NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.6) Exercise 6.6

Class 10 retains a crucial position in one’s academic career. The marks secured in matriculation board exams remain with a student throughout his or her life. Thus, a student should invest his or her maximum effort while preparing for the board exams. Every topic, if studied diligently and under the supervision of supreme mentors, seems easy. It is difficult but not impossible to find reliable resources either offline or online.

Extramarks is one such platform upon which students of Class 10 can rely. The NCERT Solutions offered by Extramarks are of the highest quality. Even a hard-to-crack topic like triangles has been prepared and explained in such a way that every student can learn this topic if they pay the required attention.

Students looking for Class 10th Maths Ex 6.6 Solutions will find the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 available on Extramarks to be extremely useful.These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 explain important concepts, provide solutions to the exercise questions, and a complete analysis of past years’ papers as well. Thus, in a single place, students can get all the vital stuff that is needed to boost their exam preparation.

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6

Extramarks provides NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 for the benefit of students of Class 10 preparing for Mathematics. Prepared by talented experts in Mathematics, these solutions are upgraded from time to time as per the latest syllabus recommended by CBSE and the latest edition of the Class 10 Mathematics book published by NCERT. These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 are available in a PDF format which can be downloaded easily. Students need to go to the website of Extramarks and click on NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. By clicking on the Download button, these NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 can be downloaded in PDF form. They can take a print of the same and kick- start their preparation. It will save their time and effort to visit different websites and compile notes. The quality of these NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 is scrutinised minutely in a timely manner. Thus, students can get authentic solutions to such a complicated chapter in a simplified manner.

Access NCERT Solutions for Class 10 Maths Chapter 6 -Triangles

In this competitive world, where one needs to work hard to secure a rank for oneself, getting a firm base of knowledge and understanding important concepts is important. It is possible to find NCERT Solutions online from a variety of sources. What matters when selecting any of these sources for preparation are their authenticity, reliability, and advantages over some other sources.

Extramarks is a place thatmatches all these prerequisites. Here, the best quality NCERT Solutions for any of the classes can be accessed. The solutions for primary classes can be found in NCERT Solutions Class 1, NCERT Solutions Class 2, NCERT Solutions Class 3, NCERT Solutions Class 4, NCERT Solutions Class 5, NCERT Solutions Class 6, NCERT Solutions Class 7 and NCERT Solutions Class 8. Apart from this, NCERT Solutions Class 9 and NCERT Solutions Class 11 are also available. The board classes need extra attention and dedication. Students of Class 10 and Class 12 can also take advantage of Extramarks’ services by accessing NCERT Solutions Class 10 and NCERT Solutions Class 12 respectively.

For all these classes, the NCERT Solutions for all the subjects are provided in a chapter-wise method. For example, students of Class 10 who want to prepare Chapter 6 of Mathematics can take the guidance of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 will give them a basic overview of the chapter as well as a detailed analysis of every conceptl. When they are done with this chapter, past years’ papers for this exercise can also be foundunder NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

Exercise 6.6 Class 10 Maths NCERT Solutions

Under this section, students of Class 10 may find all the solutions in a stepwise manner to Class 10 Maths Chapter 6 Ex 6.6. Upon clicking the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6, they will be able to access not only the NCERT solutions to the questions of Exercise 6.6 but also the detailed explanation of every concept and theorem. These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 can be used to enhance the board exam preparation. Having a thorough understanding of important concepts gives students an edge over others. They will be able to tackle cognitive questions like multiple choice questions or assertion reason-based questions with ease. Thus, students of Class 10 may opt for Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 while revising for the board exams. Practising with Extramarks helps one polish and prune one’s skills and secure more marks. For selective learning, students may visit the past years’ papers section under NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 and prepare accordingly. Systematic and well-organised notes are the need of the hour for Class 10 students. This need can be fulfilled using Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

Important Concepts Covered in Exercise 6.6 of Class 10 Maths NCERT Solutions

Triangles is a vast topic in the Class 10 Mathematics book published by NCERT. There are many theorems and concepts in this Chapter which are significant from an examination point of view. These concepts can be understood with Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 are prepared with utmost caution, keeping in mind the requisites of the examination.

