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Case Study Questions for Class 10 Maths Chapter 7 Coordinate Geometry

• Last modified on: 1 year ago
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Case Study Questions:

Question 1:

The top of a table is shown in the figure given below:

(i) The coordinates of the points H and G are respectively (a) (1, 5), (5, 1) (b) (0, 5), (5, 0) (c) (1, 5), (5, 0) (d) (5, 1), (1, 5)

(ii) The distance between the points A and B is (a) 4 units (b) 4 2 units (c) 16 units (d) 32 units

(iii) The coordinates of the mid point of line segment joining points M and Q are (a) (9, 3) (b) (5, 11) (c) (14, 14) (d) (7, 7)

(iv) Which among the following have same ordinate? (a) H and A (b) T and O (c) R and M (d) N and R

(v) If G is taken as the origin, and x, y axis put along GF and GB, then the point denoted by coordinate (4, 2) is (a) H (b) F (c) Q (d) R

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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Class 10 Maths Chapter 7 Case Based Questions - Coordinate Geometry

Study Case - 1

The class X students of a school in Rajinder Nagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Mango are planted on the boundary at the distance of 1 m from each other.

Based on the above figure, answer the following questions:

  Q4: What will be the coordinates of the vertices of ΔPQR if C is the origin? (a) (14, 3), (11, 2), (8, 6) (b) (15, 4), (12, 3), (9, 7) (c) (14, 2), (11, 1), (8, 5) (d) (15, 3), (12, 2), (9, 6) Ans:  (d) Explanation:  When C is taken as origin, we will take CB as X-axis and CD as Y-axis. Then, coordinates of points, P, Q and R are (15, 3), (12, 2)and (9, 6) respectively.

Q5: What are the coordinates of P if D is taken as the origin? (a) (−15, 5) (b) (15, 5) (c) (15, 7) (d) (15, 3) Ans:  (a) Explanation:  When D is taken as origin, we will take DA as negative X-axis and DC as positive Y-axis.   Then coordinates of point P is (−15, 5).

Case Study - 2

Based on the above information, give the answer of the following questions: 

Study Case - 3

Based on the above information give the answer of the following questions:

Q1: The coordinates of the point A are: (a) (4,9) (b) (5,9) (c) (92,9) (d) (4,8) Ans: (c) Explanation:  As the distance of point A is 9/2 units from the Y-axis and 9 units from the X-axis, its coordinates are x = 9/2, y = 9 or (9/2, 9)

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Case Study on Coordinate Geometry Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Coordinate Geometry Class 10 Maths can use this page to download the PDF file. 

The case study questions on Coordinate Geometry are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Coordinate Geometry case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Coordinate Geometry Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Coordinate Geometry, therefore, they prepared a set of solutions along with the case study questions on Coordinate Geometry.

The case study on Coordinate Geometry Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Coordinate Geometry case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Coordinate Geometry Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Coordinate Geometry case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Coordinate Geometry Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Coordinate Geometry case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Coordinate Geometry Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Coordinate Geometry as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Coordinate Geometry?

Students can choose their own way to answer Case Study on Coordinate Geometry Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Coordinate Geometry Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Coordinate Geometry questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Coordinate Geometry Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Coordinate Geometry?

 A few essential things to know to solve Case Study Questions on Class 10 Coordinate Geometry are -

• Basic Formulas of Coordinate Geometry: One of the most important things to know to solve Case Study Questions on Class 10 Coordinate Geometry is to learn about the basic formulas or revise them before solving the case-based questions on Coordinate Geometry.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Coordinate Geometry case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Coordinate Geometry Class 10 Maths?

Use Selfstudys.com to find Case Study on Coordinate Geometry Class 10 Maths. For ease, here is a step-wise procedure to download the Coordinate Geometry Case Study for Class 10 Maths in PDF for free of cost.

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Case Based Questions (MCQ)

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Question 2 - Case Based Questions (MCQ) - Chapter 7 Class 10 Coordinate Geometry

Last updated at April 16, 2024 by Teachoo

The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

This question is inspired from   Ex 7.4, 5 (Optional) - Chapter 7 Class 10 - Coordinate Geometry   Question 18 - CBSE Class 10 - Sample Paper for 2021 Board

Taking A as origin, find the coordinates of P (a) (4, 6) (b) (6, 4) (c) (0, 6) (d) (4, 0)

What will be the coordinates of R, if C is the origin? (a) (8, 6) (b) (3, 10) (c) (10, 3) (d) (0, 6)

What will be the coordinates of Q, if C is the origin? (a) (6, 13) (b) (–6, 13) (c) (–13, 6) (d) (13, 6)

Calculate the area of the triangles if A is the origin (a) 4.5 (b) 6 (c) 8 (d) 6.25

Calculate the area of the triangles if C is the origin (a) 8 (b) 5 (c) 6.25 (d) 4.5

Question The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot. Marking points on the image Niharika runs 1/4th of the distance AD on 2nd line So, Niharika’s x−cordinate = 2 Niharika’ y – cordinate = 1/4 × 100 = 25 ∴ Coordinates of Niharika = G (2, 25) Question 1 Taking A as origin, find the coordinates of P (a) (4, 6) (b) (6, 4) (c) (0, 6) (d) (4, 0) If A is the Origin Coordinates of point P = (4, 6) Coordinates of point Q = (3, 2) Coordinates of point R = (6, 5) So, the correct answer is (a) Question 2 What will be the coordinates of R, if C is the origin? (a) (8, 6) (b) (3, 10) (c) (10, 3) (d) (0, 6) If C is the Origin (and taking downward as positive) Coordinates of point P = (12, 2) Coordinates of point Q = (13, 6) Coordinates of point R = (10, 3) So, the correct answer is (c) Question 3 What will be the coordinates of Q, if C is the origin? (a) (6, 13) (b) (–6, 13) (c) (–13, 6) (d) (13, 6) Coordinates of point Q = (13, 6) So, the correct answer is (d) Question 4 Calculate the area of the triangles if A is the origin (a) 4.5 (b) 6 (c) 8 (d) 6.25 If A is the origin, We need to find Area of Δ PQR where P(4, 6) , Q(3, 2) and R(6, 5) Area of ∆PQR with vertices P (4, 6), Q (3, 2) & R (6, 5) Here, 𝒙_𝟏=4 , 〖 𝒚〗_𝟏=6 𝒙_𝟐=3 , 𝒚_𝟐=2 𝒙_𝟑=6 , 𝒚_𝟑=5 Now, ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 𝟏/𝟐 [𝟒 (𝟐−𝟓)+𝟑 (𝟓−𝟔)+ 𝟔 (𝟔−𝟐)] = 1/2 [4 (−3)+3 (−1)+6 (4)] = 1/2 [−12−3+24] = 1/2 × 9 = 4.5 sq. units So, the correct answer is (a) Question 5 Calculate the area of the triangles if C is the origin (a) 8 (b) 5 (c) 6.25 (d) 4.5 If C is the origin, We need to find Area of Δ PQR where P(12, 2) , Q(13, 6) and R(10, 3) Area of ∆PQR with vertices P (12, 2), Q (13, 6) & R (10, 3) Here, 𝒙_𝟏=12 , 𝒚_𝟏=2 𝒙_𝟐=13 , 𝒚_𝟐=6 𝒙_𝟑=10 , 𝒚_𝟑=3 Now, ar (∆PQR) = 1/2 [𝑥_1 (𝑦_2−𝑦_3 )+ 𝑥_2 (𝑦_3−𝑦_1 )+𝑥_3 (𝑦_1−𝑦_2 )] = 𝟏/𝟐 [𝟏𝟐(𝟔−𝟑)+𝟏𝟑(𝟑−𝟐)+𝟏𝟎(𝟐−𝟔)] = 1/2 [12 × 3+13 × 1+10(−4)] = 1/2 [36+13−40] = 1/2 × 9 = 4.5 sq. units So, the correct answer is (d)

