# NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.2)

Class 9 is one of the most important academic years for students. It is widely believed that Class 9 is more important than Class 10, as in Class 9 foundational concepts are taught, which are very useful with regard to learning the curricula of Class 10, Class 11, and Class 12. Often, due to this reason, students find the curriculum of Class 9 to be difficult. The academic curriculum of Class 9 is very important for students to understand vital topics in the long run. In Class 10, students appear for the board examinations for the first time. If students want to score well in their Class 10 board examinations, they should prepare well for the topics imparted to them in Class 9.

The academic curriculum of Class 9 also helps students figure out which subjects they are most interested in. The subjects taught in Class 9 help students select their future subject stream in Class 11 and Class 12. Students, therefore, must prepare for the annual examinations of Class 9 effectively. Students can explore different subject streams to narrow their interests.

Mathematics is an important subject for students to learn. It enhances their cognitive and reasoning abilities. With the use of mathematical concepts, students develop their ability to think analytically. Mathematics also aids in the development of mental discipline. Mathematics has prevalent relevance in daily life. In some way or another, concepts of Mathematics are used in everyday life. As a result, students start learning Mathematics at an early age.

Mathematical abilities can be useful for pursuing higher education in a variety of disciplines. Its principles can be used in Computer Science, Engineering, Astronomy, Statistics, Data Science, and other fields. Students who are interested in pursuing these careers should focus on the academic curriculum of Class 9 Mathematics. The academic curriculum of Class 9 Mathematics is quite beneficial in understanding the fundamentals of other topics as well.

Students frequently regard Mathematics as challenging. Students must practice a significant number of questions to overcome their inhibitions. After answering a significant number of questions, students will find Mathematics simple. It is recommended that students answer a variety of questions. Solving practice papers can also help students gain confidence in their mathematical abilities. Students must prepare thoroughly for the annual examination of Class 9 Mathematics because it will help them comprehend other subjects as well. Mathematical abilities are required for understanding concepts implicit in many scientific disciplines, including Physics and Chemistry. Mathematics can also be used in subjects such as Business and Statistics. If students are having difficulty with Mathematics, they should devote more time and effort to their studies.

**NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.2) Exercise 9.2 **

Chapter 9 of Class 9 Mathematics is on Areas of Parallelograms and Triangles. Mathematics is a subject that needs a lot of practice. Students can practice the questions of Class 9 Chapter 9 with the aid of NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, are offered by Extramarks for the benefit of students. Students can download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 from Extramarks’ website and mobile application. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, can be downloaded in PDF format for offline access.

With the help of the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 students will be able to solve Class 9 Maths Ex 9.2 thoroughly. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are written in a stepwise manner for the clarity of students. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 also contain figures and diagrams to explain the solutions better.

## Access NCERT Solutions For Class 9 Maths Chapter 9 – Areas of Parallelograms and Triangles

Students can access the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 from the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, are written by expert subject specialists for the benefit of students. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, will help students score well in the Class 9 Maths examination. Students are advised to download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, in PDF format.

**NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.2**

Chapter 9 of Class 9 Maths, “Areas of Parallelograms and Triangles,” is part of the unit Mensuration, which carries a topic weightage of 14 marks. A total of 14 marks will be given for the unit Mensuration, consisting of two multiple-choice questions worth 2 marks each, two short-type questions worth 4 marks each, and one long-type question worth 6 marks.

Areas of Parallelograms and Triangles, Figures On The Same Base And Between The Same Parallels, and Triangles On The Same Base And Between the Same Parallels are all topics covered in Chapter 9: Areas of Parallelograms and Triangles.

Class 9 Maths Chapter 9 Exercise 9.2 deals with the subtopic “Parallelograms on the Same Base and Between the Same Parallels.” This subtopic explains to students the theorem that states that parallelograms on the same base and between the same parallels are equal in area. It is very important for students to understand and retain the theorems and their proofs in Chapter 9 Class 9 Mathematics. Many times, students are asked to prove the theorems in examinations. Moreover, understanding theorems is very important for having a concrete conceptual foundation for the chapter. Students are advised to first understand the theorem properly and then attempt questions based on the theorem. If they find any difficulty in answering the questions, then they can refer to the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, are available on the website and mobile application of Extramarks and can be downloaded in PDF format. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are the best resources to prepare for this chapter.

