# NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.5) Exercise 6.5

The Central Board of Secondary Education in New Delhi provides secondary education to both public and private schools. Governed and administered by the Indian government, CBSE is a national education system. The National Council of Educational Research and Training was founded in 1961 with the aim of improving the quality of school education in India. These organisations provide state and federal governments with advisory and assistance services. Moreover, they produce and publish educational kits, educational manuals, textbooks, supplementary materials, newsletters, journals, and multimedia products. For promotion to Class 11 and Class 12, a minimum score of 33% must be achieved by students on both the theory and practical examinations.

Students can improve their mathematical skills, as well as their logical and arithmetic abilities by studying, NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. A simplified explanation of solutions is provided for difficult problems in order to aid in the learning process. In order to prepare for the Class 10 Mathematics examination, students should practise the solutions to the Class 10 Mathematics textbook. It is the purpose of the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 to provide students with answers to all questions in the textbook to assist them in better comprehending the questions. There are many competitive examinations and other college entrance tests that are based on textbooks.Students are advised to regularly practise these NCERT Solutions for Class 10 Maths Chapter 6 Exercises 6.5 in order to achieve higher test scores. On the Extramarks website, students are also encouraged to solve the exemplar problems and provide solutions. It is with the assistance of these solutions that students are able to gain a better understanding of the significance of studying Class 10 Mathematics. Students can access a variety of study materials provided by Extramarks in order to become familiar with the concepts covered in this chapter.

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## NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.5) Exercise 6.5

Students can gain a greater understanding of the topics and concepts covered in Class 10 Mathematics Chapter 6 through NCERT Solutions. Chapter 6 NCERT Textbook questions from the Class 10 textbook should be practised to ensure that students are familiar with the topics covered in the chapter. In order to overcome any difficulties they may encounter when studying Chapter 6 for examinations, students can take assistance from the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. It is recommended that students follow the steps outlined in NCERT Solutions in order to improve their exam performance. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 covers all questions from the CBSE textbook. A step-by-step explanation of each question can be found throughout these solutions.

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### Importance of NCERT Solutions for Class 10 Maths Chapter 6 Triangles (Ex 6.5) Exercise 6.5

Triangles are the subject of Chapter 6 of the Mathematics syllabus for Class 10. The subject of triangles is one of the most important chapters in Class 10 and is quite useful throughout a students’ academic career in various fields. A discussion of the important aspects of triangles is provided in this chapter, which is complemented by expert-curated solutions. These NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 assist students in gaining a good understanding of the various types of problems that they are going to encounter during their examinations. There is also a discussion of the important properties of triangles in the solutions.

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### Triangles – Important Properties

It is important for students in Class 10 to understand the concept of Triangles. In Chapter 6 of the NCERT Textbook for Class 10, various topics are discussed. The fact that all the students have already studied this chapter in their previous classes makes it easier for them to become familiar with it. Triangles, as they are aware, have a number of distinguishing characteristics that set them apart from one another.Throughout the chapter, all of these properties, including sides, angles, and much more, are discussed in detail. In preparation for their Class 10 Board Examination, students are advised to study all the properties carefully.

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### Access NCERT Solutions for Class 10 Maths Chapter – 6 – Triangles

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### CBSE Class 10 Maths Chapter 6 Exercise 6.5 Solutions

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### NCERT Solutions for Class 10 Maths

To facilitate concept-based learning, NCERT Solutions For Class 10 Mathematics Chapter 6 provides detailed explanations of all questions. As a result of the simple language used in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, the questions are more easily retained and understood. Students are also able to track their progress by completing additional practise questions in addition to those provided for each subject. In accordance with the syllabus and examination format, Extramarks provides detailed responses from subject experts. Extramarks provides chapter-by-chapter solutions to all the subjects in Class 10.

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### NCERT Solutions for Class 10 Maths Chapter 6 Exercises

In order to prepare for the examination, students should become familiar with the syllabus. There are a total of six exercises in Chapter 6 of the textbook. The examination includes both short- and long-answer questions.The following table can be used by students to review the same:

### Class 10 Maths Exercise 6.5 Solutions

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 are available at Extramarks. Students are advised to use these solutions to answer all the questions in the NCERT Class 10 Mathematics textbooks. Each question is thoroughly explained in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. A team of highly qualified professionals curates these solutions to help students understand and comprehend them thoroughly. This section presents solutions in a step-by-step format that is easy to understand. For students to ensure that they understand the questions, NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 are available in PDF format. TAll questions from the textbook are answered in these NCERT Solutions for Maths Class 10 Chapter 6 Exercise 6.5.The students in Class 10 should refer to the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5.

