In ΔPDC and ΔBEC,          PDC=BEC=90°and DCP=ECBTherefore, by AA similarity critarion, ΔPDC~ΔBEC.

Q.8

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CDat F. Show that ΔABE~ΔCFB.

Ans.

In the above figure, ABCD is a parallelogram in which ABDC.Opposite angles in parallelogram are equal. Therefore,EAB=BCFAlso, ABE=CFB [Alternate interior angles]Therefore, by AA similarity critarion, ΔABE~ΔCFB.

Q.9

In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:(i) ΔABC~ΔAMP(ii) CAPA=BCMP

Ans.

(i)In ΔABC and ΔAMP, ABC=AMP=90°and, CAB=PAM Therefore, by AA similarity critarion, ΔABC~ΔAMP.(ii)Corresponding sides are proportional in similar triangles.Therefore,      ΔABC~ΔAMPCAPA=BCMP

Q.10

CD and GH are respectively the bisectors of ΔACB and ΔEGF such that D and H lie on sides AB and FEof ΔABC and ΔEFG respectively. If ΔABC~ΔFEG, show that:(i)    CDGH=ACFG(ii)   ΔDCB~ΔHGE(iii)  ΔDCA~ΔHGF

Ans.

It is given that ΔABC~ΔFEG.A=F, B=E, BCA=EGFAlso,      12BCA=12EGFACD=FGH and DCB=HGE By AA similarity critarion,      ΔDCA~ΔHGF and ΔDCB~ΔHGECDGH=ACFG

Q.11

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB=AC. If ADBC and EFAC, prove that ΔABDΔECF. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeqabeqadiWaceGabeqabeWaaqaafeaakqaabeqaaiaabMeacaqGUbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGMbGaae4BaiaabYgacaqGSbGaae4BaiaabEhacaqGPbGaaeOBaiaabEgacaqGGaGaaeOzaiaabMgacaqGNbGaaeyDaiaabkhacaqGLbGaaeilaiaabccacaqGfbGaaeiiaiaabMgacaqGZbGaaeiiaiaabggacaqGGaGaaeiCaiaab+gacaqGPbGaaeOBaiaabshacaqGGaGaae4Baiaab6gacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabccacaqGdbGaaeOqaiaabccacaqGWbGaaeOCaiaab+gacaqGKbGaaeyDaiaabogacaqGLbGaaeizaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeOBaiaabccacaqGPbGaae4Caiaab+gacaqGZbGaae4yaiaabwgacaqGSbGaaeyzaiaabohacaqGGaGaaeiDaiaabkhacaqGPbGaaeyyaiaab6gacaqGNbGaaeiBaiaabwgacaqGGaGaaeyqaiaabkeacaqGdbaabaGaae4DaiaabMgacaqG0bGaaeiAaiaabccacaqGbbGaaeOqaiabg2da9iaabgeacaqGdbGaaeOlaiaabccacaqGjbGaaeOzaiaabccacaqGbbGaaeiraiabgwQiEjaabkeacaqGdbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGfbGaaeOraiabgwQiEjaabgeacaqGdbGaaeilaiaabccacaqGWbGaaeOCaiaab+gacaqG2bGaaeyzaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaeiiaiabgs5aejaabgeacaqGcbGaaeiraiablYJi6iabgs5aejaabweacaqGdbGaaeOraiaab6caaaaa@B14C@

Ans.

In ΔABD and ΔECF,          BDA=CFE=90°and ABD=ECF [AB=ACB=C]Therefore, by AA similarity critarion, ΔABD~ΔECF.

Q.12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR andmedian PM of ΔPQR (see the following figure). Show that ΔABC~ΔPQR.

Ans.

It is given that AD and PM are medians. Therefore,12BC=BD=DCand12QR=QM=MRIt is given that         ABPQ=BCQR=ADPM    ABPQ=12BC12QR=ADPM    ABPQ=BDQM=ADPM Therefore, by SSS similarity critarion, ΔABD~ΔPQM and soin ΔABC and ΔPQR, B=Q and ABPQ=BCQR.Therefore, by SAS similarity critarion, ΔABC~ΔPQR.

