# NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.4

The National Council of Educational Research and Training (NCERT) has been given the responsibility of creating and providing elementary and secondary school students’ textbooks. While discussing the importance of NCERT books for CBSE board exam preparation, it should be mentioned that they emphasize the basics to assist students in understanding the essential ideas. There is no doubt that NCERT texts are comprehensive and successful in their own right. These textbooks are crucial for examinations, but they also aim to enhance classroom instruction. By completing the exercises in the book for self-evaluation and learning the fundamental strategies for resolving challenging situations, students may benefit themselves. Students should access NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 for a better understanding of the exercise.

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.4) Exercise 7.4

CBSE students of Class 10 can easily download solutions of Class 9 Maths Chapter 7 Exercise 7.4. They can download the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 from the Extramarks website or mobile application in PDF format so that they can study NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 even in offline mode.

Exercise 7.4 of Class 9 Chapter 7 – Triangles can be solved with the help of NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4. As a stepping stone for students in Class 10, Class 9 is significant in a student’s academic career. Students would not get hindered in their learning process with regard to any theme if they utilise NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4. Extramarks follows the format used in the NCERT Mathematics book. Students’ academic goals are shaped in the 9th grade. Considering the curriculum of Class 11, students must understand the benefits and drawbacks of the academic stream they strive to select. Ex 7.4 Class 9 Maths NCERT Solutions are the key material for students who are facing problems with regard to the exercise.

Mathematics is very relevant in day-to-day life, and understanding its applications is important. As a result, selecting mathematics as their stream in Class 11 is always the proper choice. As the stream determines a major part of one’s academic future, a student must choose the stream most suited for them. Students need to be quite clear about what they want to pursue after Class 10. Furthermore, Extramarks provides students with support throughout their academic journey.

### Access NCERT Solutions for Class 9 Maths Chapter 7 – Triangles

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### Exercise 7.4

Mathematical problems in Ex 7.4 Class 9 Maths NCERT Solutions are answered using an understanding-based approach. All of the questions from the Triangles chapter have been answered in the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4. Students will also have access to the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 could prove to be indispensable in aiding students in self-evaluation and self-guided learning for forthcoming examinations.

### An Overview: NCERT Maths Class 9 Chapter 7 Exercise 7.4

The theme of Inequalities of the Triangles is discussed in the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 for Class 9 students. Equipped with descriptive examples and sample papers available on the Extramarks learning portal, these solutions would aid in clarifying various topics in a clear and concise manner. These examples should be read in depth by students in order to fully grasp implicit concepts. They will be able to learn how to apply these concepts. Additionally, prior to the practice problems, there are theorems based on triangle-based inequalities.

### How NCERT Solutions for Class 9 Maths Exercise 7.4 Help to Make Exam Preparations?

Students in Class 9 have studied the Congruence of Sides and Angles of a Triangle in the previous exercises of Mathematics Chapter 7. They must have been able to recognise the similarities between the two triangles as a result of this. In the event that two triangles are not identical, there could be a query regarding the ideal method of comparison to be used for comparing them. The concept of triangular inequalities must be understood in order to achieve this goal. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4, provide an excellent explanation of this concept through the solutions of the related exercise. It must be read by students in order to fully comprehend this subject. The theorems based on Triangle Inequalities had been explained before the practice questions.

### List out Some of the Properties of Triangles Maths Class 9 Chapter 7 Exercise 7.4.

Some of the themes that have been mentioned in the chapter have been listed. Students must go through each and every exercise of Chapter 7- Triangles. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 are easily available on the Extramarks website in PDF format.

• Triangles
• Introduction
• Congruence Of Triangles
• Criteria For Congruence Of Triangles
• Some Properties Of A Triangle
• Some More Criteria For Congruence Of Triangles

The properties of Triangles could be written as follows:

• A triangle’s (all varieties) total sum of angles is 180°.
• The length of a triangle’s two longest sides added together is longer than the third side.
• Similar to this, the length of the third side of a triangle third side is shorter than the difference between its two sides.
• The longest of a triangle’s three sides is the side that faces the larger angle.
• A triangle’s exterior angle is always equal to the product of its inner and exterior opposing angles. The outer angle attribute of a triangle refers to this characteristic.
• When the corresponding angles of two triangles matchup and the side lengths are proportional, two triangles are said to be comparable.
• Area of a triangle = ½ × Base × Height
• The perimeter of a triangle = sum of all its three sides

### NCERT Solutions for Class 9

Along with NCERT Solutions for Class 9, students have access to a variety of study resources on the Extramarks digital learning portal. A complete set of course materials is available on the Extramarks website and mobile application. Therefore, it is recommended that students search the Extramarks website for any relevant questions they may have regarding the chapter. In addition, NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4, can be accessed by students. There are notes, important problems, and sample questions available for Class 9 Maths Chapter 7 Exercise 7.4. Students may access questions from previous years and get help with their responses by using the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4. All the other chapters’ NCERT solutions for Class 9 are also available on Extramarks’ website and mobile application.