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

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First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
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• Coordinate Geometry Case Study Question
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• Some Applications of Trigonometry Case Study Question
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Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry

August 6, 2019 by Sastry CBSE

Coordinate Geometry Class 10 Important Questions Very Short Answer (1 Mark)

Question 2. If the points A(x, 2), B(-3, 4) and C(7, -5) are collinear, then find the value of x. (2014D) Solution: When the points are collinear, x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 ) = 0 x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0 x(1) + 21 + 42 = 0 x + 63 = 0 ∴ x = -63

Question 3. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.? (2016OD) Solution: As we know, a 2 – a 1 = a 3 – a 2 2k – 1 – (k + 9) = 2k + 7 – (2k – 1) 2k – 1 – k – 9 = 2k + 7 – 2k + 1 k – 10 = 8 ∴ k = 8 + 10 = 18

Coordinate Geometry Class 10 Important Questions Short Answer-I (2 Marks)

Question 7. Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7). (2011D) Solution: Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). ∴ AP = BP …[Given AP 2 = BP 2 …[Squaring both sides (x – 2) 2 + (y – 5) 2 = (x + 3) 2 + (y – 7) 2 ⇒ x 2 – 4x + 4 + y 2 – 10y + 25 ⇒ x 2 + 6x + 9 + y 2 – 14y + 49 ⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25 ⇒ -10x + 4y = 29 ∴ 10x + 29 = 4y is the required relation.

Question 8. Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear. (2015OD) Solution: When points are collinear, ∴ Area of ∆ABC = 0 = (x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )) = 0 = x (7 – 5) – 5 (5 – y) -4 (y – 7) = 0 = 2x – 25 + 5y – 4y + 28 = 0 ∴ 2x + y + 3 = 0 is the required relation.

Question 15. If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y. (2015OD) Solution: We have A(4, 3), B(-1, y) and C(3, 4). In right angled triangle ABC, (BC) 2 = (AB) 2 + (AC)…. [Pythagoras theorem ⇒ (-1 – 3) 2 + (y – 4) 2 = (4 + 1) 2 + (3 – y) 2 + (4 – 3) 2 + (3 – 4) 2 …(using distance formula ⇒ (-4) 2 + (y 2 – 8y + 16) ⇒ (5) 2 + (9 – 6y + y 2 ) + (1) 2 + (-1) 2 ⇒ y 2 – 8y + 32 = y 2 – 6y + 36 = 0 ⇒ -8y + 6y + 32 – 36 ⇒ -2y – 4 = 0 ⇒ -2y = 4 ∴ y = -2

Question 17. Find the area of the triangle whose vertices are (1, 2), (3, 7) and (5, 3).(2011OD) Solution: Area of Triangle = \(\frac{1}{2}\) [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] = \(\frac{1}{2}\)(1(7 – 3) + 3(3 – 2) + 5(2 – 7)] = \(\frac{1}{2}\)[4 + 3 + 5(-5)] = \(\frac{18}{2}\) = 9 sq. unit

Question 18. The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p. (2015OD) Solution: Similar to Question 16, Page 102.

Coordinate Geometry Class 10 Important Questions Short Answer-II (3 Marks)

Question 20. Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units. (2011D) Solution: PQ = 10 …Given PQ 2 = 10 2 = 100 … [Squaring both sides (9 – x) 2 + (10 – 4) 2 = 100…(using distance formula (9 – x) 2 + 36 = 100 (9 – x) 2 = 100 – 36 = 64 (9 – x) = ± 8 …[Taking square-root on both sides 9 – x = 8 or 9 – x = -8 9 – 8 = x or 9+ 8 = x x = 1 or x = 17

Question 21. Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units. (2011OD) Solution: AB = 10 units … [Given AB 2 = 10 2 = 100 … [Squaring both sides (11 – 3) 2 + (y + 1) 2 = 100 8 2 + (y + 1) 2 = 100 (y + 1) 2 = 100 – 64 = 36 y + 1 = ±6 … [Taking square-root on both sides y = -1 ± 6 ∴ y = -7 or 5

Question 22. The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. (2011D) Solution: PA = QA …[Given PA 2 = QA 2 … [Squaring both sides (3 – 6) 2 + (y – 5) 2 = (3 – 0) 2 + (y + 3) 2 9 + (y – 5) 2 = 9 + (y + 3) 2 (y – 5) 2 = (y + 3) 2 y – 5 = ±(y + 3) … [Taking sq. root of both sides y – 5 = y + 3 y – 5 = -y – 3 0 = 8 … which is not possible ∴ y = 1

Question 23. If a point A(0, 2) is equidistant from the points B(3, p) and C(p, 5), then find the value of p. (2012D) Solution: Similar to Question 24, Page 103.

Question 25. If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k. (2014OD) Solution: PA = PB …Given PA 2 = PB 2 … [Squaring both sides ⇒ (k – 1 – 3) 2 + (2 – k) 2 = (k – 1 – k) 2 + (2 – 5) 2 ⇒ (k – 4) 2 + (2 – k) 2 = (-1) 2 + (-3) 2 k 2 – 8k + 16 + 4 + k 2 – 4k = 1 + 9 2k 2 – 12k + 20 – 10 = 0 2k 2 – 12k + 10 = 0 ⇒ k 2 – 6k + 5 = 0 …[Dividing by ⇒ k 2 – 5k – k + 5 = 0 ⇒ k(k – 5) – 1(k – 5) = 0 ⇒ (k – 5) (k – 1) = 0 ⇒ k – 5 = 0 or k – 1 = 0 ∴ k = 5 or k = 1

Question 27. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay. (2016OD) Solution: PA = PB … [Given PA 2 = PB 2 … [Squaring both sides ⇒ [(a + b) – x] 2 + [(b a) – y) 2 = [(a – b) – x] 2 + [(a + b) – y] 2 ⇒ (a + b) 2 + x 2 – 2(a + b)x + (b – a) 2 + y 2 – 2(b – a)y = (a – b) 2 + x 2 – 2(a – b)x + (a + b) 2 + y 2 – 2(a + b)y …[∵ (a – b) 2 = (b – a) 2 ⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y ⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a) ⇒ -2bx = – 2ay ⇒ bx = ay (Hence proved)

Question 29. If the point P(2, 2) is equidistant from the points A(-2, k) and B(-2k, -3), find k. Also find the length of AP. (2014D) Solution: PA = PB … [Given PA 2 = PB 2 … [Squaring both sides ⇒ (2 + 2) 2 + (2 – k) 2 = (2 + 2k) 2 + (2 + 3) 2 16 + 4 + k 2 – 4k = 4 + 4k 2 + 8k + 25 4k 2 + 8k + 25 – k 2 + 4k – 16 = 0 3k 2 + 12k + 9 = 0 ⇒ k 2 + 4k + 3 = 0 …[Dividing by 3 k 2 + 3k + k + 3 = 0 k(k + 3) + 10k + 3) = 0 (k + 1) (k + 3) = 0 k + 1 = 0 ork + 3 = 0 i k = -1 or k = -3 AP = \(\sqrt{(2+2)^{2}+(2-k)^{2}}\) When k = -1 AP =\(\sqrt{16+(2+1)^{2}}=\sqrt{16+9}\) = 5 units When k = -3 AP = \(\sqrt{16+(2+3)^{2}}=\sqrt{16+25}\) units

Question 30. Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right angled isosceles triangle. (2016OD) Solution: Similar to Question 19, Page 102.