Exercise 9.2 of Chapter 9 Class 9 Mathematics had five short-type questions and one long-type question. The solutions to all the questions can be found in the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, contain the most accurate answers for all the exercises. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, must be referred to by students.

**NCERT Solutions for Class 9**

Since Class 9 is such an important academic year for students, students should plan their studies effectively. Given below are a few tips for students to prepare for their Class 9 examinations:

- Consolidation of study material is important. There is a lot of study material accessible both online and offline for the academic syllabus of Class 9 Mathematics, but students should only choose the best study material for their preparation. They should select study materials that give them the most simple and accurate solutions. They should also consider whether the study material they choose covers the entire syllabus. Students should always acquire updated versions of study materials because CBSE issues new guidelines each year. The most essential book for students in Class 9 Mathematics should be the Class 9 NCERT textbook. For all CBSE topics, NCERT books are recommended. The NCERT books cover the entire syllabus in depth and provide numerous practice questions of varying difficulty levels for students to use.
- Along with the NCERT books, students should also refer to the study material provided by Extramarks. For instance, to practice questions of Class 9 Chapter 9 Exercise 9.2, students can refer to the Class 9 Maths Exercise 9.2 Solutions provided by Extramarks. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, will benefit students a lot. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, can be downloaded from the website and mobile application of Extramarks. Students can trust the resources provided by Extramarks, as they are expert verified and proofread regularly. For other subjects in the academic curriculum of Class 9, students should use the study materials provided by Extramarks.
- Making a timetable is crucial. When students have a plan and deadlines for completing topics, they are more likely to complete them on time. Topics with a higher weightage and difficulty level may necessitate more study time. As a result, students must schedule time for various topics and chapters. Many students find Mathematics to be a challenging subject, so they should devote more time to practising it. Students in Class 9 Mathematics should devote attention to various chapters and set weekly and monthly goals for themselves. This will allow them to keep track of their progress and guarantee that they are covering the entire course. Chapter 9 on Areas of Parallelograms and Triangles is part of the unit on Mensuration. The unit carries a total of 14 marks. Therefore, students should prepare for this topic thoroughly and devote enough time to it. To prepare for this chapter efficiently, students should use the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, have the most accurate answers for Chapter 9.
- Solving NCERT exercises is a prerequisite. The questions encompassed in the NCERT textbook are the most significant resource for Class 9 school examinations. These questions provide learners with an overview of the types of questions that will be asked in the exams. The exercises included in the NCERT textbooks include questions of varying difficulty levels. Since questions in the CBSE Board exams frequently come directly from these exercises, students must practise the embedded questions. If students encounter any difficulties, they can seek assistance from the resources provided by Extramarks. Extramarks’ resources are trustworthy and legitimate. For Class 9 Chapter 9, students can refer to the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. Students are advised to download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, in PDF format from the website and mobile application of Extramarks.
- Mock tests and Unit tests: Mock tests assist students in self-evaluating their preparation. After completing a chapter, students are encouraged to take unit examinations. They should try to answer questions to see if they understand the chapter properly. Unit exams must be analysed to identify weak areas, which students can then work on. Students should also take full-length mock exams for the term-end exam. Comprehensive mock examinations will assist them in efficiently managing their time during examinations. Mock tests will also show students how much information they can retain at one time. Students can use the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, to answer mock examination questions. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, are available on the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, can be downloaded in PDF format for easy access.
- Revision is mandatory. The most critical aspect of examination preparation is revision. Revision assists students in remembering a large amount of material for a longer period of time. With adequate revision, students should be able to recall the majority of the formulas from their Class 9 Mathematics textbooks. Revision must be done on a frequent basis. Students should go over their brief notes again, as well as the mock tests and unit tests they took. Students will develop confidence after several revisions. The greatest technique for revising Mathematics is to practice questions on a regular basis. Regularly practising questions will reduce the time it takes students to solve each question, allowing them to finish their papers on time. Students should regularly revise the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are very useful for effective preparation for Class 9 Chapter 9.
- CBSE sample papers and past years’ papers must be practised: CBSE sample papers are really useful to students. Every year, before the board exams, CBSE publishes sample papers. The sample papers are very important because they are updated in accordance with the latest CBSE criteria for examination. Students must complete CBSE sample papers. Along with sample papers, students must also solve past years’ papers on various subjects. Past years’ papers give students an insight as to what kinds of questions might be on the question paper during the examination. If students face any challenges in solving sample papers or past years’ papers, they can refer to the study resources provided by Extramarks. For Class 9 Chapter 9, students can refer to the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2
- Short notes could be of incredible assistance. Short notes help in revising formulas and theorems before the examination. Students must consolidate all the formulas and make short notes out of them. This will help them revise all important formulas and their applications before the examination, when students face a paucity of time. Students should revise the formulas as many times as they can. Students can note down all the formulas and important theorems of this chapter in the form of formulas from the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, are provided by Extramarks so that students can score well in their exams.
- Time for extracurricular activities is crucial. Extracurricular activities are an essential component of academic life. Students must participate in extracurricular activities. They assist students in developing practical skills and rejuvenating their minds. As a result, students must commit time to extracurricular activities. Extracurricular activities contribute to students’ overall personality development. Students can participate in activities such as athletics, discussions, theatrical productions, dance, and so on.