These concepts can be brushed up and polished easily by following the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. Using these solutions, students in Class 10 will be able to master the topics related to the similarity of triangles and other related theorems.

The following are the concepts that students should be familiar with after reading this chapter:

Similar Figures
CPCT – ‘Corresponding parts of congruent triangles are equal
AA Similarity
AAA Congruence Rule
SSS Congruence Rule
SAS Congruence Rule

These concepts are available on NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 in an elaborate way. Students may choose these NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 to understand these concepts in depth.

Apart from this, the following theorems and their derivations are also available in NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6:

Similar Triangles
Thales’ Theorem
Theorems related to Areas of Similar Triangles
Pythagoras’ Theorem
Converse of Pythagoras Theorem

These theorems are extremely important for exams. Thus, being thoroughly aware of these will help students attempt the questions based on these theorems in an easy way. As a result, NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 give a complete and detailed explanation of these theorems.

Besides, some questions in this exercise are based on the RHS criterion of similarity. According to the RHS criterion, the two triangles are said to be similar if the hypotenuse and one side of a triangle are proportional to the hypotenuse and corresponding side of the other triangle.

Basic Facts about Triangle

A triangle is a shape that a student learns to identify at the age of two. This shows the importance of the topic of triangles. This is the reason CBSE has included this chapter in almost every Class until the 10th. As per the standard definition of a triangle, any closed figure with three sides all joined end to end is known as a triangle. A triangle has three sides, three vertices, and three angles. The sum of all the angles in a triangle is 180 degrees. This is known as the angle sum property of a triangle. These are some basic facts which can be found on Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 to build a solid foundation.

Students in Class 10 are assigned the topic of triangles by CBSE.Class 10 Maths Chapter 6 Ex 6.6 is based on the various properties of triangles. Students are advised to practise the questions related to this with sheer dedication. For better understanding, they can opt for NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 available on the Extramarks website.

Types of Triangles

In the Mathematics book for Class 10 published by the National Council for Educational Research and Training (NCERT), students of the CBSE Board encounter various types of triangles.One can expect fact-based questions in the final exams based on the simple concept of types of triangles. In general, there are six major types of triangles based on angles and three based on sides. Here is a brief discussion about a few of them:

Right-Angle Triangle: In a triangle, if one of the three angles is 90 degrees, it can be called a right-angled triangle. The other two angles can both be 45 degrees each or might vary in their values. However, the total sum of these degrees of angles will remain the same i.e., 180 degrees.
Equilateral Triangle: If all sides of a particular triangle are equal in measurement, it is known as an equilateral triangle. All of the angles in an equilateral triangle are 60 degrees.
Acute-Angle Triangle: A triangle whose sum of all its angles is less than 90 degrees is termed an acute-angled triangle. All equilateral triangles are acute-angle triangles, but not vice versa. However, it cannot be said that all acute-angle triangles will be equilateral triangles.
Obtuse-Angle Triangle: When, in a triangle, one of the three angles is greater than 90 degrees and the other two angles are acute angles, this is known as an obtuse-angled triangle. In the case of the obtuse-angle triangle, the length of the hypotenuse is the greatest of all the three sides of a triangle.

Only four of the many types have been discussed here, and only briefly.If students want to learn more about all of these types, they can go to Extramarks NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.6.Not only will they find a description of all the existing types of triangles on Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6, but they will also find the desired solutions to Chapter 6 of Class 10 Math.

Some Problems Related to Triangles

Class 10 Maths Chapter 6 Exercise 6.6 has some typical problems whose details and explanations are given on Extramarks in a simplified manner. Students can access the detailed explanations of these problems on NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. Some of the critical topics have been hinted at below:

Congruency of triangles: Congruency is a special type of topic related to triangles that one comes across while solving the questions from the exercises in Chapter 6, entitled Triangles. In these problems, students are expected to use different criteria like SAS, ASA, etc. Two of the triangles are supposed to be congruent. Solving these questions using proper steps can fetch a lot of marks for the students in their board exams. In case of a lack of guidance, students can learn these steps using NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. These NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 discuss every criterion of congruency in an easy way. Thus, students may access Extramarks’ NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 if they wish to strengthen this weak area.