#### Exercise 6.5 Class 10 Question 1

As part of Ex 6.5 Class 10, the first question asks the student to measure the sides of four triangles. Pythagoras’ theorem must be applied to each of them, and for further assistance, students can refer to the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5.

#### Exercise 6.5 Class 10 Question 2

The question 2 of 6.5 Exercise Class 10 presents a pictorial representation of a triangle PQR. By using the similarity criterion and Pythagoras theorem, students can try to find the answer. Still, if they face a problem, they can take assistance from the various study materials provided to them.

#### Exercise 6.5 Class 10 Question 3

The third question in Mathematics Class 10, Exercise 6.5, presents a triangle ABC with three statements regarding its sides. The students are required to use various properties of triangles, which they are going to study in the chapter.

#### Exercise 6.5 Class 10 Question 4

The third question in Mathematics Class 10, 6.5, presents a triangle ABC with three statements regarding its sides. These statements must be supported by the triangle properties of similarity.

#### Exercise 6.5 Class 10 Question 5

In order to prove that it is a right-angle triangle, the converse of Pythagoras theorem must be applied. To answer this question, students must apply the knowledge they gained in class 10 about triangles.

#### Exercise 6.5 Class 10 Question 6

An ABC triangle is presented in a diagram. The students are required to determine the length of each altitude. To answer this question, they need to use the Pythagorean theorem in addition to this feature of an equilateral triangle.

#### Exercise 6.5 Class 10 Question 7

In this case, the students need to demonstrate that the square of the sides of a rhombus is equal to the square of the diagonals of the rhombus. To do so, they mustapply the property of equilateral triangles and the Pythagoras theorem.

#### Exercise 6.5 Class 10 Question 8

Question 8 in exercise 6.5 of Class 10 Mathematics requires the use of the basic concepts of Triangles in order to prove two relations. By approaching this problem in a step-by-step manner, the students should be able to accomplish it easily.

#### Exercise 6.5 Class 10 Question 9

The problem involves a ladder, and there is a need to determine the distance between two objects. Using the Pythagoras theorem, the students must calculate the height and distance using Trigonometric Derivatives.

#### Exercise 6.5 Class 10 Question 10

This question can be answered if the students have a solid understanding of Pythagoras theorem. Since it is a practical problem, simply knowing the theorem is not sufficient. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 assist them in understanding the derivation.

#### Exercise 6.5 Class 10 Question 11

In question 10, the students are required to combine their knowledge of Trigonometry and Pythagoras’ theorem to determine the distance between the two planes.

#### Exercise 6.5 Class 10 Question 12

The Pythagoras theorem and trigonometry concepts are both applied in this problem. Practising such problems and gaining a solid foundation in both of these aspects of Mathematics makes them more comfortable with the concepts during the examination.

#### Exercise 6.5 Class 10 Question 13

In question thirteen of Exercise 6.5, there are equations involving the Pythagorastheorem and the interrelationship between triangles that must be proven using the sides of a right-angled triangle.

#### Exercise 6.5 Class 10 Question 14

In this question, the students are given the figure of a triangle ABC, and they must implement Pythagoras’ theorem in detail to get an accurate answer.

#### Exercise 6.5 Class 10 Question 15

In question 15 of this exercise, the students are given various features of an equilateral triangle. In spite of the fact that it is a complicated problem, once they have gone through the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, they can easily understand it.

#### Exercise 6.5 Class 10 Question 16

In order to answer this question, the students must be familiar with some facts about an equilateral triangle. The questions can be easily solved if the students understand the concepts thoroughly.

#### Exercise 6.5 Class 10 Question 17

Last but not least, there are multiple-choice questions in which the students have to determine the angle of a triangle. The students must study diligently to solve these questions.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.5

With the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, students can study the entire chapter at their own pace. By practising the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, students can gain an understanding of the importance of studying Class 10 Mathematics Chapter 6 Exercise 6.5. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 can be used to study for a variety of competitive exams. Extramarks provides a variety of study materials to assist students in developing an in-depth understanding of the concepts presented in this chapter. Furthermore, the study materials include NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 as well as all the key concepts outlined in the CBSE-recommended syllabus. The solutions, developed by qualified experts, are intended to assist students in better preparing for CBSE exams. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 can be used to study for other entrance exams in addition to competitive exams. Prior to the Class 10 board examination, students should practise questions on the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 at least twice. NCERT Solutions are readily available for each chapter to assist students in gaining a thorough understanding of the subject.