Question 32. Show that the points (-2, 3), (8, 3) and (6, 7) are the vertices of a right triangle. (2013D) Solution: Let A (-2, 3), B(8,3), C(6, 7). (AB) 2 = (8 + 2) 2 + (3 – 3) 2 = 10 2 + 0 2 = 100 (BC) 2 = (6 – 8) 2 + (7 – 3) 2 = (-2) 2 + 4 2 = 20 (AC) 2 = (6 + 2) 2 + (7 – 3) 2 = 8 2 + 4 2 = 80 Now, (BC) 2 + (AC) 2 = 20 + 80 = 100 = (AB) 2 …[By converse of Pythagoras’ theorem Therefore, Points A, B, C are the vertices of a right triangle.

Question 36. Find that value of k for which point (0, 2), is equidistant from two points (3, k) and (k, 5). (2013OD) Solution: Similar to Question 29, Page 104.

Question 37. If the points A(-2, 1), B(a, b) and C(4, -1) are collinear and a – b = 1, find the values of a and (2014D) Solution: Since A(-2, 1), B(a, b) and C(4, -1) are collinear. ∴ x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 ) = 0 ⇒ -2[b – (-1)] + a(-1 – 1) + 4(1 – b) = 0 ⇒ -2b – 2 – 2a + 4 – 4b = 0 ⇒ -2a – 6b = -2 ⇒ a + 3b = 1 … [Dividing by (-2) ⇒ a = 1 – 3 We have, a – b = 1 …[Given (1 – 3b) – b = 1 …[From (i) 1 – 3b – b = 1 -4b = 1 – 1 = 0 ∴ b = \(\frac{0}{-4}\) = 0 From (i), a = 1 – 3(0) = 1 – 0 = 1 ∴ a = 1, b = 0

Question 38. If the points A(-1, -4), B(b, c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c. (2014D) Solution: A(-1, -4), B(b, c), C(5, -1) are collinear. ∴ x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 ) = 0 ⇒ -1(c + 1) + b(-1 + 4) + 5(-4 – c) = 0 ⇒ -c – 1 – b + 4b – 20 – 5c = 0 ⇒ 3b – 6c = 21 ⇒ – b – 2c = 7 …[Dividing by 3 ⇒ b = 7 + 2c We have, 2b + c = 4 … [Given 2(7 + 2c) + c = 4 … [From (i) ⇒ 14 + 4c + c = 4 ⇒ 5c = 4 – 14 = -10 ⇒ c = -2 ⇒ b = 7 + 2(-2) = 3 … [From (i) ∴ b = 3, c = -2

Question 46. If the points P(-3, 9), Q(a, b) and R(4, -5) are collinear and a + b = 1, find the values of a and b. (2014D) Solution: Similar to Question 37, Page 105.

Question 48. Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the coordinates of the point of division. (2014OD) Solution: Similar to Question 41, Page 106.

Coordinate Geometry Class 10 Important Questions Long Answer (4 Marks)

Question 50. If the points A(x, y), B(3, 6) and C(-3, 4) are collinear, show that x – 3y + 15 = 0. (2012OD) Solution: Pts. A(x, y), B(3, 6), C(-3, 4) are collinear. ∴ Area of ∆ = 0 As area of ∆ = \(\frac{1}{2}\)[x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] ∴Area of ∆ABC = x(6 – 4) + 3(4 – y) + (-3) (y – 6) = 0 = 2x + 12 – 3y – 3y + 18 = 0 = 2x – 6y + 30 = 0 ∴ x – 3y + 15 = 0 … [Dividing both sides by 2

Question 51. Find the values of k for which the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear. (2015OD) Solution: A (k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k). When points are collinear, area of ∆ is 0. ∴ Area of triangle = 0 ⇒ [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] = 0 ⇒ [(k + 1) (2k + 3 – 5k) + 3k (5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0 ⇒ [(k + 1) (3 – 3k) + 3k(3k) + (5k – 1)(-3)] = 0 ⇒ [3k – 3k 2 + 3 – 3k + 9k 2 – 15k + 3) = 0 ⇒ 6k 2 – 15k + 6 = 0 ⇒ 2k 2 – 5k + 2 = 0 ⇒ 2k 2 – 4k – 1k + 2 = 0 ⇒ 2k(k – 2) – 1(k – 2) = 0 ⇒ (k – 2)(2k – 1) = 0 ⇒ k – 2 = 0 or 2k – 1 = 0 ⇒ k = 2 or k = \(\frac{1}{2}\) We get, k = 2, \(\frac{1}{2}\)

Question 54. If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. (2012D) Solution: Area of ∆ = 15 sq. units \(\frac{1}{2}\) [1(p – 7) + 4(7 + 3)) + (-9)(-3 – p)] = ±15 p – 7 + 40 + 27 + 9p = ±30 10p + 60 = ± 30 10p = -60 ± 30 p = \(\frac{-60 \pm 30}{10}\) ∴ Taking +ve sign, p = \(\frac{-60+30}{10}=\frac{-30}{10}\) = -3 Taking -ve sign, p = \(\frac{-60-30}{10}=\frac{-90}{10}\) = -9

Question 55. For the triangle ABC formed by the points A(4, -6), B(3,-2) and C(5, 2), verify that median divides the triangle into two triangles of equal area. (2013OD) Solution: Let A(4, -6), B(3, -2) and C(5, 2) be the vertices of ∆ABC. Since AD is the median ∴ D is the mid-point of BC. ⇒ D\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)\) ⇒ D(4,0) Area of ∆ABD = \(\frac{1}{2}\) [4(-2 – 0) + 3(0 + 6) + 4(-6 + 2)] = \(\frac{1}{2}\) [-8 + 18 – 16) = \(\frac{1}{2}\) [-6] = -3 But area of A cannot be negative. ∴ ar(∆ABD) = 3 sq.units …(i) Area of ∆ADC = \(\frac{1}{2}\) [4(0 – 2) + 4(2 + 6) + 5(-6 – 0)] = \(\frac{1}{2}\)(-8 + 32 – 30] = \(\frac{1}{2}\) [-6] = -3 But area of ∆ cannot be negative. ∴ ar(∆ADC) = 3 sq.units From (i) and (ii), ∴ Median AD of AABC divides it into two ∆s of equal area.

Question 57. Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units. (2015OD) Solution: Let A(k + 1, 1), B(4, -3) and C(7, -k). We have, Area of ∆ABC = 6 … [Given 6 = \(\frac{1}{2}\) [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] 6 = \(\frac{1}{2}\)[(k + 1)(-3 + k) + 4(-k – 1) + 7(1 + 3)] 12 = (-3k + k 2 – 3 + k – 4k – 4 + 28] 12 = [k 2 – 6k + 21] ⇒ k 2 – 6k + 21 – 12 = 0 ⇒ k 2 – 6k + 9 = 0 ⇒ k 2 – 3k – 3k + 9 = 0 ⇒ k(k – 3) – 3(k – 3) = 0 = ⇒ (k – 3) (k – 3) = 0 ⇒ k – 3 = 0 or k – 3 = 0 ⇒ k = 3 or k = 3 Solving to get k = 3.

Question 58. Prove that the area of a triangle with vertices (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t. (2016D) Solution: Let A(t, t – 2), B(t + 2, + + 2), C(t + 3, t). Area of ∆ABC = \(\frac{1}{2}\)[x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] = \(\frac{1}{2}\) [t(t + 2 – t) + (t + 2)(t – (t – 2)) +(t + 3)((t – 2) – (t + 2))] = \(\frac{1}{2}\) [t(2) + (t + 2)(2) + (t + 3)(-4)] = \(\frac{1}{2}\) (2+ + 2+ + 4 – 46 – 12] = \(\frac{1}{2}\) [-8] = -4 Area of ∆ is always positive. ∴ Area of ∆ = 4 sq. units, which is independent of t.