**CBSE Study Materials for Class 9**

The core subjects in Class 9 are Mathematics, Social Science, Science, English, and Hindi or Sanskrit. All of the subjects are necessary for passing the Class 9 annual exams. The Extramarks website and Learning App provide access to study resources for all subjects in Class 9. Based on their needs, students can obtain revision notes, past years’ papers and their solutions, and NCERT solutions. All of the resources will help them comprehend the subjects better and answer questions from the NCERT textbooks. Students should save the resources in PDF format for later use. Students can improve their grades by utilising the Extramarks resources. Students must read the resources frequently.

**CBSE Study Materials**

All CBSE-affiliated schools recommend NCERT books. Students should consider NCERT books as their primary source of preparation. Along with NCERT books, students can use the study resources provided by Extramarks. Extramarks provides study material in both Hindi and English. The resources provided by Extramarks are very beneficial for students as they encapsulate the entire syllabus holistically.

**Q.1** In the given figure, ABCD is parallelogram, AE ⊥DC and CF ⊥AD.

If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.

**Ans**

$\begin{array}{l}\text{In parallelogram ABCD},\text{CD}=\text{AB}=\text{16 cm}\\ \mathrm{}\left[\text{Opposite sides of a parallelogram are equal}\right]\mathrm{}\\ \text{We know that}\mathrm{}\text{Area of a parallelogram}=\text{Base}\times \text{Corresponding altitude}\\ \mathrm{}\text{Area of parallelogram ABCD}=\text{CD}\times \text{AE}=\text{AD}\times \text{CF}\\ \mathrm{}\text{16 cm}\times \text{8 cm}=\text{AD}\times \text{1}0\text{cm}\\ \mathrm{AD}=\frac{16\times 8}{10}=12.8\text{\hspace{0.17em}}\mathrm{m}\\ \text{Thus},\text{\hspace{0.17em}the\hspace{0.17em}length\hspace{0.17em}of\hspace{0.17em}}\mathrm{AD}\text{\hspace{0.17em}}\mathrm{is}\text{\hspace{0.17em}}2.8\text{\hspace{0.17em}}\mathrm{cm}.\end{array}$

**Q.2 **

$\begin{array}{l}\mathrm{If}\mathrm{E},\mathrm{F},\mathrm{G}\mathrm{and}\mathrm{H}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}\u2013\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{sides}\\ \mathrm{ofa}\mathrm{parallelogram}\mathrm{ABCD}\mathrm{show}\mathrm{that}\mathrm{ar}\left(\mathrm{EFGH}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right).\end{array}$