Using NCERT textbooks and solutions, students can gain a thorough understanding of each concept. Using these NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, students can understand all concepts covered in the textbook, including advanced concepts. Extramarks provides students with easy access to the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. These NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 are designed to meet the needs of students studying commerce as a major field of expertise in schools. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 are useful because they are written in simple language. One of the most significant advantages of these NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 is that they are available for download both online and offline. This feature is especially useful for students who do not have constant access to the internet. Students studying NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 may find these resources useful. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 can be used to prepare for exams and revise for finals.

All the topics covered in the CBSE Class 10 Mathematics Syllabus are thoroughly covered in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. All major concepts, terminologies, questions, and formulas are explained to students so that they are able to comprehend the material thoroughly. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 can help students improve their performance on competitive examinations, aptitude tests, and college entrance exams. To help students gain a comprehensive understanding of the concepts, Extramarks offers NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5. There are detailed explanations of the problems in Class 10 NCERT Solutions. Extramarks’ website provides students with revision notes, sample problems, past years’ papers, assignments, additional questions, worksheets, and, most importantly, NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 to prepare well for the board examination.

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- By following the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5, the students can clarify their doubts at a deeper level so that they can successfully attempt similar problems independently.
- Answers to each question are provided in NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5 and nothing is omitted.
- In order to provide the learners with high-quality solutions, highly qualified professionals have taken into account the CBSE pattern as well as the NCERT guidelines.
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**Q.1 ** Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

**Ans.**

(i) Given sides of the triangle are 7 cm, 24 cm and 25 cm.

Squares of the given sides of the triangle are 49 cm^{2}, 576 cm^{2} and 625 cm^{2}.

Now,

49 cm^{2 }+ 576 cm^{2} = 625 cm^{2}

Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.

Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 25 cm.

(ii)

$\begin{array}{l}\text{Given sides of the triangle are 3 cm},\text{8 cm and 6 cm}.\\ {\text{Squares of the given sides of the triangle are 9 cm}}^{\text{2}},{\text{64 cm}}^{2}\\ {\text{and 36 cm}}^{2}\text{.}\\ \text{Now,}\\ 9+36\ne 64\\ \text{We find that sum of the squares of two sides is not equal to}\\ \text{the square of third side.}\\ \text{Therefore, triangle of the given sides is not a right triangle.}\\ \\ \text{(iii)}\\ \text{Given sides of the triangle are 50 cm},\text{80 cm and 100 cm}.\\ {\text{Squares of the given sides of the triangle are 2500 cm}}^{\text{2}},\\ {\text{6400 cm}}^{2}{\text{and 10000 cm}}^{2}\text{.}\\ \text{Now,}\\ 2500+6400\ne 10000\\ \text{We find that sum of the squares of two sides is not equal to}\\ \text{the square of third side.}\\ \text{Therefore, triangle of the given sides is not a right triangle.}\end{array}$

(iv) Given sides of the triangle are 13 cm, 12 cm and 5 cm.

Squares of the given sides of the triangle are 169 cm^{2}, 144 cm^{2} and 25 cm^{2}.

Now, 144 c m^{2 }+ 25 cm^{2} = 169 cm^{2}

Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.

Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 13 cm.

**Q.2 ** PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that ( PM )^{2 }=

**Ans.**

$\begin{array}{l}\text{Let}\angle \text{MPR}=\mathrm{\theta}\\ \text{In}\mathrm{\Delta}\text{MPR,}\\ \angle \text{MRP}=180\mathrm{\xb0}-90\mathrm{\xb0}-\mathrm{\theta}=90\mathrm{\xb0}-\mathrm{\theta}\\ \text{Similarly, in}\mathrm{\Delta}\text{MPQ,}\\ \angle \text{MPQ}=90\mathrm{\xb0}-\angle \text{MPR}=90\mathrm{\xb0}-\mathrm{\theta}\\ \angle \text{MQP}=180\mathrm{\xb0}-90\mathrm{\xb0}-(90\mathrm{\xb0}-\mathrm{\theta})=\mathrm{\theta}\\ \text{Now, in}\mathrm{\Delta}\text{MPR and}\mathrm{\Delta}\text{MQP, we have}\\ \angle \text{MQP}=\angle \text{MPR,}\angle \text{MPQ}=\angle \text{MRP and}\angle \text{PMQ}=\angle \text{PMR.}\\ \text{Therefore, by AAA similarity critarion,}\mathrm{\Delta}\text{MPR}~\mathrm{\Delta}\text{MQP}.\\ \text{Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{QM}}{\text{PM}}=\frac{\text{MP}}{\text{MR}}\\ \Rightarrow {\text{PM}}^{2}=\text{QM}\times \text{MR}\end{array}$