Question 60. If the area of triangle ABC formed by A(x, y), B(1, 2) and C(2, 1) is 6 square units, then prove that x + y = 15 or x + y = -9. (2013D) Solution: Let A(x, y), B(1, 2), C(2, 1). Area of ∆ABC = 6 sq. units …[Given As \(\frac{1}{2}\) [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] = 6 ∴ x(2 – 1) + 1(1 – y) + 2(y – 2) = ±12 x + 1 – y + 2y – 4 = ±12 Taking +ve sign x + y = 12 + 4 – 1 ∴ x + y = 15

Taking -ve sign x + y = -12 + 4 – 1 ∴ x + y = -9

Question 69. Find the value of k, if the points P(5, 4), Q(7, k) and R(9, – 2) are collinear. (2011D) Solution: Given points are P(5, 4), Q(7, k) and R(9, -2). x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 ) = 0 …[∵ Points are collinear ∴ 5 (k + 2) + 7 (- 2 – 4) + 9 (4 – k) = 0 5k + 10 – 14 – 28 + 36 – 9k = 0 4 = 4k ∴ k=1

Question 71. Find the point of y-axis which is equidistant from the points (-5, -2) and (3, 2). (2011D) Solution: Similar to Question 26, Page 103.

Question 72. Point M(11, y) lies on the line segment joining the points P(15, 5) and Q(9, 20). Find the ratio in which point M divides the line segment PQuestion Also find the value of y. (2011OD) Solution: Similar to Question 62, Page 110.

Question 74. Find the value of k, for which the points A(6,-1), B(k, -6) and C(0, -7) are collinear. (2012OD) Solution: Similar to Question 69, Page 112.

Question 75. Find the value of p, if the points A(1, 2), B(3, p) and C(5, -4) are collinear. (2012OD) Solution: Similar to Question 69, Page 112.

Question 76. Find the value of x for which the points (x, -1), (2, 1) and (4, 5) are collinear. (2013D) Solution: Similar to Question 69, Page 112.

Question 78. If A(-3,5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area. (2014OD0 Solution: Similar to Question 66, Page 111.

Question 79. A(4, -6), B(3, -2) and C(5, 2) are the vertices of a ∆ABC and AD is its median. Prove that the median AD divides AABC into two triangles of equal areas. (2014OD) Solution: Similar to Question 56, Page 109.

Question 80. If A(-, 8), B(-3,-4), C(0, -5) and D(5, 6) are the vertices of a quadrilateral ABCD, find its area. (2015D) Solution: Similar to Question 66, Page 111.

Question 81. If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area. (2015D) Solution: Similar to Question 66, Page 111.

Question 82. Find the values of k so that the area of the triangle with vertices (1, -1), (-4, 2k) and (-k, -5) is 24 sq. units. (2015OD) Solution: Similar to Question 57, Page 109.

Important Questions for Class 10 Maths

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CBSE 10th Maths: Case Study Questions With Answers

Students taking the 10th board examinations will see new kinds of case study questions in class. The board initially incorporated case study questions into the board exam. The chapter-by-chapter case study question and answers are available here.

Chapterwise Case Study Questions for Class 10 Mathematics

Case study questions are an essential component of the Class 10 Mathematics curriculum. They provide students with real-world scenarios where they can apply mathematical concepts and problem-solving skills. By analyzing and solving these case study questions, students develop a deeper understanding of the subject and improve their critical thinking abilities.

The above  Case studies for Class 10 Maths  will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of schools.studyrate.in for the benefit of Class 10 students.

• Class 10th Science Case Study Questions

Benefits of Case Study Questions for Class 10 Mathematics

Case study questions offer several benefits to both students and teachers. Here are some key advantages:

• Practical Application : Case study questions bridge the gap between theory and real-life situations, allowing students to apply mathematical concepts in practical scenarios.
• Analytical Thinking : By solving case study questions, students enhance their analytical thinking and problem-solving skills.
• Conceptual Clarity : Case study questions help reinforce the fundamental concepts of mathematics, leading to improved conceptual clarity.
• Exam Preparation : Practicing case study questions prepares students for their Class 10 Mathematics exams, as they become familiar with the question formats and types.
• Comprehensive Assessment : Teachers can use case study questions to assess students’ understanding of various mathematical concepts in a comprehensive manner.

How to Use Case Study Questions Effectively

To make the most out of the case study questions, follow these effective strategies:

• Read the question carefully : Understand the given scenario and identify the mathematical concepts involved.
• Analyze the problem : Break down the problem into smaller parts and determine the approach to solve it.
• Apply relevant formulas and concepts : Utilize your knowledge of the subject to solve the case study question.
• Show your working : Clearly demonstrate the steps and calculations involved in reaching the solution.
• Check your answer : Always verify if your solution aligns with the given problem and recheck calculations for accuracy.

Tips for Solving Case Study Questions

Here are some useful tips to excel in solving case study questions:

• Practice regularly : Regular practice will enhance your problem-solving skills and familiarity with different question formats.
• Understand the concepts: Ensure you have a strong foundation in the underlying mathematical concepts related to each chapter.
• Work on time management : Practice solving case study questions within a stipulated time to improve your speed and efficiency during exams.
• Seek clarification : If you encounter any doubts or difficulties, don’t hesitate to seek guidance from your teacher or peers.

Case study questions are an invaluable resource for Class 10 Mathematics students. They provide practical application opportunities and strengthen conceptual understanding. By utilizing the chapter-wise case study questions provided in this article, students can enhance their problem-solving skills, prepare effectively for exams, and develop a deeper appreciation for the subject.

FAQs on Class 10 Maths Case Study Questions

Q1: can i download the class 10 maths case study questions in pdf format.

Yes, you can download the Class 10 Maths case study questions in PDF format from our site free of cost.

Q2: Are the case study questions aligned with the latest curriculum?

Yes, the case study questions presented in this article are designed to align with the latest Class 10 Mathematics curriculum.

Q3: How can case study questions improve my exam preparation?

Case study questions help you understand the practical application of mathematical concepts, enabling you to approach exam questions with greater confidence and clarity.

Q5: Where can I find more resources for Class 10 Mathematics preparation?

Download more resources of Class 10th Maths from schools.studyrate.in, we offer additional resources and practice materials for Class 10 Mathematics.

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Class 10 Case study of Chapter 7 coordinate geometry

 Case study:1

To conduct Annual day activities, in a rectangular shaped school ground PQRS, lines have been drawn with chalk powder at a distance of 1 m each.  10 banners have been placed at a distance of 10 m from each other along PS. The first three banners have been shown in the graph along PS.

Three students Amit(A), Bhanu(B) and Chirag(C). Amit runs 1/5 th the distance PS on the first line and posts a blue flag. Bhanu runs 1/10 th the distance PS on the third line and posts a green flag. Chirag runs half the distance PS on the fifth line and posts a red flag.

(A) Find the coordinates of positions of blue, green and red flags.

(B) Find the distance between the blue and green flags.

(A) Since, 10 banners have been placed at a distance of 10 m from each other along PS,

∴ Distance PS = 10×10 = 100 m

(B) The coordinates of the blue and green flags are (1, 20) and (3, 10) respectively.

Applying the distance formula, the distance between the blue and green flags is:

Case study:2

A farmer has plot of land in the shape of a quadrilateral as shown below:

(A) Find the image of the vertex A on the y-axis.

(B) Find the distances BC and CD.

(A) The image of any point on the y-axis will have the same y-coordinate, but its x-coordinate will be negative of its earlier value. As coordinates of its image on y-axis wiill be (4, 5).