Triangles with the Parallel Line: There are some questions where a parallel line can intersect with the triangle, and one is supposed to find the missing angle. These measurement-related sums help one increase their knowledge regarding the calculation of angles and can be helpful in civil engineering during higher education. The NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 offer a complete analysis of such questions.

Finding the Missing Angle: In Class 10 Maths Chapter 6 Ex 6.6, there are some complicated questions on triangles where one has to find angles at the point of intersection of lines. For this, knowing the angles formed along transversal and parallel lines is important. This information can be accessed using NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

Apart from these, other important topics are available on the Extramarks website. Students can access the same using NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

Tips to Solve the Sums on Triangle Easily

Triangles is a topic that commands a good quality of questions in the examination. Having the appropriate knowledgeto solve the questions related to triangles is necessary. Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 come up with some basic tips to solve the questions related to triangles.These hints, found in NCERT Solutions for Class 10 Maths, Chapter 6 Exercise 6.6, teach students how to approach triangle questions in the most presentable manner. Here are a few tips to follow which solving these questions:

Be well-versed in all the properties of triangles.
Learn all the theorems related to triangles.
Go through the question multiple times and try to comprehend what is given and what is being asked.
While attempting the question, write down everything that is given.
Use the appropriate method to solve the questions.
Do not change the order of steps to be followed. Solve the problems in a step-by-step fashion.The proper methods and steps to solve these questions can be learnt using NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.
Draw diagrams wherever necessary.

For complete details, students may visit NCERT Solutions Class 10 on the Extramarks website. As part of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6, a comprehensive list of tips and guidelines can be found. Visiting these NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 can prove beneficial to the students of Class 10 in their board exam preparation.

How to Practice Numerical Problems on Triangle

Class 10 Maths Chapter 6 Exercise 6.6 deals with the questions based on triangles and their properties. Thus, it is important to use the correct approach while solving these questions. The most important thing that is required to be practised is drawing the appropriate triangle diagrams as per the instructions given in the question. Once students can draw the triangle diagram correctly, it will be easier and faster to derive the answer. However, if the diagram is not made properly, the entire solution may go wrong. Thus, the students of Class 10 must learn all the properties of triangles as well as the appropriate techniques to draw a diagram as mentioned in the question. It is also important to label the diagram correctly. This can be mastered by students of Class 10 using the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. A complete description of the approaches one should use while dealing with such questions can be accessed using Extramarks NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

With guidance from NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6, and multiple revisions, one may become able to tackle such questions very comfortably.

NCERT Solutions for Class 10 Maths Chapter 6 Other Exercises

Extramarks provides NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. Apart from this, the NCERT Solutions for the rest of the exercises can also be accessed on Extramarks. These exercises should be prepared with concentration and selective learning, as the number of questions to be expected from each exercise varies. For example, only short questions can be expected from Exercise 6.1 whereas four long questions can be expected from Exercise 6.2. A complete analysis of the questions to be asked from each exercise is given in a tabular form on NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6.

NCERT Solutions for Class 10 Maths

Mathematics is a subject that seems daunting to everyone. Learners can understand this daunting subject if it is practised appropriately.

Extramarks is taking this step to help students overcome their fear of mathematicsThus, it provides complete and well-compiled solutions to every chapter of Class 10 Math. Students can access every chapter individually. It smoothstheir learning process, as they can choose and select whichever Chapter they wish to access. In total, there are 15 chapters in the book. Extramarks provides solutions to all these chapters in a sequential format. In this way, it becomes a great benefactor for the students of Class 10.

Q.1

$In the following figure, PS is the bisector of ∠ QPR of Δ PQR. Prove that \frac{QS}{SR} = \frac{PQ}{PR} .$

Ans.