**Q.3 ** ** **

$\begin{array}{l}\text{In the following figure, ABD is a triangle right angled\hspace{0.17em}at A and AC\hspace{0.33em}}\perp \text{\hspace{0.33em}}\mathrm{B}\text{D.}\\ \text{Show that}\\ \left(\text{i}\right){\text{AB}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}}.\text{\hspace{0.33em}BD}\\ \left(\text{ii}\right){\text{AC}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}.\hspace{0.33em}DC}\\ \left(\text{iii}\right){\text{AD}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BD\hspace{0.33em}.\hspace{0.33em}CD}\end{array}$

**Ans.**

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta}\text{ADB and}\mathrm{\Delta}\text{CAB, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{DAB}=\angle \text{ACB}=90\mathrm{\xb0}\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABD}=\angle \text{CBA [Common angle]}\\ \\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta}\text{ADB}~\mathrm{\Delta}\text{CAB}.\\ \text{Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AB}}{\text{CB}}=\frac{\text{BD}}{\text{AB}}\\ \Rightarrow {\text{AB}}^{2}=\text{BC}\cdot \text{BD}\\ \text{(ii)}\\ \text{Let}\angle \text{CAB}=\mathrm{\theta}\\ \text{In}\mathrm{\Delta}\text{CBA, we have}\\ \angle \text{CBA}=180\mathrm{\xb0}-90\mathrm{\xb0}-\mathrm{\theta}=90\mathrm{\xb0}-\mathrm{\theta}\\ \text{In}\mathrm{\Delta}\text{CAD, we have}\\ \angle \text{CAD}=90\mathrm{\xb0}-\angle \text{CAB}=90\mathrm{\xb0}-\mathrm{\theta}\\ \angle \text{CDA}=180\mathrm{\xb0}-90\mathrm{\xb0}-(90\mathrm{\xb0}-\mathrm{\theta})=\mathrm{\theta}\\ \\ \text{In}\mathrm{\Delta}\text{CBA and}\mathrm{\Delta}\text{CAD, we have}\\ \angle \text{CBA}=\angle \text{CAD},\text{\hspace{0.17em}}\angle \text{CAB}=\angle \text{CDA and}\angle \text{ACB}=\angle \text{DCA}=90\mathrm{\xb0}\\ \\ \text{Therefore, by AAA similarity critarion,}\mathrm{\Delta}\text{CBA}~\mathrm{\Delta}\text{CAD}.\\ \text{Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AC}}{\text{DC}}=\frac{\text{BC}}{\text{AC}}\\ \Rightarrow {\text{AC}}^{2}=\text{BC}\cdot \text{DC}\\ \text{(iii)}\\ \text{In}\mathrm{\Delta}\text{DCA and}\mathrm{\Delta}\text{DAB,}\\ \text{}\angle \text{DCA}=\angle \text{DAB}=90\mathrm{\xb0},\text{}\angle \text{CDA}=\angle \text{ADB}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta}\text{DCA}~\mathrm{\Delta}\text{DAB}.\\ \text{Therefore, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{DC}}{\text{DA}}=\frac{\text{DA}}{\text{DB}}\\ \Rightarrow {\text{AD}}^{2}=\text{BD}\cdot \text{CD}\end{array}$

**Q.4 ** ABC is an isosceles triangle right angled at C. Prove that AB^{2} = 2AC^{2}.

**Ans.**

Given that ABC is an isosceles triangle right angled at C.

Therefore, AC = BC

Using Pythagoras theorem in the given triangle,

we have

AB^{2 }= AC^{2} + BC^{2} = AC^{2} + AC^{2 }= 2AC^{2}

**Q.5 ** ABC is an isosceles triangle with AC = BC. If AB^{2} = 2AC^{2}, prove that ABC is a right triangle.

**Ans.**

Given that ABC is an isosceles triangle with AC = BC and AB^{2} = 2AC^{2}.

Therefore,

AB^{2} = 2AC^{2} = AC^{2} + BC^{2}

Therefore, by converse of Pythagoras theorem, ABC is a right triangle.