(B) From the graph,

B = (0, 7), C = (5, -5), D = (-4, -2)

∴ Using distance formula.

Case study:3

Class X students of a secondary school in Krishnagar have been alloted a rectangular plot of a land for gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

(A) Considering A as the origin, what are the coordinates of A ?

(a) (0, 1)              (b) (1, 0)

(c) (0, 0)              (d) (-1, -1)

(B) What are the coordinates of P ?

(a) (4, 6)               (b) (6, 4)

(c) (4, 5)                (d) (5, 4)

(C) What are the coordinates of R ?

(a) (6, 5)                  (b) (5, 6)

(c) (6, 0)                   (d) (7, 4)

(D) What are the coordinates of D ?

(a) (16, 0)                  (b) (0, 0)

(c) (0, 16)                   (d) (16, 1)

(E) What are the coordinates of P, If D is taken as the origin ?

(a) (12, 2)                    (b) (-12, 6)

(c) (12, 3)                      (d) (6, 10)

(A) Answer (c) (0, 0)

(B) Answer (a) (4, 6)

(C) Answer (a) (6, 5)

(D) Answer (a) (16, 0)

(E) Answer (b) (-12, 6)

Some other Case study question:

5: arithmetic progression.

Class 10 Case based problem of Chapter 5 A.P. 1

Class 10 Case based problem of Chapter 5 A.P. 2

6: Triangle

Class 10 Case based problem of Chapter 6 Triangles 1

Class 10 Case based problem of Chapter 6 Triangles 2

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• NCERT Solutions
• NCERT Solutions for Class 10
• NCERT Solutions for Class 10 Maths
• Chapter 7 Coordinate Geometry

NCERT Solutions For Class 10 Maths Chapter 7 - Coordinate Geometry

Ncert solutions for class 10 maths chapter 7 – cbse free pdf download.

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry covers all the exercises provided in the NCERT textbook. These NCERT solutions, prepared by experts at BYJU’S, are comprehensive study material for the students preparing for the CBSE Class 10 board examination. These solutions are available for easy access and download by the students. Here, you can get detailed stepwise answers to different types of questions provided in the NCERT textbook. Practising the Solutions of NCERT will help you attain perfection on the topics involved in the coordinate geometry chapter.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 7 Coordinate Geometry

Download most important questions for class 10 maths chapter – 7 coordinate geometry.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

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Access Answers to NCERT Solutions for Class 10 Maths Chapter 7

Exercise 7.1 page no: 161.

1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1)

(ii) (-5, 7), (-1, 3)

(iii) (a, b), (- a, – b)

Distance formula to find the distance between two points (x 1 , y 1 ) and (x 2 , y 2 ) is, say d,

2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns, A and B, discussed in Section 7.2?

Let us consider town A at point (0, 0). Therefore, town B will be at point (36, 15).

Distance between points (0, 0) and (36, 15)

In section 7.2, A is (4, 0) and B is (6, 0) AB 2 = (6 – 4) 2 – (0 – 0) 2 = 4

The distance between towns A and B will be 39 km. The distance between the two towns, A and B, discussed in Section 7.2, is 4 km.

3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.

Solution:   If the sum of the lengths of any two line segments is equal to the length of the third line segment, then all three points are collinear.

Consider, A = (1, 5) B = (2, 3) and C = (-2, -11)

Find the distance between points: say AB, BC and CA

Since AB + BC ≠ CA

Therefore, the points (1, 5), (2, 3), and (- 2, – 11) are not collinear.

4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Since two sides of any isosceles triangle are equal, to check whether given points are vertices of an isosceles triangle, we will find the distance between all the points.

Let the points (5, – 2), (6, 4), and (7, – 2) represent the vertices A, B and C, respectively.

This implies whether given points are vertices of an isosceles triangle.

5. In a classroom, 4 friends are seated at points A, B, C and D, as shown in Fig. 7.8. Champa and Chameli walk into the class, and after observing for a few minutes, Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct.

From the figure, the coordinates of points A, B, C and D are (3, 4), (6, 7), (9, 4) and (6,1).

Find the distance between points using the distance formula, we get

All sides are of equal length. Therefore, ABCD is a square, and hence, Champa was correct.

6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)

(ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

(i) Let the points (- 1, – 2), (1, 0), ( – 1, 2), and ( – 3, 0) represent the vertices A, B, C, and D of the given quadrilateral, respectively.

Side length = AB = BC = CD = DA = 2√2

Diagonal Measure = AC = BD = 4

Therefore, the given points are the vertices of a square.

(ii) Let the points (- 3, 5), (3, 1), (0, 3), and (- 1, – 4) represent the vertices A, B, C, and D of the given quadrilateral, respectively.

It’s also seen that points A, B and C are collinear. So, the given points can only form 3 sides, i.e. a triangle and not a quadrilateral which has 4 sides. Therefore, the given points cannot form a general quadrilateral.

(iii) Let the points (4, 5), (7, 6), (4, 3), and (1, 2) represent the vertices A, B, C, and D of the given quadrilateral, respectively.

Opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9).

To find a point on the x-axis.

Therefore, its y-coordinate will be 0. Let the point on the x-axis be (x,0).

Consider A = (x, 0); B = (2, – 5) and C = (- 2, 9).

Simplify the above equation,

Remove the square root by taking square on both sides, we get

(2 – x) 2 + 25 = [-(2 + x)] 2  + 81

(2 – x) 2 + 25 = (2 + x) 2  + 81

x 2  + 4 – 4x + 25 = x 2  + 4 + 4x + 81

8x = 25 – 81 = -56

Therefore, the point is (- 7, 0).

8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.

Given: Distance between (2, – 3) and (10, y) is 10.

Using the distance formula,

Simplify the above equation and find the value of y.

Squaring both sides,

64 + (y + 3) 2 = 100

(y + 3) 2 = 36

y + 3 = +6 or y + 3 = −6

y = 6 – 3 = 3 or y = – 6 – 3 = -9

Therefore, y = 3 or -9.

9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also, find the distance QR and PR.

Given: Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), which means PQ = QR

Step 1: Find the distance between PQ and QR using the distance formula,

Squaring both sides to omit square root

41 = x 2  + 25

x = 4 or x = -4

Coordinates of Point R will be R (4, 6) or R (-4, 6),

If R (4, 6), then QR

10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Point (x, y) is equidistant from (3, 6) and (- 3, 4).

Squaring both sides, (x – 3) 2 +(y – 6) 2 = (x + 3) 2 +(y – 4) 2

x 2 + 9 – 6x + y 2 + 36 – 12y = x 2 + 9 + 6x + y 2 +16 – 8y

36 – 16 = 6x + 6x + 12y – 8y

20 = 12x + 4y

3x + y – 5 = 0

Exercise 7.2 Page No: 167

1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3.

Let P(x, y) be the required point. Using the section formula, we get

x = (2×4 + 3×(-1))/(2 + 3) = (8 – 3)/5 = 1

y = (2×-3 + 3×7)/(2 + 3) = (-6 + 21)/5 = 3

Therefore, the point is (1, 3).

2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Let P (x 1 , y 1 ) and Q (x 2 , y2) be the points of trisection of the line segment joining the given points, i.e. AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

x 1 = (1×(-2) + 2×4)/3 = (-2 + 8)/3 = 6/3 = 2

y 1 = (1×(-3) + 2×(-1))/(1 + 2) = (-3 – 2)/3 = -5/3

Therefore: P (x 1 , y 1 ) = P(2, -5/3)

Point Q divides AB internally in the ratio 2:1.

x 2 = (2×(-2) + 1×4)/(2 + 1) = (-4 + 4)/3 = 0

y 2 = (2×(-3) + 1×(-1))/(2 + 1) = (-6 – 1)/3 = -7/3

The coordinates of the point Q are (0, -7/3)

3. To conduct sports day activities in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

From the given instruction, we observed that Niharika posted the green flag at 1/4 th of the distance AD, i.e. (1/4 ×100) m = 25 m from the starting point of the 2nd line. Therefore, the coordinates of this point are (2, 25).