$\begin{array}{l} We draw a line segment RT parallel to SP which \\ intersects extended line segment QP at point T. \\ It is given that PS is angle bisector of ∠ QPR. \\ Therefore, \\ ∠ QPS = ∠ SPR (1) \\ Also, \\ ∠ SPR = ∠ PRT (As PS ∥ TR) (2) \\ ∠ QPS = ∠ QTR (As PS ∥ TR) (3) \\ Using these equations we get, \\ ∠ PRT = ∠ QTR \\ So, PT = PR \\ Now in Δ QTR and Δ QPS, \\ ∠ QSP = ∠ QRT (As PS ∥ TR) \\ ∠ QPS = ∠ QTR (As PS ∥ TR) \\ ∠ Q is common \\ ∴ Δ QTR ~ Δ QPS \\ \Rightarrow \frac{QS}{SR} = \frac{QP}{PT} \\ \Rightarrow \frac{QS}{SR} = \frac{QP}{PR} [PT = PR] \end{array}$

Q.2

$\begin{array}{l} In the following figure, D is a point on hypotenuse AC of Δ ABC, DM ⊥ BC and DN ⊥ AB. Prove that: \\ (i) {DM}^{2} = DN . MC (ii) {DN}^{2} = DM . AN \end{array}$

Ans.
(i)

$\begin{array}{l} We have, DN ∥ CB, DM ∥ AB and ∠ B = 90 ° \\ So, DNMB is a rectangle. \\ ∴ DN = MB and DM = NB \\ ∠ 2 + ∠ 3 = 90 ° . .. (1) \\ In Δ CDM, \\ ∠ 1 + ∠ 2 + ∠ DMC = 180 ° \\ ∠ 1 + ∠ 2 = 90 ° . .. (2) \\ In Δ DMB \\ ∠ 3 + ∠ DMB + ∠ 4 = 180 ° \\ ∠ 3 + ∠ 4 = 90 ° . .. (3) \\ From equation (1) and (2), \\ ∠ 1 = ∠ 3 \\ From equation (1) and (3), \\ ∠ 2 = ∠ 4 \\ ∴ Δ DCM ~ Δ BDM \\ \Rightarrow \frac{DM}{BM} = \frac{MC}{DM} \\ \Rightarrow {DM}^{2} = BM \times MC = DN \times MC [BM = DN] \\ (ii) \\ In Δ DBN, \\ ∠ 5 + ∠ 7 = 90 ° . .. (4) \\ In Δ DAN, \\ ∠ 6 + ∠ 8 = 90 ° . .. (5) \\ In Δ DAB, \\ ∠ 5 + ∠ 6 = 90 ° . .. (6) \\ From equation (4) and (6), we get \\ ∠ 6 = ∠ 7 \\ From equation (5) and (6), we get \\ ∠ 5 = ∠ 8 \\ ∴ Δ BND ~ Δ DNA [By AA similarity critarion] \\ \Rightarrow \frac{AN}{DN} = \frac{DN}{NB} \\ \Rightarrow {DN}^{2} = AN \times NB = AN \times DM [As NB = DM] \end{array}$

Q.3

$\begin{array}{l} In the following figure, ABC is a triangle in which ∠ ABC > 90° and AD ⊥ CB produced. Prove that \\ {AC}^{2} = {AB}^{2} + {BC}^{2} + 2 BC . BD . \end{array}$

Ans.
Applying Pythagoras theorem in ΔADB, we get
AB² = AD² + DB² ….(1)
Applying Pythagoras theorem in ΔACD, we get
AC² = AD² + DC²
⇒AC² = AD² + ( BD + BC )²
⇒AC² = AD² + DB² + BC² + 2BD x BC
Now using equation ( 1 ), we get
AC² = AB² + BC² + 2BD × BC

Q.4

$\begin{array}{l} In the following figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC . Prove that \\ {AC}^{2} = {AB}^{2} + {BC}^{2} - 2BC .BD . \end{array}$

Ans.

$\begin{array}{l} A pplying Pythagoras theorem i n ΔADB, we get \\ {AD}^{2} + {BD}^{2} = {AB}^{2} \\ \Rightarrow {AD}^{2} = {AB}^{2} - {BD}^{2} . .. (1) \\ A pplying Pythagoras theorem i n ΔADC, we get \\ {AD}^{2} + {DC}^{2} = {AC}^{2} \\ Now using equation (1), we get \\ {AB}^{2} - {BD}^{2} + {DC}^{2} = {AC}^{2} \\ \Rightarrow {AB}^{2} - {BD}^{2} + {(BC - BD)}^{2} = {AC}^{2} \\ \Rightarrow {AC}^{2} = {AB}^{2} - {BD}^{2} + {BC}^{2} + {BD}^{2} - 2 BC \times BD \\ \Rightarrow {AC}^{2} = {AB}^{2} + {BC}^{2} - 2 BC \times BD \end{array}$