**Q.6 ** ABC is an equilateral triangle of side 2a. Find each of its altitudes.

**Ans.**

$\begin{array}{l}\text{It is given that ABC is an equilateral triangle of side 2a}.\\ \text{Let AD is an altitude}.\\ \text{We know that altitude bisects opposite side in an}\\ \text{equilateral triangle}.\\ \text{Therefore},\text{BD}=\text{CD}=\text{a}.\\ \text{Using Pythagoras theorem in}\mathrm{\Delta}\text{ADB, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^{2}={\text{BD}}^{2}+{\text{AD}}^{2}\\ \Rightarrow {\text{AD}}^{2}={\text{AB}}^{2}-{\text{BD}}^{2}={\left(2\mathrm{a}\right)}^{2}-{\mathrm{a}}^{2}=4{\mathrm{a}}^{2}-{\mathrm{a}}^{2}=3{\mathrm{a}}^{2}\\ \Rightarrow \text{AD}=\mathrm{a}\sqrt{3}\\ \text{We know that all the altitudes in an equilateral triangle}\\ \text{are of same length. Therefore, length of each altitude is a}\sqrt{3}\text{.}\end{array}$

**Q.7** Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

**Ans.**

$\begin{array}{l}\text{Let ABCD is a rhombus.}\\ \text{We know that diagonals of a rhombus bisect each other}\\ \text{at right angles.}\\ \text{Using Pythagoras Theorem in}\mathrm{\Delta}\text{AOB,}\mathrm{\Delta}\text{BOC,}\mathrm{\Delta}\text{COD and}\mathrm{\Delta}\text{AOD,}\\ \text{we get}\\ {\text{AB}}^{2}={\text{OA}}^{2}+{\text{OB}}^{2}\text{,}\\ {\text{BC}}^{2}={\text{OB}}^{2}+{\text{OC}}^{2},\\ {\text{CD}}^{2}={\text{OC}}^{2}+{\text{OD}}^{2},\\ {\text{DA}}^{2}={\text{OA}}^{2}+{\text{OD}}^{2}\\ \text{Adding all these equations, we get}\\ {\text{AB}}^{2}+{\text{BC}}^{2}+{\text{CD}}^{2}+{\text{DA}}^{2}=2({\text{OA}}^{2}+{\text{OB}}^{2}+{\text{OC}}^{2}+{\text{OD}}^{2})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=2\left[2{\left(\frac{\text{AC}}{2}\right)}^{2}+2{\left(\frac{\text{BD}}{2}\right)}^{2}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{AC}}^{2}+{\text{BD}}^{2}\end{array}$

**Q.8 **

$\begin{array}{l}\text{In\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}figure,\hspace{0.33em}O\hspace{0.33em}is\hspace{0.33em}a\hspace{0.33em}point\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}interior\hspace{0.33em}of\hspace{0.33em}a\hspace{0.33em}triangle\hspace{0.33em}ABC,}\\ \text{OD\hspace{0.33em}}\perp \text{\hspace{0.33em}BC,\hspace{0.33em}OE\hspace{0.33em}}\perp \text{\hspace{0.33em}AC\hspace{0.33em}and\hspace{0.33em}OF\hspace{0.33em}}\perp \text{\hspace{0.33em}AB.\hspace{0.33em}}\\ {\text{(i) OA}}^{\text{2}}{\text{+OB}}^{\text{2}}{\text{+OC}}^{\text{2}}-{\text{\hspace{0.33em}OD}}^{\text{2}}-{\text{OE}}^{\text{2}}-{\text{OF}}^{\text{2}}{\text{\hspace{0.33em}=\hspace{0.33em}AF}}^{\text{2}}{\text{+BD}}^{\text{2}}{\text{+CE}}^{\text{2}}\text{,}\\ {\text{(ii)\hspace{0.33em} AF}}^{\text{2}}{\text{+BD}}^{\text{2}}{\text{+CE}}^{\text{2}}{\text{\hspace{0.33em}=\hspace{0.33em}AE}}^{\text{2}}{\text{+CD}}^{\text{2}}{\text{+BF}}^{\text{2}}\text{.}\end{array}$

**Ans.**

$\begin{array}{l}\text{We join O to A},\text{B and C}.\\ \text{(i)}\\ \text{In}\mathrm{\Delta}\text{AOF,}\\ {\text{OA}}^{2}={\text{OF}}^{2}+{\text{AF}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{In}\mathrm{\Delta}\text{BOD,}\\ {\text{OB}}^{2}={\text{OD}}^{2}+{\text{BD}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{In}\mathrm{\Delta}\text{COE,}\\ {\text{OC}}^{2}={\text{OE}}^{2}+{\text{EC}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{Adding these equations, we get}\\ {\text{OA}}^{2}+{\text{OB}}^{2}+{\text{OC}}^{2}={\text{OF}}^{2}+{\text{AF}}^{\text{2}}+{\text{OD}}^{2}+{\text{BD}}^{\text{2}}+{\text{OE}}^{2}+{\text{EC}}^{\text{2}}\\ \Rightarrow {\text{OA}}^{2}+{\text{OB}}^{2}+{\text{OC}}^{2}-{\text{OD}}^{2}-{\text{OE}}^{2}-{\text{OF}}^{2}={\text{AF}}^{\text{2}}+{\text{BD}}^{\text{2}}+{\text{EC}}^{\text{2}}\\ \text{(ii)}\\ \text{We have form above result,}\\ {\text{AF}}^{\text{2}}+{\text{BD}}^{\text{2}}+{\text{EC}}^{\text{2}}={\text{OA}}^{2}+{\text{OB}}^{2}+{\text{OC}}^{2}-{\text{OD}}^{2}-{\text{OE}}^{2}-{\text{OF}}^{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=({\text{OA}}^{2}-{\text{OE}}^{2})+({\text{OB}}^{2}-{\text{OF}}^{2})+({\text{OC}}^{2}-{\text{OD}}^{2})\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}={\text{AE}}^{2}+{\text{BF}}^{2}+{\text{CD}}^{2}\end{array}$