Similarly, Preet posted a red flag at 1/5 of the distance AD, i.e. (1/5 ×100) m = 20 m from the starting point of the 8th line. Therefore, the coordinates of this point are (8, 20).

Distance between these flags can be calculated by using the distance formula,

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let’s say this point is P(x, y).

x = (2 + 8)/2 = 10/2 = 5 and y = (20 + 25)/2 = 45/2

Hence, P( x , y ) = (5, 45/2)

Therefore, Rashmi should post her blue flag at 45/2 = 22.5m on the 5th line.

4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

Consider the ratio in which the line segment joining (-3, 10) and (6, -8) is divided by point (-1, 6) be k :1.

Therefore, -1 = ( 6 k -3)/( k +1)

– k – 1 = 6 k -3

Therefore, the required ratio is 2: 7.

5. Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also, find the coordinates of the point of division.

Let the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis be k:1. Therefore, the coordinates of the point of division, say P(x, y) is ((-4 k +1)/( k +1), (5 k -5)/( k +1)).

We know that the y-coordinate of any point on the x-axis is 0.

Therefore, ( 5k – 5)/(k + 1) = 0

So, the x -axis divides the line segment in the ratio 1:1.

Now, find the coordinates of the point of division:

P (x, y) = ((-4(1)+1)/(1+1) , (5(1)-5)/(1+1)) = (-3/2 , 0)

6. If (1, 2), (4, y ), ( x , 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y .

Let A, B, C and D be the points of a parallelogram: A(1, 2), B(4, y ), C( x , 6) and D(3, 5).

Since the diagonals of a parallelogram bisect each other, the midpoint is the same.

To find the value of x and y, solve for the midpoint first.

Midpoint of AC = ( (1+x)/2 , (2+6)/2 ) = ((1+x)/2 , 4)

Midpoint of BD = ((4+3)/2 , (5+y)/2 ) = (7/2 , (5+y)/2)

The midpoint of AC and BD are the same, this implies

(1+x)/2 = 7/2 and 4 = (5+y)/2

x + 1 = 7 and 5 + y = 8

x = 6 and y = 3

7. Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).

Let the coordinates of point A be ( x , y ).

Mid-point of AB is (2, – 3), which is the centre of the circle.

Coordinate of B = (1, 4)

(2, -3) =((x+1)/2 , (y+4)/2)

(x+1)/2 = 2 and (y+4)/2 = -3

x + 1 = 4 and y + 4 = -6

x = 3 and y = -10

The coordinates of A(3,-10).

8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB.

The coordinates of points A and B are (-2,-2) and (2,-4), respectively.

Since AP = 3/7 AB

Therefore, AP:PB = 3:4

Point P divides the line segment AB in the ratio 3:4.

9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts.

Draw a figure, line dividing by 4 points.

From the figure, it can be observed that points X, Y, and Z are dividing the line segment in a ratio 1:3, 1:1, and 3:1, respectively.

10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4), and (-2,-1) taken in order.

[Hint: Area of a rhombus = 1/2 (product of its diagonals)

Let A(3, 0), B (4, 5), C( – 1, 4) and D ( – 2, – 1) are the vertices of a rhombus ABCD.

Exercise 7.3 Page No: 170

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (-1, 0), (2, -4)

(ii) (-5, -1), (3, -5), (5, 2)

x 1 = 2, x 2 = -1, x 3 = 2, y 1 = 3, y 2 = 0 and y 3 = -4

Substitute all the values in the above formula, we get

= 1/2 {8 + 7 + 6}

So, the area of the triangle is 21/2 square units.

x 1 = -5, x 2 = 3, x 3 = 5, y 1 = -1, y 2 = -5 and y 3 = 2

= 1/2{35 + 9 + 20} = 32

Therefore, the area of the triangle is 32 square units.

2. In each of the following, find the value of ‘k’, for which the points are collinear.

(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

(i) For collinear points, the area of triangle formed by them is always zero.

Let points (7, -2), (5, 1), and (3, k) are vertices of a triangle.

Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0

7 – 7k + 5k +10 -9 = 0

-2k + 8 = 0

(ii) For collinear points, the area of triangle formed by them is zero.

Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0

1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0

8 – 6k + 10 = 0

3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Let the vertices of the triangle be A (0, -1), B (2, 1), and C (0, 3).

Let D, E, and F be the mid-points of the sides of this triangle.

Coordinates of D, E, and F are given by

D = (0+2/2, -1+1/2 ) = (1, 0)

E = ( 0+0/2, -1+3/2 ) = (0, 1)

F = ( 0+2/2, 3+1/2 ) = (1, 2)

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)} = 1/2 (1+1) = 1

The area of ΔDEF is 1 square unit

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)] = 1/2 {8} = 4

The area of ΔABC is 4 square units

Therefore, the required ratio is 1:4.

4. Find the area of the quadrilateral whose vertices, taken in order, are

(-4, -2), (-3, -5), (3, -2) and (2, 3).

Let the vertices of the quadrilateral be A (- 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3).

Join AC and divide the quadrilateral into two triangles.

We have two triangles, ΔABC and ΔACD.

= 1/2 (12 + 0 + 9)

= 21/2 square units

= 1/2 (20 + 15 + 0)

= 35/2 square units

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units = 28 square units

5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC, whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = Midpoint of BC = ((3+5)/2, (-2+2)/2) = (4, 0)

= 1/2 (-8 + 18 – 16)

= -3 square units

However, the area cannot be negative. Therefore, the area of ΔABD is 3 square units.

= 1/2 (-8 + 32 – 30) = -3 square units

However, the area cannot be negative. Therefore, the area of ΔACD is 3 square units.

The area of both sides is the same. Thus, median AD has divided ΔABC into two triangles of equal areas.

Exercise 7.4 Page No: 171

1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7).

Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k:1 ratio.

Coordinates of point of division can be given as follows:

x = (2 + 3k)/(k + 1) and y = (-2 + 7k)/(k + 1)

Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have

2{(2 + 3k)/(k + 1)} + {(-2 + 7k)/(k + 1)} – 4 = 0

(4 + 6k)/(k + 1) + (-2 + 7k)/(k + 1) = 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

Hence, the ratio is 2:9.

2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

If given points are collinear, then the area of the triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

Area of a triangle = 1/2 × [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )] = 0

2x – y + 7y – 14 = 0

2x + 6y – 14 = 0

x + 3y – 7 = 0.

Which is the required result.

3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).

Let A = (6, -6), B = (3, -7), and C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y), then

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly, OB = OC

(x – 3) 2 + (y + 7) 2 = (x – 3) 2 + (y – 3) 2

(y + 7) 2 = (y – 3) 2

y 2 + 14y + 49 = y 2 – 6y + 9

Substituting the value of y in equation (1), we get

-6x = 2y – 14

-6x = -4 – 14 = -18

Hence, the centre of the circle is located at point (3,-2).

4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.

Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD.

To Find: Coordinate of points B and D.

Step 1: Find the distance between A and C and the coordinates of point O.

We know that the diagonals of a square are equal and bisect each other.