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate Geometry

Chapter 4 Linear Equations in Two Variables

Chapter 5 Introduction to Euclids Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 9 Areas of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics

Chapter 15 Probability

### CBSE Study Materials for Class 9

Through NCERT Solutions for Class 9, students get access to many different study materials. The Extramarks website and mobile apps both offer access to all course materials. As a result, it is encouraged that students look up any pertinent questions they may have about the chapter on the Extramarks website. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4, are also available to students. Accompanied by Class 9 Maths Chapter 7 Exercise 7.4, they may locate notes, significant issues, and practice questions. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4, provide students with access to past years’ papers and assistance with their answers.

The panel of experts that has developed NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4 addressed all the concerns of students to make sure that all topics were grasped properly and thoroughly. These solutions will provide students with the information they need to address various kinds of questions methodically. Additionally, it will help students perform well on their Mathematics examinations.

### CBSE Study Materials

The nation’s top management and research organisation is NCERT. While it is possible to pass the CBSE exams with sufficient marks by studying the NCERT books, doing so for pupils who adhere to the CBSE board curriculum may occasionally be challenging. It could be difficult for students to finish the tasks in the NCERT Mathematics Book. Students should be made aware that NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.4, are available on Extramarks, where they can also get other quality study resources. Comprehensive solutions for Class 9 Maths Ex 7.4, are available with the aid of Extramarks.

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Q.1 Show that in a right angled triangle, the hypotenuse is the longest side.

Ans

$\begin{array}{l}\text{Given}:\mathrm{In}\text{}\mathrm{\Delta }\text{ABC,}\angle \text{B}=\text{90°}\\ \mathrm{To}\text{prove: AC is the longest side of}\mathrm{\Delta }\text{ABC}.\\ \text{Proof:In\hspace{0.17em}}\mathrm{\Delta }\text{ABC,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{A}+90\mathrm{°}+\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{C}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{C}=90\mathrm{°}\\ \text{Hence},\text{the other two angles have to be acute}\left(\text{i}.\text{e}.,\text{less than 9}0\mathrm{º}\right).\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}\text{is the largest angle of the triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{A}\text{and}\angle \mathrm{B}>\angle \mathrm{C}\\ \therefore \mathrm{AC}>\mathrm{BC}\text{and AC}>\mathrm{A}\text{B}\\ \left[\begin{array}{l}\text{In any triangle},\text{}\\ \text{the side opposite to the larger}\left(\text{greater}\right)\text{angle is longer}.\end{array}\right]\\ \text{Therefore,​ AC is the largest side of the triangle.}\end{array}$

Q.2 In Fig. shown below, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB and AC are produced upto P and Q respectively.}\angle \text{PBC}<\angle \text{QCB.}\\ \text{To prove: AC}>\text{AB}\\ \text{Proof: Since,}\angle \text{PBC}+\angle \text{ABC}=\text{180°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Linear}\text{Pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{PBC}=\text{180°}-\angle \text{ABC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\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}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{QCB}+\angle \text{ACB}=\text{180°\hspace{0.17em}\hspace{0.17em}\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[\mathrm{Linear}\text{Pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{QCB}=\text{180°}-\angle \text{ACB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{It}\text{​​ is given that, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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{PBC}<\angle \text{QCB}\\ \text{180°}-\angle \text{ABC}<\text{180°}-\angle \text{ACB}\\ ⇒\text{\hspace{0.17em}}-\angle \text{ABC}<-\angle \text{ACB}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ACB}<\angle \text{ABC}\\ ⇒\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}AB}<\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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[\begin{array}{l}\text{Side opposite to smaller}\\ \text{angle is smaller.}\end{array}\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}AC}>\text{AB}\\ \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}Hence​ proved.}\end{array}$

Q.3 In Fig. shown, ∠B<∠A and ∠C<∠D. Show that AD
<bc< p=””></bc<>

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{​}\mathrm{\Delta }\text{OAB,}\angle \text{B}<\angle \text{A and in}\mathrm{\Delta }\text{OCD,}\angle \text{C}<\angle \text{D}.\\ \mathrm{To}\text{prove:\hspace{0.17em}AD}<\text{BC}\\ \text{Proof}:\text{\hspace{0.17em}In​}\mathrm{\Delta }\text{OAB,}\\ \text{}\angle \text{B}<\angle \text{A\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ ⇒\mathrm{OA}<\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Side}\text{opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \mathrm{In}\text{​}\mathrm{\Delta }\text{OCD,}\\ \text{}\angle \text{C}<\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ ⇒\mathrm{OD}<\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\text{Side opposite to smaller angle is}\\ \text{smaller.}\end{array}\right]\\ \text{Adding relation}\left(\mathrm{i}\right)\text{​ and}\left(\mathrm{ii}\right)\text{, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}+\mathrm{OD}<\mathrm{OB}+\mathrm{OC}\\ ⇒\mathrm{AD}<\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence Proved.}\end{array}$

Q.4 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure below). Show that ∠A > ∠C and ∠B > ∠D.