**Q.9** A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

**Ans.**

$\begin{array}{l}\text{Let AB is a ladder and AC is wall}.\text{A represents the window}\\ \text{and BC is the distance of the foot of the ladder from the}\\ \text{base of the wall}.\\ \text{ABC is a right angle triangle. Therefore, using Pythagoras}\\ \text{theorem, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^{2}={\text{BC}}^{2}+{\text{AC}}^{2}\\ \Rightarrow \text{\hspace{0.17em}}{10}^{2}={\text{BC}}^{2}+{8}^{2}\\ \Rightarrow {\text{BC}}^{2}=100-64=36\\ \Rightarrow \text{BC}=6\\ \text{Therefore, the distance of the foot of the ladder from the}\\ \text{base of the wall is 6 m.}\end{array}$

**Q.10** A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

**Ans.**

$\begin{array}{l}\text{Let AC is a pole and AB is a guy wire with stake at B}.\text{}\\ \text{Then, ABC is a right angle triangle. Therefore, using Pythagoras}\\ \text{theorem, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^{2}={\text{BC}}^{2}+{\text{AC}}^{2}\\ \Rightarrow \text{\hspace{0.17em}}{24}^{2}={\text{BC}}^{2}+{18}^{2}\\ \Rightarrow {\text{BC}}^{2}={24}^{2}-{18}^{2}=576-324=252\\ \Rightarrow \text{BC}=\sqrt{252}=\sqrt{4\times 9\times 7}=6\sqrt{7}\\ \text{Therefore, the distance of the foot of the ladder from the}\\ \text{base of the wall is 6}\sqrt{7}\text{m.}\end{array}$

**Q.11 ** ** **

$\begin{array}{l}\text{An aeroplane leaves an airport and flies due north\hspace{0.17em}at a speed of 1000 km per hour.}\\ \text{At the same time,\hspace{0.33em}another aeroplane leaves the same airport and flies\hspace{0.17em}due west at}\\ \text{a speed of 1200 km per hour. How far\hspace{0.33em}apart will be the two planes after 1}\frac{\text{1}}{\text{2}}\text{\hspace{0.17em}\hspace{0.17em}hours?}\end{array}$

**Ans.**

$\begin{array}{l}\text{Distance travelled by the plane flying towards north in}\\ \text{1}\frac{1}{2}\text{hours}=\text{1}000\times 1\frac{1}{2}\text{km}=1500\text{km}\\ \text{Distance travelled by the plane flying towards west in}\\ \text{1}\frac{1}{2}\text{hours}=\text{12}00\times 1\frac{1}{2}\text{km}=1800\text{km}\\ \text{We represent thsese distances by OA and OB.}\\ \text{Also, AB represents the distance between the two planes.}\\ \text{Using Pythagoras theorem, we get}\\ {\text{AB}}^{2}={\text{OA}}^{2}+{\text{OB}}^{2}={1500}^{2}+{1800}^{2}=22,50,000+32,40,000\\ \Rightarrow {\text{AB}}^{2}=54,90,000\\ \Rightarrow \text{AB}=\sqrt{54,90,000}=\sqrt{9\times 6,10,000}=300\sqrt{61}\\ \text{Therefore, distance between the two planes}=300\sqrt{61}\text{km}\end{array}$

**Q.12 **

$\begin{array}{l}\text{Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet}\\ \text{of the poles is 12 m, find the distance between their tops.}\end{array}$