Q.5

$\begin{array}{l} In the following figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that: \\ {(i) AC}^{2} = {AD}^{2} + BC.DM + {(\frac{BC}{2})}^{2} \\ {(ii) AB}^{2} = {AD}^{2} - BC . DM + {(\frac{BC}{2})}^{2} \\ {(iii) AC}^{2} + {AB}^{2} = {2AD}^{2} + \frac{1}{2} {BC}^{2} \end{array}$

Ans.
We have the following figure.

$\begin{array}{l} (i) \\ Applying Pythagoras theorem i n Δ AMD, we get \\ {AM}^{2} + {DM}^{2} = {AD}^{2} . .. (1) \\ Applying Pythagoras theorem i n Δ AMC, we get \\ {AM}^{2} + {MC}^{2} = {AC}^{2} \\ \Rightarrow {AM}^{2} + {(DM + DC)}^{2} = {AC}^{2} \\ \Rightarrow ({AM}^{2} + {DM}^{2}) + {DC}^{2} + 2 DM . DC = {AC}^{2} \\ Using equation (1), we ge t \\ {AD}^{2} + {DC}^{2} + 2 DM . DC = {AC}^{2} \\ \Rightarrow {AD}^{2} + {(\frac{BC}{2})}^{2} + 2 DM . \frac{BC}{2} = {AC}^{2} \\ \Rightarrow {AD}^{2} + {(\frac{BC}{2})}^{2} + DM \cdot BC = {AC}^{2} \\ (ii) \\ Applying Pythagoras theorem in Δ ABM, we get \\ {AB}^{2} = {AM}^{2} + {MB}^{2} \\ = ({AD}^{2} - {DM}^{2}) + {MB}^{2} \\ = ({AD}^{2} - {DM}^{2}) + {(BD - DM)}^{2} \\ = {AD}^{2} - {DM}^{2} + {BD}^{2} + {DM}^{2} - 2 BD \times DM \\ = {AD}^{2} + {BD}^{2} - 2 BD \times DM \\ = {AD}^{2} + {(\frac{BC}{2})}^{2} - 2 \frac{BC}{2} \times DM \\ = {AD}^{2} + {(\frac{BC}{2})}^{2} - BC \times DM \\ (iii) \\ Applying Pythagoras theorem in Δ AMB, we get \\ {AM}^{2} + {MB}^{2} = {AB}^{2} . .. (1) \\ Applying Pythagoras theorem in ΔAMC, we get \\ {AM}^{2} + {MC}^{2} = {AC}^{2} . .. (2) \\ Adding equations (1) and (2), we get \\ 2 {AM}^{2} + {MB}^{2} + {MC}^{2} = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + {(BD - DM)}^{2} + {(MD + DC)}^{2} = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + {BD}^{2} + {DM}^{2} - 2 BD . DM + {MD}^{2} + {DC}^{2} + 2 DM . DC = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AM}^{2} + 2 {DM}^{2} + {BD}^{2} + {DC}^{2} + 2 DM (- BD + DC) = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 ({AM}^{2} + {DM}^{2}) + {(\frac{BC}{2})}^{2} + {(\frac{BC}{2})}^{2} + 2 DM (- \frac{BC}{2} + \frac{BC}{2}) = {AB}^{2} + {AC}^{2} \\ \Rightarrow 2 {AD}^{2} + \frac{{BC}^{2}}{2} = {AB}^{2} + {AC}^{2} \end{array}$

Q.6 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Ans.
Let ABCD be a parallelogram. We draw perpendiculars AF on CD and DE on extended side BA.