Ans

$\begin{array}{l}\mathrm{Given}:\text{\hspace{0.17em}}\mathrm{In}\text{​}\mathrm{quadrilateral}\text{ABCD,}\mathrm{A}\text{B is the smallest side and CD is the largest side.}\\ \mathrm{To}\text{prove:\hspace{0.17em}}\angle \text{A}>\angle \text{C and}\angle \mathrm{B}>\angle \mathrm{D}.\end{array}$

$\begin{array}{l}\text{Proof}:\text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABC,}\\ \text{BC}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}>\angle \text{ACB\hspace{0.17em}}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\end{array}$

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACD,}\\ \text{CD}>\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{CAD}>\angle \text{ACD\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{i}\right)\text{and relation}\left(\mathrm{ii}\right),\text{we get}\\ \angle \text{BAC}+\angle \text{CAD}>\angle \text{ACB}+\angle \text{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAD}>\angle \text{BCD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{A}>\angle \text{C\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{ABD,}\\ \text{AD}>\text{AB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABD}>\angle \text{ADB\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iv}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD,}\\ \text{CD}>\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{CBD}>\angle \text{BDC\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{v}\right)\left[\begin{array}{l}\mathrm{Angle}\text{opposite to greater side is}\\ \text{greater.}\end{array}\right]\\ \mathrm{Adding}\text{relation}\left(\mathrm{iv}\right)\text{and relation}\left(\mathrm{v}\right),\text{we get}\\ \angle \text{ABD}+\angle \text{CBD}>\angle \text{ADB}+\angle \text{BDC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABC}>\angle \text{ADC}\\ ⇒\text{\hspace{0.17em}}\angle \text{B}>\angle \text{D\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{vi}\right)\\ \mathrm{From}\text{relation}\left(\mathrm{iii}\right)\text{​ and relation}\left(\mathrm{vi}\right),\text{we have}\\ \angle \text{A}>\angle \text{C and}\angle \text{B}>\angle \text{D.\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.5 In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans

$\begin{array}{l}\text{\hspace{0.17em}Given}:\mathrm{In}\text{​}\mathrm{\Delta }\text{PQR, PR > PQ and PS bisects}\angle \text{QPR.}\\ \text{To prove:\hspace{0.17em}}\angle \text{PSR}>\angle \text{PSQ}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In}\mathrm{\Delta }\text{PQR},\text{\hspace{0.17em}PS bisects}\angle \text{QPR}\\ \mathrm{i}.\mathrm{e}.,\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}}\angle \text{QPS}=\angle \text{RPS\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{In}\mathrm{\Delta }\text{PQS},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQS}+\angle \mathrm{QPS}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSR}=\angle \mathrm{PQR}+\angle \mathrm{QPS}\\ ⇒\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}}\angle \mathrm{PQR}=\angle \mathrm{PSR}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{In}\text{}\mathrm{\Delta }\text{PRS},\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRS}+\angle \mathrm{SPR}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Ex.}\angle \mathrm{PSQ}=\angle \mathrm{PRQ}+\angle \mathrm{RPS}\\ ⇒\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}}\angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{RPS}\\ ⇒\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}}\angle \mathrm{PRQ}=\angle \mathrm{PSQ}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{From}\text{equation}\left(\mathrm{i}\right)\right]\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PR}>\text{PQ}\\ \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}}\angle \mathrm{PQR}>\angle \mathrm{PRQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angle of greater side}\\ \text{is greater.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{PSR}-\angle \mathrm{QPS}>\angle \mathrm{PSQ}-\angle \mathrm{QPS}\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{From}\text{equation}\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right)\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}}\angle \mathrm{PSR}>\angle \mathrm{PSQ}\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}}\mathrm{Hence}\text{proved.}\end{array}$

Q.6 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans

$\begin{array}{l}\text{Given}:\text{A line l and a point A which is away from line l.}\\ \text{To prove: Perpendicular line segment from A to line l is the shortest.}\\ \text{Contruction: Draw AB}\perp \stackrel{↔}{\mathrm{l}}\text{\hspace{0.17em}},\text{AC and AD.}\\ \text{Proof: In}\mathrm{\Delta ABC},\text{}\angle \text{B=90°}\\ \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}}\mathrm{Then},\text{}\angle \text{A}+\angle \text{C}=\text{90°}\\ ⇒\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}}\angle \mathrm{B}>\angle \mathrm{C}\\ ⇒\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}}\mathrm{AC}>\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Similarly,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}>\angle \mathrm{D}\\ ⇒\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}}\mathrm{AD}>\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Side}\text{opposite to greater angle is greater.}\right]\\ \text{Therefore},\text{it can be observed that of all line segments drawn}\\ \text{from a given point not on it},\text{the perpendicular line segment is}\\ \text{the shortest}.\mathrm{}\end{array}$

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