AC = √[(3 + 1) 2  + (2 – 2) 2 ] = 4

Coordinates of O can be calculated as follows:

x = (3 – 1)/2 = 1 and y = (2 + 2)/2 = 2

Step 2: Find the side of the square using the Pythagoras theorem

Let a be the side of the square and AC = 4

From the right triangle, ACD,

Hence, each side of the square = 2√2

Step 3: Find the coordinates of point D

Equate the length measure of AD and CD

Say, if the coordinates of D are (x 1 , y 1 )

AD 2 = (x 1 + 1) 2  + (y 1 – 2) 2

Similarly, CD 2 = (x 1  – 3) 2  + (y 1 – 2) 2

Since all sides of a square are equal, which means AD = CD

(x 1 + 1) 2  + (y 1 – 2) 2 = (x 1  – 3) 2  + (y 1 – 2) 2

x 1 2 + 1 + 2x 1 = x 1 2 + 9 – 6x 1

The value of y 1 can be calculated as follows by using the value of x.

From step 2: each side of the square = 2√2

CD 2 = (x 1  – 3) 2  + (y 1 – 2) 2

8 = (1 – 3) 2  + (y 1 – 2) 2

8 = 4 + (y 1 – 2) 2

y 1 – 2 = 2

Hence, D = (1, 4)

Step 4: Find the coordinates of point B

From line segment, BOD

Coordinates of B can be calculated using coordinates of O, as follows:

Earlier, we had calculated O = (1, 2)

Say B = (x 2 , y 2 )

1 = (x 2 + 1)/2

And 2 = (y 2 + 4)/2

=> y 2 = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

5. The class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular lawn in the plot, as shown in fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot.

(i) Taking A as the origin, find the coordinates of the vertices of the triangle.

(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?

Also, calculate the areas of the triangles in these cases. What do you observe?

(i) Taking A as the origin, the coordinates of the vertices P, Q and R are,

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here, AD is the x-axis and AB is the y-axis.

(ii) Taking C as the origin,

The coordinates of vertices P, Q and R are ( 12, 2), (13, 6) and (10, 3), respectively.

Here, CB is the x-axis and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

= ½ (- 12 – 3 + 24 )

= 9/2 sq unit

(ii) Area of triangle PQR in case of origin C:

= ½ ( 36 + 13 – 40)

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

The area is the same in both cases because the triangle remains the same no matter which point is considered as the origin.

6. The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E, respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the ∆ ADE and compare it with the area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

Given: The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2)

AD/AB = AE/AC = 1/4

AD/(AD + BD) = AE/(AE + EC) = 1/4

Point D and Point E divide AB and AC, respectively, in ratio 1:3.

Coordinates of D can be calculated as follows:

x = (m 1 x 2 + m 2 x 1 )/(m 1 + m 2 ) and y = (m 1 y 2 + m 2 y 1 )/(m 1 + m 2 )

Here, m 1 = 1 and m 2 = 3

Consider line segment AB which is divided by point D at the ratio 1:3.

x = [3(4) + 1(1)]/4 = 13/4

y = [3(6) + 1(5)]/4 = 23/4

Similarly, the coordinates of E can be calculated as follows:

x = [1(7) + 3(4)]/4 = 19/4

y = [1(2) + 3(6)]/4 = 20/4 = 5

Find the area of triangle:

The area of triangle ∆ ABC can be calculated as follows:

= ½ (12 – 4 + 7) = 15/2 sq unit

The area of ∆ ADE can be calculated as follows:

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

Hence, the ratio of the area of triangle ADE to the area of triangle ABC = 1:16.

7. Let A (4, 2), B (6, 5), and C (1, 4) be the vertices of ∆ ABC.

(i) The median from A meets BC at D. Find the coordinates of point D.

(ii) Find the coordinates of the point P on AD such that AP:PD = 2:1.

(iii) Find the coordinates of points Q and R on medians BE and CF, respectively, such that BQ:QE = 2:1 and CR:RF = 2:1.

(iv) What do you observe?

[Note: The point which is common to all the three medians is called the centroid,

and this point divides each median in the ratio 2:1.]

(v) If A (x 1 , y 1 ), B (x 2 , y 2 ) and C (x 3 , y 3 ) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.

(i) Coordinates of D can be calculated as follows:

Coordinates of D = ( (6+1)/2, (5+4)/2 ) = (7/2, 9/2)

So, D is (7/2, 9/2)

(ii) Coordinates of P can be calculated as follows:

Coordinates of P = ( [2(7/2) + 1(4)]/(2 + 1), [2(9/2) + 1(2)]/(2 + 1) ) = (11/3, 11/3)

So, P is (11/3, 11/3)

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2 , 3)

So, E is (5/2, 3)

Points Q and P would be coincident because the medians of a triangle intersect each other at a common point called the centroid. Coordinate of Q can be given as follows:

Coordinates of Q =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)

F is the midpoint of the side AB

Coordinates of F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)

Point R divides the side CF in ratio 2:1

Coordinates of R = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)

(iv) Coordinates of P, Q and R are the same, which shows that medians intersect each other at a common point, i.e. centroid of the triangle.

(v) If A (x 1 , y 1 ), B (x 2 , y 2 ) and C (x 3 , y 3 ) are the vertices of triangle ABC, the coordinates of the centroid can be given as follows:

x = (x 1 + x 2 + x 3 )/3 and y = (y 1 + y 2 + y 3 )/3

8. ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively. Is the quadrilateral PQRS a square, a rectangle or a rhombus? Justify your answer.

P id the midpoint of side AB,

Coordinate of P = ( (-1 – 1)/2, (-1 + 4)/2 ) = (-1, 3/2)

Similarly, Q, R and S are (As Q is the midpoint of BC, R is the midpoint of CD and S is the midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5, 3/2)

Coordinate of S = (2, -1)

Length of PQ = √[(-1 – 2) 2 + (3/2 – 4) 2 ] = √(61/4) = √61/2

Length of SP = √[(2 + 1) 2 + (-1 – 3/2) 2 ] = √(61/4) = √61/2

Length of QR = √[(2 – 5) 2 + (4 – 3/2) 2 ] = √(61/4) = √61/2

Length of RS = √[(5 – 2) 2 + (3/2 + 1) 2 ] = √(61/4) = √61/2

Length of PR (diagonal) = √[(-1 – 5) 2 + (3/2 – 3/2) 2 ] = 6

Length of QS (diagonal) = √[(2 – 2) 2 + (4 + 1) 2 ] = 5

The above values show that PQ = SP = QR = RS = √61/2, i.e. all sides are equal.

But PR ≠ QS, i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.

This chapter comes under Unit-Coordinate Geometry and has a weightage of 6 marks in the board examination. There will be one mark MCQ questions, 2 marks reasoning questions, and 3 marks short answer questions. This chapter has fundamental concepts that lay the foundation for your future studies.

Sub-topics of Class 10 Chapter 7 Coordinate Geometry

• Introduction to Coordinate Geometry
• Distance Formula
• Section Formula
• Area of the Triangle

List of Exercises from Class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1 – 10 Questions, which includes 8 practical-based questions, 2 reasoning questions Exercise 7.2 – 10 Questions, which includes 8 long answer questions, 2 short answer questions Exercise 7.3 – 5 Questions, which includes 3 long answer questions, 2 practical-based questions Exercise 7.4 – 8 Questions, which includes 6 long answer questions, 1 practical-based question, 1 reasoning question

NCERT Solutions are carefully drafted to assist the student in scoring good marks in the examination. It provides you with much-needed problem-solving practice.

This chapter deals with finding the area between two points whose coordinate values are provided. For instance, the area of a triangle. This chapter has some basic concepts like the area of a triangle, rhombus, the distance between sides, and intersections. This chapter teaches you the relationship between numerical and geometry and their application in our daily lives.

These  NCERT Solutions for Class 10 Maths have different types of questions and their answers which will help you with alternate solutions and diagrammatic representation. The solutions provided here are written in simple language with apt information. By studying these NCERT Solutions thoroughly, students will be able to solve complex problems easily.