$\begin{array}{l} Applying Pythagoras theorem in Δ DEA, we get \\ {DE}^{2} + {EA}^{2} = {DA}^{2} . .. (i) \\ Applying Pythagoras theorem in ΔDEB, we get \\ {DE}^{2} + {EB}^{2} = {DB}^{2} \\ {DE}^{2} + {(EA + AB)}^{2} = {DB}^{2} \\ ({DE}^{2} + {EA}^{2}) + {AB}^{2} + 2 EA \times AB = {DB}^{2} \\ {DA}^{2} + {AB}^{2} + 2 EA \times AB = {DB}^{2} . .. (ii) \\ Applying Pythagoras theorem in ΔADF, we get \\ {AD}^{2} = {AF}^{2} + {FD}^{2} \\ Applying Pythagoras theorem in ΔAFC, we get \\ {AC}^{2} = {AF}^{2} + {FC}^{2} \\ = {AF}^{2} + {(DC - FD)}^{2} \\ = {AF}^{2} + {DC}^{2} + {FD}^{2} - 2 DC \times FD \\ = ({AF}^{2} + {FD}^{2}) + {DC}^{2} - 2 DC \times FD \\ {AC}^{2} = {AD}^{2} + {DC}^{2} - 2 DC \times FD . .. (iii) \\ Since ABCD is a parallelogram \\ AB = CD . .. (iv) \\ and, BC = AD . .. (v) \\ In Δ DEA and Δ ADF, \\ ∠ DEA = ∠ AFD = 90 °, ∠ EAD = ∠ ADF (EA ∥ DF) \\ Therefore, ∠ FAD = ∠ EDA \\ AD is common in both triangles. \\ Therefore, by ASA congruence critarion, we get \\ ΔEAD ≅ ΔFDA \\ \Rightarrow EA = DF . .. (vi) \\ Adding equation (ii) and (iii), we get \\ {DA}^{2} + {AB}^{2} + 2 EA \times AB + {AD}^{2} + {DC}^{2} - 2 DC \times FD = {DB}^{2} + {AC}^{2} \\ \Rightarrow {DA}^{2} + {AB}^{2} + {AD}^{2} + {DC}^{2} + 2 EA \times AB - 2 DC \times FD = {DB}^{2} + {AC}^{2} \\ \Rightarrow {BC}^{2} + {AB}^{2} + {AD}^{2} + {DC}^{2} + 2 EA \times AB - 2 AB \times EA = {DB}^{2} + {AC}^{2} \\ (Using equations (iv), (v) and (vi)) \\ \Rightarrow {AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} \\ Hence, the sum of the squares of the diagonals of a parallelogram is \\ equal to the sum of the squares of its sides . \end{array}$

Q.7

$\begin{array}{l} In the following figure, two chords AB and CD intersect each other at the point P. Prove that : \\ (i) Δ APC ~ Δ DPB (ii) AP.PB = CP.DP \end{array}$

Ans.

$\begin{array}{l} Let us join CB. \\ (i) In Δ APC and Δ DPB, \\ ∠ APC = ∠ DPB [Vertically opposite angles] \\ ∠ CAP = ∠ BDP [Angles in same segment for chord CB] \\ ∴ Δ APC ~ Δ DPB [By AA similarity criterion] \\ (ii) \\ We know that corresponding sides of similar triangles \\ are proportional. \\ Therefore, \\ Δ APC ~ Δ DPB \\ \Rightarrow \frac{AP}{DP} = \frac{PC}{PB} \\ \Rightarrow AP \cdot PB = PC \cdot DP \end{array}$

Q.8

$\begin{array}{l} In the following figure, two chords AB and CD of a circle intersect each other at the point P \\ (when produced) outside the circle. Prove that \\ (i) Δ PAC ~ Δ PDB (ii) PA \cdot PB = PC \cdot PD \end{array}$

Ans.

$\begin{array}{l} (i) \\ In Δ PAC and Δ PDB, \\ ∠ P = ∠ P (common) \\ ∠ PAC = ∠ PDB \\ (Exterior angle of a cyclic quadrilateral is \\ equal to opposite interior angle) \\ Δ PAC ~ Δ PDB \\ (ii) \\ We know that corresponding sides of similar triangles \\ are proportional. \\ ∴ Δ PAC ~ Δ PDB \\ \Rightarrow \frac{PA}{PD} = \frac{PC}{PB} \\ \Rightarrow PA \cdot PB = PC \cdot PD \end{array}$

Q.9

$\begin{array}{l} In the following figure, D is a point on side BC of ΔABC such that \frac{BD}{CD} = \frac{AB}{AC} . \\ Prove that AD is the bisector of ∠ BAC . \end{array}$

Ans.