**Ans**

$\begin{array}{l}\text{Given}:\text{ABCD is a parallelogram in which E, F, G and H are}\\ \text{mid-points of AB, BC, CD and DA respectively}\text{.}\\ \text{To Prove:}ar\left(EFGH\right)=\frac{1}{2}ar\left(ABCD\right)\\ \text{Contruction}:\text{Join HF}\text{.}\\ \text{Proof: Since, ABCD is a parallelogram}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So, AD}\parallel BC\text{and AD}=\text{BC}\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{2}AD=\frac{1}{2}BC\text{and AD}\parallel BC\\ \Rightarrow \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}AH=BF\text{and AH}\parallel BF\\ \text{So, ABFH is a parallelogram}\text{.}\\ \text{Then, AB}\parallel HF.\\ \text{Since}\Delta \text{HEF and parallelogram ABFH are on the same base}\\ \text{HF and between the same}\text{parallel lines AB and HF},\\ \therefore \text{Area (}\Delta \text{HEF)}=\frac{1}{2}\text{Area (ABFH})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\text{\hspace{0.17em}}\left(i\right)\\ \text{Similarly, it can be proved that}\\ \therefore \text{Area (}\Delta \text{HGF)}=\frac{1}{2}\text{Area (DCFH})\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\text{\hspace{0.17em}}\left(ii\right)\\ \text{On adding equations (1) and (2), we get}\\ \text{Area (}\Delta \text{HEF)}+\text{Area (}\Delta \text{HGF)}=\frac{1}{2}\text{Area (ABFH})+\frac{1}{2}\text{Area (DCFH})\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left\{\text{Area (ABFH})+\text{Area (DCFH})\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area (}\Delta \text{EFGH)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{Area (ABCD})\end{array}$

**Q.3 **P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC).

**Ans**

Given: ABCD is a parallelogram. Points P and Q lies on sides CD and AD respectively. To prove: area(BQC) = area(APB) Proof: It can be observed that ∆BQC and parallelogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC. So, area(BQC) = (1/2) area ABCD …(i) Similarly, ∆APB and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.

So, area(APB) = (1/2) area ABCD …(ii) From equation (i) and equation(ii), we have area(BQC) = area(APB). Hence proved.

**Q.4 **

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{P}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{in}\mathrm{the}\mathrm{interior}\mathrm{of}\mathrm{a}\\ \mathrm{parallelogram}\mathrm{ABCD}.\mathrm{Show}\mathrm{that}\mathrm{}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{APB}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)\mathrm{}\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{APD}\right)+\mathrm{ar}\left(\mathrm{PBC}\right)=\mathrm{ar}\left(\mathrm{APB}\right)+\mathrm{ar}\left(\mathrm{PCD}\right)\mathrm{}\end{array}$