**Ans.**

$\begin{array}{l}\text{Let AB and CD be two poles of heights 6 m and 11m}\\ \text{respectively}.\text{Let their feet are on the ground at points B}\\ \text{and D}.\text{}\\ \text{It is given that distance between feet is 12 m}.\text{Therefore},\\ \text{BD}=\text{12 m}.\\ \text{Also},\\ \text{AB}=\text{DE}=\text{6m}\\ \text{and}\\ \text{CE}=\text{CD}-\text{DE}=\text{11 m}-\text{6 m}=\text{5 m}\\ \text{Now},\text{AEC is a right triangle}.\text{Therefore},\text{by Pythagoras theorem},\\ \text{AC}=\sqrt{{\text{AE}}^{2}+{\text{CE}}^{2}}=\sqrt{{\text{12}}^{2}+{\text{5}}^{2}}=\sqrt{\text{1}44+2\text{5}}=\sqrt{\text{169}}=13\text{m}\\ \text{Therefore,}\\ \text{Distance between tops of the two poles}=13\text{m}\end{array}$

**Q.13 **

$\begin{array}{l}\text{D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.}\\ \text{Prove that}{\mathrm{AE}}^{2}+{\mathrm{BD}}^{2}={\mathrm{AB}}^{2}+{\mathrm{DE}}^{2}\text{.}\end{array}$

**Ans.**

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ACE,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AE}}^{2}={\text{AC}}^{2}+{\text{CE}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{In}\mathrm{\Delta}\text{DCB,}\\ {\text{BD}}^{2}={\text{CD}}^{2}+{\text{BC}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{On adding (1) and (2), we have}\\ {\text{AE}}^{2}+{\text{BD}}^{2}={\text{AC}}^{2}+{\text{CE}}^{2}+{\text{CD}}^{2}+{\text{BC}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(3\right)\\ \text{In}\mathrm{\Delta}\text{ACB,}\\ {\text{AB}}^{2}={\text{AC}}^{2}+{\text{BC}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(4\right)\\ \text{In}\mathrm{\Delta}\text{DCE,}\\ {\text{DE}}^{2}={\text{CD}}^{2}+{\text{CE}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(5\right)\\ \text{On adding (4) and (5), we have}\\ {\text{AB}}^{2}+{\text{DE}}^{2}={\text{AC}}^{2}+{\text{BC}}^{2}\text{\hspace{0.17em}}+{\text{CD}}^{2}+{\text{CE}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(6\right)\\ \text{From (3) and (6), we find that}\\ {\text{AE}}^{2}+{\text{BD}}^{2}={\text{AB}}^{2}+{\text{DE}}^{2}\end{array}$

**Q.14 **

$\begin{array}{l}\text{The perpendicular from A on side BC of a}\mathrm{\Delta}\text{ABC intersects BC at D such that}\mathrm{DB}=3\mathrm{CD}\text{(see the}\\ \text{following figure). Prove that}2{\mathrm{AB}}^{2}=2{\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}.\end{array}$

**Ans.**

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ADC,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^{2}={\text{CD}}^{2}+{\text{AD}}^{2}\\ \Rightarrow {\text{AD}}^{2}={\text{\hspace{0.17em}AC}}^{2}-{\text{CD}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(1)}\\ \text{In}\mathrm{\Delta}\text{ADB,}\\ {\text{AB}}^{2}={\text{AD}}^{2}+{\text{DB}}^{2}\\ \Rightarrow {\text{AD}}^{2}={\text{AB}}^{2}-{\text{DB}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \text{From (1) and (2), we have}\\ {\text{AC}}^{2}-{\text{CD}}^{2}={\text{AB}}^{2}-{\text{DB}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(3\right)\\ \text{Now, it is given that DB}=\text{3CD}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC}=\text{CD}+\text{DB}\\ \Rightarrow \text{BC}=\text{CD}+\text{3CD}=4\text{CD}\\ \Rightarrow \text{CD}=\frac{\text{BC}}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(4\right)\\ \text{Again,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC}=\text{CD}+\text{DB}\\ \Rightarrow \text{BC}=\frac{\text{DB}}{3}+\text{DB}=\frac{4\text{DB}}{3}\\ \Rightarrow \text{DB}=\frac{3\text{BC}}{4}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(5\right)\\ \text{From (3), (4) and (5), we have}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^{2}-{\text{CD}}^{2}={\text{AB}}^{2}-{\text{DB}}^{2}\\ \Rightarrow {\text{AC}}^{2}-\frac{{\text{BC}}^{2}}{16}\text{\hspace{0.17em}}={\text{\hspace{0.17em}AB}}^{2}-\frac{9{\text{BC}}^{2}}{16}\\ \Rightarrow 16{\text{AC}}^{2}-{\text{BC}}^{2}=16{\text{AB}}^{2}-9{\text{BC}}^{2}\\ \Rightarrow 16{\text{AB}}^{2}=16{\text{AC}}^{2}-{\text{BC}}^{2}+9{\text{BC}}^{2}=16{\text{AC}}^{2}+8{\text{BC}}^{2}\\ \Rightarrow 2{\text{AB}}^{2}=2{\text{AC}}^{2}+{\text{BC}}^{2}\end{array}$