$\begin{array}{l} Let us extend BA to P such that AP = AC. Join PC. \\ It is given that \\ \frac{BD}{CD} = \frac{AB}{AC} \\ \Rightarrow \frac{BD}{CD} = \frac{AB}{AP} \\ Therefore, by converse of basic proportinality theorem, AD ∥ PC. \\ Therefore, \\ ∠ BAD = ∠ APC . .. (1) \\ ∠ DAC = ∠ ACP . .. (2) \\ Also, \\ AP = AC \\ \Rightarrow ∠ APC = ∠ ACP \\ \Rightarrow ∠ APC = ∠ ACP = ∠ DAC [From (2)] \\ \Rightarrow ∠ BAD = ∠ DAC [From (1)] \\ Therefore, AD is the bisector of angle BAC. \end{array}$

Q.10 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the following figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Ans.

$\begin{array}{l} Let AB be the height of the tip of fishing rod from water \\ surface. \\ Let BC be the horizontal distance of fly from the tip of \\ fishing rod. \\ AC is the length of string. \\ On applying Pythagoras theorem in Δ ABC, we get \\ {AC}^{2} = {AB}^{2} + {BC}^{2} \\ \Rightarrow {AB}^{2} = {(1.8)}^{2} + {(2.4)}^{2} \\ \Rightarrow {AB}^{2} = 3.24 + 5.76 \\ \Rightarrow {AB}^{2} = 9.00 \\ \Rightarrow AB = \sqrt{9} = 3 \\ Therefore, length of the string is 3 m. \\ It is given that string is pulled at the rate of 5 cm per second. \\ So, string pulled in 12 seconds = 12 \times 5 = 60 cm = 0 .6 m \end{array}$

$\begin{array}{l} Let after 12 second Fly be at point D. \\ Length of string out after 12 second is AD \\ AD = AC – string pulled by Nazima in 12 seconds \\ = 3.00 – 0.6 \\ = 2.4 \\ In ΔADB, \\ {AB}^{2} {+ BD}^{2} {= AD}^{2} \\ \Rightarrow {(1.8)}^{2} {+ BD}^{2} = {(2.4)}^{2} \\ \Rightarrow {BD}^{2} = 5.76 - 3.24 = 2.52 \\ ⇒ BD = 1.587 \\ Horizontal distance of fly = BD + 1.2 \\ = 1.587 + 1.2 \\ = 2.787 \\ = 2.79 m \end{array}$

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FAQs (Frequently Asked Questions)

1. During past years was there any question asked from Exercise 6.6 of Chapter 6 of Class 10 Maths in the final Board Exams?

Class 10 Maths Chapter 6 Ex 6.6 is optional and therefore is ignored by most of the students. However, since the whole exercise is based on the concepts from the entire chapter, making this exercise a part of exam preparation will be helpful. It will be a sort of test to judge the knowledge one has gained about this chapter. Proving theorems like Basic Proportionality Theorem, Pythagoras theorem and solving questions which are based on these theorems can seem difficult at the first instance. But with Extramarks, it becomes easier. The NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 available on the Extramarks website explain every solution in the easiest possible terms. The past years’ papers available on NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6 can give the students a complete analysis of the frequency of questions from this exercise in the board exams.

2. How to score well in the upcoming Mathematics examination?

To achieve higher marks in the Mathematics examination, students of Class 10 must practice exercises consistently. When solving questions, all NCERT solutions should be referred to. When attempting to improve problem-solving skills, students can use NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6. Students are also advised to solve sample papers and past years’ papers regularly. By doing so, students can practice questions that may appear on the Mathematics board examination. For students to understand derivations and theorems regarding the chapters of Mathematics, they must thoroughly revise the theories of Mathematics.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.6) Exercise 6.6

NCERT Solutions Class 10 Maths Chapter 6 Exercise 6.6

Access NCERT Solutions for Class 10 Maths Chapter 6 -Triangles

Exercise 6.6 Class 10 Maths NCERT Solutions