Q.13

D is a point on the side BC of a triangle ABC such that ADC=BAC. Show that CA2=CB·CD.

Ans.

In ΔBAC and ΔADC,ADC=BAC (Given)ACD=BCA (Common angle)ΔADC~ΔBAC (By AA similarity critarion)We know that corresponding sides of similar trianglesare proportional.   CACD=CBCA  CA2=CBCD

Q.14

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR

Ans.

It is given that,ABPQ=ACPR=ADPMWe extend AD and PM up to point E and L respectively,such that AD=DE and PM=ML. We join B to E, C to E, Q to L and R to L.

It is given that AD and PM are medians. Therefore,12BC=BD=DC    and    12QR=QM=MRIn quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram.AC=BE and AB=EC Similarly, we can prove that quadrilateral PQLR isa parallelogram and PR=QL, PQ=LR.It is given that         ABPQ=ACPR=ADPM    ABPQ=BEQL=2AD2PM    ABPQ=BEQL=AEPL Therefore, by SSS similarity critarion, ΔABE~ΔPQL and soin ΔABE and ΔPQL, BAE=QPL ...(1) Similarly, we can show that ΔAEC~ΔPLR and CAE=RPL ...(2) Adding (1) and (2), we getBAE+CAE=QPL+RPLCAB=RPQAlso, we haveABPQ=ACPRTherefore, by SAS similarity critarion, ΔABC~ΔPQR.

Q.15

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a towercasts a shadow 28 m long. Find the height of the tower.

Ans.

Let CD is a pole and DF is its shadow.Let AB is a tower and BE is its shadow.At the same time in a day, the sun rays will fall on pole and tower at the same angle.Therefore,DCF=BAEAlso, CDF=ABE=90°ΔABE~ΔCDF [By AA similarity critarion]Corresponding sides are proportional in similar triangles.Therefore,ABCD=BEDFAB6 =284=7AB=42Therefore, the height of the tower is 42 m.

Q.16

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC~ΔPQR, prove thatABPQ=ADPM.

Ans.

It is given that AD and PM are medians. Therefore,12BC=BD=DC and 12QR=QM=MRAlso, ΔABC~ΔPQRTherefore,        ABPQ=BCQR=ACPR and A=P, B=Q,  C=RIn ΔABD and ΔPQM,ABPQ=12BC12QR=BDQM and B=Q ΔABD~ΔPQM [By SAS similarity critarion]ABPQ=ADPM

Please register to view this section

FAQs (Frequently Asked Questions)

1. Where can I find NCERT solutions for Class 10 Mathematics Exercise 6.3?

Students can find NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 on the Extramarks website or application. Students can download NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 from the website as they are written in an easy-to-understand, detailed format with topic-specific explanations. All NCERT questions are provided by Extramarks experts. They can browse through these solutions to answer any questions or concerns they may have.

2. Do students have to practice all the questions from the NCERT Solutions?

Yes, it is important to practice all questions in the Class 10 Mathematics Exercise 6.3 NCERT Answers. Each question provided is an application of a different concept, and practising these questions will help students to fully understand each concept. NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  can be downloaded for free from his website or app at Extramarks. Extramarks provide NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7 and NCERT Solutions Class 6

3. How many examples are there of Class 10 Mathematics Chapter 6 Exercise 6.3?

Exercise 6.3 in Mathematics Class 10 has a total of 16 sample questions. These examples focus on problems that require the application of various triangle theorems and triangle congruence. The answers to these questions can be found in the NCERT solution for Class 10 Mathematics, available at Extramarks. Extramarks team of experts has structured their answers to NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 concisely and in detail. Also, the NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3 are fully based on the CBSE curriculum.

4. How long does it take students to complete Class 10 Maths Chapter 6 Exercise 6.3?

The time to complete NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  will vary for each student. It may take several hours or a day or two to complete the entire exercise. Students may be able to complete the exercises in less time if the concepts are very clear to them. Therefore, to assist students, Extramarks provides access to handpicked answers of NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.3  to these questions. Students can examine them and clear their doubts. All resources are available in the Extramarks app.

.