**Q.15 **

**Ans.**

$\begin{array}{l}\text{It is given that ABC is an equilateral triangle.}\\ \text{Let length of each side of}\mathrm{\Delta}\text{ABC be a.}\\ \text{It is given that D is a point on side BC such that}\\ \text{BD}=\frac{1}{3}\text{BC}=\frac{\mathrm{a}}{3}\text{.}\\ \text{Let AE is an altitude of equilateral triangle ABC. So,}\\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}=\frac{\mathrm{a}}{2}\text{.}\\ \text{Also, AE}=\sqrt{{\text{AC}}^{2}-{\text{EC}}^{2}}=\sqrt{{\mathrm{a}}^{2}-\frac{{\mathrm{a}}^{2}}{4}}=\frac{\sqrt{3}}{2}\mathrm{a}\\ \text{Again, DE}=\text{BE}-\text{BD}=\frac{\mathrm{a}}{2}-\frac{\mathrm{a}}{3}=\frac{\mathrm{a}}{6}\\ \text{In}\mathrm{\Delta}\text{ADE,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}}^{2}={\text{AE}}^{2}+{\text{DE}}^{2}\text{[By Pythagoras Theorem]}\\ \Rightarrow {\text{AD}}^{2}={\left(\frac{\sqrt{3}}{2}\mathrm{a}\right)}^{2}+\frac{{\mathrm{a}}^{2}}{36}=\frac{3}{4}{\mathrm{a}}^{2}+\frac{{\mathrm{a}}^{2}}{36}=\frac{27{\mathrm{a}}^{2}+{\mathrm{a}}^{2}}{36}=\frac{28}{36}{\mathrm{a}}^{2}=\frac{7}{9}{\mathrm{a}}^{2}\\ \Rightarrow {\text{AD}}^{2}=\frac{7}{9}{\mathrm{a}}^{2}=\frac{7}{9}{\text{AB}}^{2}\text{[AB}=\text{a]}\\ \Rightarrow 9{\text{AD}}^{2}=7{\text{AB}}^{2}\end{array}$

**Q.16 ** In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

**Ans.**

$\begin{array}{l}\text{Let ABC is an equilateral triangle and AE is its altitude.}\\ \text{Let length of each side of}\mathrm{\Delta}\text{ABC be a.}\\ \text{So,}\\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}=\frac{\mathrm{a}}{2}\text{.}\\ \text{Using Pythagoras theorem in}\mathrm{\Delta}\text{ACE, we have}\\ {\text{AE}}^{2}={\text{AC}}^{2}-{\text{EC}}^{2}={\mathrm{a}}^{2}-\frac{{\mathrm{a}}^{2}}{4}=\frac{3}{4}{\mathrm{a}}^{2}\\ \Rightarrow 3{\mathrm{a}}^{2}=4{\text{AE}}^{2}\\ \text{Therefore, three times the square of one side of an}\\ \text{equilateral triangle is equal to four times the square}\\ \text{of one of its altitude.}\end{array}$

**Q.17 **

$\begin{array}{l}\text{Tick the correct answer and justify: In}\mathrm{\Delta}\text{ABC, AB}=\text{6}\sqrt{\text{3}}\text{cm, AC}=\text{12 cm and BC}=\text{6 cm.}\\ \text{The angle B is :}\\ \text{(A) 120\xb0 \hspace{0.17em} (B) 60\xb0}\\ \text{(C) 90\xb0 (D) 45\xb0}\end{array}$

**Ans.**

$\begin{array}{l}\text{It is given that in}\mathrm{\Delta}\text{ABC},\text{}\\ \text{AB}=\text{6}\sqrt{\text{3}}\text{cm, AC}=\text{12 cm and BC}=\text{6 cm}\\ \Rightarrow {\text{AB}}^{2}={\text{108 cm}}^{2}{\text{, AC}}^{2}={\text{144 cm}}^{2}{\text{and BC}}^{2}=36{\text{cm}}^{2}\\ \text{Now,}\\ {\text{AB}}^{2}+{\text{BC}}^{2}={\text{108 cm}}^{2}+36{\text{cm}}^{2}=144{\text{cm}}^{2}={\text{AC}}^{2}\\ \text{Therefore, by converse of Pythagoras theorem, we find that}\\ \text{in}\mathrm{\Delta}\text{ABC, angle B is 90\xb0.}\\ \text{Therefore, correct answer is (C).}\end{array}$

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