**Ans**

$\begin{array}{l}\text{Given}:ABCD\text{is a parallelogram}\text{. P is any point inside of it}\text{.}\\ \mathrm{To}\text{prove:}\left(i\right)ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(ii\right)ar\left(APD\right)+ar\left(PBC\right)=ar\left(APB\right)+ar\left(PCD\right)\\ Construction:\text{}\text{}Draw\text{EF}\parallel \text{AB and QR}\parallel AD\text{through point P}\text{.}\\ \text{Proof: In parallelogram ABCD,}\\ \text{AD}\parallel BC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Opposite\text{sides of parallelogram}\text{.}\right]\\ \Rightarrow \text{AE}\parallel BF\\ \text{and\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AB}\parallel EF\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{construction}\right]\\ \text{Thus},\text{ABFE is a prallelogram}\text{.}\\ \text{Since,}\Delta \text{APB and parallelogram ABFE are lying on the same base}\\ \text{AB and between the same parallel lines AB and EF}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}area}\left(\Delta \text{APB}\right)=\frac{1}{2}area\left(ABFE\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(i\right)\\ \text{Now EF}\parallel CD\text{because EF}\parallel AB\text{and AB}\parallel \text{CD}\text{.}\\ \text{So,}\Delta \text{DPC and parallelogram ABFE are lying on the same base}\\ \text{DC and between the same parallel lines AB and EF}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{DPC}\right)=\frac{1}{2}area\left(ABFE\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(ii\right)\\ \text{Adding equation}\left(i\right)\text{and equation}\left(ii\right),\text{we get}\\ area\left(\Delta \text{APB}\right)+area\left(\Delta \text{DPC}\right)=\frac{1}{2}area\left(ABFE\right)+\frac{1}{2}area\left(ABFE\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}area\left(ABCD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(A\right)\\ \left(ii\right)\text{In parallelogram ABCD,}\\ \text{AB}\parallel DC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{Opposite sides of parallelogram}\text{.}\right]\\ \Rightarrow \text{AQ}\parallel DR\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{AD}\parallel QR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{construction}\right]\\ \text{Thus},\text{AQRD is a prallelogram}\text{.}\\ \text{Since,}\Delta \text{APD and parallelogram AQRD are lying on the same base}\\ \text{AD and between the same parallel lines AD and QR}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{APD}\right)=\frac{1}{2}area\left(AQRD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(iii\right)\\ \text{Now QR}\parallel BC\text{because QR}\parallel BC\text{and AD}\parallel B\text{C}\text{.}\\ \text{So,}\Delta \text{BPC and parallelogram QBCR are lying on the same base}\\ \text{BC and between the same parallel lines QR and BC}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}area\left(\Delta \text{BPC}\right)=\frac{1}{2}area\left(\text{QBCR}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(iv\right)\\ \text{Adding equation}\left(iii\right)\text{and equation}\left(iv\right),\text{we get}\\ \text{area}\left(\Delta \text{APD}\right)+area\left(\Delta \text{BPC}\right)=\frac{1}{2}area\left(AQRD\right)+\frac{1}{2}area\left(\text{QBCR}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}area\left(ABCD\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(B\right)\\ \text{From equatin}\left(A\right)\text{and}\left(B\right),\text{we have}\\ \text{area}\left(\Delta \text{APB}\right)+area\left(\Delta \text{DPC}\right)=area\left(\Delta \text{APD}\right)+area\left(\Delta \text{BPC}\right).\end{array}$

**Q.5 **

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{given}\mathrm{figure},\mathrm{PQRS}\mathrm{and}\mathrm{ABRS}\mathrm{are}\mathrm{parallelograms}\mathrm{and}\\ \mathrm{X}\mathrm{is}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{side}\mathrm{}\mathrm{BR}.\mathrm{Show}\mathrm{that}\mathrm{}\\ \left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PQRS}\right)=\mathrm{ar}\left(\mathrm{ABRS}\right)\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{\Delta PXS}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{PQRS}\right)\end{array}$

**Ans**

$\begin{array}{l}\text{Given:}\mathrm{PQRS}\mathrm{and}\mathrm{ABRS}\mathrm{are}\mathrm{parallelograms}\mathrm{and}\mathrm{X}\mathrm{is}\mathrm{any}\mathrm{point}\mathrm{on}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \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{side}\text{\hspace{0.17em}}\mathrm{}\mathrm{BR}.\\ \text{To prove:\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{ar}\left(\mathrm{PQRS}\right)=\mathrm{ar}\left(\mathrm{ABRS}\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}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta PXS}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{PQRS}\right)\\ \text{Proof:}\left(\mathrm{i}\right)\text{Since, the parallelogram PQRS and ABRS lie on the}\\ \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}same base SR}\mathrm{}\text{and also},\text{these lie in between the same}\\ \text{parallel lines SR and PB}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area}\left(\text{PQRS}\right)\text{}=\text{Area}\left(\text{ABRS}\right)\text{}...\text{}\left(\text{i}\right)\mathrm{}\\ \left(\mathrm{ii}\right)\text{Since,}\mathrm{\Delta}\text{AXS and parallelogram ABRS lie on the same base}\\ \text{and are between the same parallel lines AS and BR.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta AXS}\right)=\frac{1}{2}\mathrm{area}\left(\mathrm{ABRS}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{}\mathrm{we}\text{have}\\ \text{\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta AXS}\right)=\frac{1}{2}\mathrm{area}\left(\text{PQRS}\right).\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

**Q.6** A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided? What are the shapes of these parts? The farmer wants to sow wheat and pulses in equal portions of the field separately. How should she do it?

**Ans**

$\begin{array}{l}\text{In the figure},\text{point A divides the field into three parts}.\\ \text{These parts are triangular in shape:}\Delta \text{PSA},\text{}\Delta \text{PAQ},\text{and}\Delta S\text{RA}\\ \text{Area of}\Delta \text{PSA}+\text{Area of}\Delta \text{PAQ}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}\mathrm{\dots}\text{}\left(\text{i}\right)\\ \text{We know that if a parallelogram and a triangle are on}\text{the same}\\ \text{base and between the same parallels},\text{then the area of the}\\ \text{triangle is half the area of the}\text{parallelogram}.\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Area}\left(\Delta \text{PAQ}\right)=\frac{\text{1}}{2}\text{Area}\left(\text{PQRS}\right)\text{}\text{\hspace{0.17em}}\mathrm{\dots}\text{}\left(\text{ii}\right)\\ \text{From equation}\left(\text{i}\right)\text{and equation}\left(\text{ii}\right),\text{we get}\\ \text{Area of}\Delta \text{PSA}+\frac{1}{2}\text{Area of}\Delta \text{PAQ}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}\\ \text{Area of}\Delta \text{PSA}+\text{Area of}\Delta S\text{RA}=\text{Area of PQRS}-\frac{1}{2}\text{Area of}\Delta \text{PAQ}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\text{Area of}\Delta \text{PAQ}\\ \text{Therefore},we\text{can say that the farmer must sow wheat in}\\ \text{triangular part PAQ and pulses in other two triangular parts PSA}\\ \text{and SRA or wheat in triangular parts PSA and SRA and pulses in}\\ \text{triangular parts PAQ}.\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Is Class 9 Mathematics challenging?

It is true that the curriculum of Class 9 has an increased difficulty level in all subjects but students can make it easier for themselves by studying regularly and practising various kinds of questions. The curriculum of Class 9 Mathematics must also be prepared in this manner. Students can solve exercises given within the textbook with the help of resources provided by Extramarks. To solve the questions of Chapter 9 Class 9, students can take the help of the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2.

### 2. From where can students download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2?

Students can download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 from the website and mobile application of Extramarks.

### 3. Can the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 be downloaded in PDF format?

Yes, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are available in PDF format. Students are advised to download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 in PDF format for convenient access.

### 4. Is the Class 9 Mathematics syllabus important for Class 10 board examinations?

Yes, the Class 9 Mathematics syllabus is very important for Class 10 board examinations as students learn the foundations of various mathematical themes in Class 9. If they understand the basics well they will be able to learn the more complex topics easily.

### 5. Will the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 help students to solve mock papers?

Yes, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 will help students in solving the mock examination questions. Students must download the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 for preparing the chapter efficiently.

### 6. Are Extramarks’ study materials reliable?

Yes, the study resources provided by Extramarks are reliable and authentic. The study resources provided by Extramarks cover the whole syllabus holistically and can be used by students to prepare for their exams. Further, the study materials are updated regularly according to the latest CBSE guidelines.

### 7. Are solutions of other Chapter 9 Exercises also provided by Extramarks?

Yes, along with the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, solutions for Exercise 9.1, Exercise 9.3, and Exercise 9.4 are available on the website and mobile application of Extramarks for the benefit of students. Students can refer to the solutions in case of any challenges.

### 8. Are the Class 9 Maths Exercise 9.2 Solutions also provided in Hindi by Extramarks?

Extramarks does provide study materials in Hindi for students. They can visit the website and mobile application to download the study material in Hindi.

### 9. Are Areas of Parallelograms and Triangles Class 9 Exercise 9.2 Solutions easy to understand?

Yes, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 provided by Extramarks are written in an easy-to-understand language for the benefit of students. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 must be downloaded from the website and mobile application of Extramarks. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 can be downloaded in PDF format.