Key Features of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

• The solution has all the exercise questions provided in the NCERT textbook.
• Diagrammatic representations and alternate methods will help you understand the concepts thoroughly.
• Solving these NCERT Solutions will make you acquainted with important formulas and standards.
• This NCERT Solution has different examples that will help you relate to real-life examples to relate geometry and numerical.

BYJU’S is India’s top online education provider with the country’s best teachers on board. You can get NCERT study materials on BYJU’S site and application. To get access to all the study materials, visit BYJU’S website or download BYJU’S learning App.

To provide students with more problems for practice, we at BYJU’S also provide solutions for other textbooks which can be accessed absolutely free of cost.

• RD Sharma Solutions for Class 10 Maths Chapter 14 Coordinate Geometry

Disclaimer – 

Dropped Topics –  7.4 Area of a triangle

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• CBSE Class 10

Important Case Study Questions CBSE Class 10 Maths

Cbse class 10 maths important case study questions: cbse class 10 maths exam 2024 is just around the corner. case study questions can be a hard nut to track if not prepared well. check here important case study questions from class 10th maths curriculum  for cbse class 10 maths board exam 2024..

CBSE Class 10 Maths Question Paper Structure

• Section A : 20 Multiple Choice Questions (MCQs) carrying 1 mark each. 
• Section B : 5 Short Answer-I (SA-I) type questions carrying 2 marks each. 
• Section C : 6 Short Answer-II (SA-II) type questions carrying 3 marks each. 
• Section D : 4 Long Answer (LA) type questions carrying 5 marks each.
• Section E : 3 Case Based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1 and 2 marks each respectively.

CBSE Class 10 Maths Important Case Study Questions

Related:  CBSE Class 10 Maths Important Formulas for Last Minute Revision for Board Exam 2024

1 Two Friends Geeta and Sita were playing near the river. So, they decide to play a game in which they have to throw the stone in the river, and whoever will throw the stone at maximum distance, win the game. Geeta Starts first and throws the stone in the river. During her throw, her hand was making an angle of 60° with the Horizontal plane. Sita throws at 45°.

• Straight Line
• Semi circle
• Bi-Quadratic
• Parabola Open Upward
• Parabola Open Downward
• Hyperbola Open Upward
• Hyperbola Open downward
• Two Real Points
• One Real Point
• Three Real Points
• Putting y=0 in given Polynomial
• Putting y=1 in the given Polynomial
• Putting x=0 in the given Polynomial.
• Putting x=1 in the given Polynomial.

2 The department of Computer Science and Technology is conducting an International Seminar. In the seminar, the number of participants in Mathematics, Science and Computer Science are 60, 84 and 108 respectively. The coordinator has made the arrangement such that in each room, the same number of participants are to be seated and all of them being in the same subject. Also, they allotted the separate room for all the official other than participants.

(i) Find the total number of participants.

(a) 60 

(b) 84 

(c) 108 

(d) none of these

(ii) Find the LCM of 60, 84 and 108.

(a) 12 

(b) 504 

(c) 544320 

(iii) Find the HCF of 60, 84 and 108.

(iv) Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

(b) 20 

(c) 21 

(v) Based on the above (iv) conditions, find the minimum number of rooms required for all the participants and officials.

• In the standard form of quadratic polynomial, ax2 + bx + c, what are a, b and c ?
• If the roots of the quadratic polynomial are equal, what is the discriminant D ?
• If α and 1/α are the zeroes of the quadratic polynomial 2x2 – x + 8k, then find the value of k ?
• Represent the first situation algebraically.

a) 12x+10y=11200

b) 10x+12y=11200

c) 12x-10y=11200

d) 10x-12y=1120

2 Represent the second situation algebraically

a) 46x+55y=51400

b) 55x+46y=51400

c) 55x-46y=51400

d) 46x-55y=51400

3 The system of linear equations representing both the situations will have.

a) Infinite number of solutions

b) Unique solution

c) No Solutions

d) Exactly two solutions

4 The graph of the system of linear equations representing both the situations will be

a) Parallel lines

b) Coincident lines

c) Intersecting lines

d) None of these

• Represent algebraically the situation in hall “Rose”.

a) 50x + y = 10000

b) 50x − y = 10000

c) x + 50y = 10000

d) x − 50y = 10000

2 Represent algebraically the situation in hall “Jasmine”

a) x + 25y = 7500

b) x − 25y = 7500

c) 25x + y = 7500

d) 25x − y = 7500

3 What is the fixed rent of the halls?

4 Find the amount the hotel charged per person.

6 Riya has a field with a flowerbed and grassland. The grassland is in the shape of a rectangle while the flowerbed is in the shape of a square. The length of the grassland is found to be 3 m more than twice the length of the flowerbed. Total area of the whole land is 1260m2

(a)If the length of the square is x m then find the total length of the field i

(b) What will be the perimeter of the whole figure in terms of x?

(c )Find the value of x if the area of total field is 1260 m2

(d) Find the area of grassland and the flowerbed separately.

7 Your friend Veer wants to participate in a 200m race. He can currently run that distance in 51 seconds and with each day of practice it takes him 2 seconds less. He wants to do it in 31 seconds.

1 Write first four terms are in AP for the given situation.

2 What is the minimum number of days he needs to practice till his goal is achieved.

3 How many seconds it takes after the 5th day .

8 Helicopter Patrolling: A helicopter is hovering over a crowd of people watching a police standoff in a parking garage across the street. Stewart notices the shadow of the helicopter is lagging approximately 57 m behind a point directly below the helicopter. If he is 160 cm tall and casts a shadow of 38 cm at this time,

(i) what is the altitude of the helicopter?

(ii) What will be length of shadow of Stewart at 12:00 pm

(iii) Write the name of triangles formed for this situation.

9 Seema has a 10 m × 10 m kitchen garden attached to her kitchen. She divides it into a 10 ×10 grid and wants to grow some vegetables and herbs used in the kitchen. She puts some soil and manure in that and sow a green chilly plant at A, a coriander plant at B and a tomato plant at C. Her friend Kusum visited the garden and praised the plants grown there. She pointed out that they seem to be in a straight line. See the below diagram carefully and answer the following questions:

(i) Find the distance between A and B is

(ii) Find the mid- point of the distance AB

(iii) Find the distance between B and C

10 A heavy-duty ramp is used to winch heavy appliances from street level up to a warehouse loading dock. If the ramp is 2 meter high and the incline is 4 meter long.

(Use √3 = 1.73)

a What angle does the dock make with the street?

b How long is the base of the ramp? ( In round figure)

• If the length of the base is 12 cm and the height is 5 cm then the length of the hypotenuse of that sandwich is:
• If he increases the base length to 15 cm and the hypotenuse to 17 cm, then the height of the sandwich is :
• The value of tan 45° + cot 45°

(a) 1 (b) 2 (c) 3 (d) 4

12 A flag pole ‘h’ metres is on the top of the hemispherical dome of radius ‘r’ metres. A man is standing 7 m away from the dome. Seeing the top of the pole at an angle 45° and moving 5 m away from the dome and seeing the bottom of the pole at an angle 30°.

(i) the height of the pole

(ii) radius( height) of the dome

(iii) Is it possible to see the pole at the angle of 60 0

(iv) If the height of pole is increased, the angle elevation will .....

14 John had a farm with many animals like cows, dogs, horses etc. He had sufficient grass land for the cows and horses to graze, One day Three of his horses were tied with 7 metre long ropes at the three corners of a triangular lawn having sides 20m, 34m and 42m.

(a) Find the area of the triangular lawn .

(b) Find the area of the field that can be grazed by the horses.

(c) The area that cannot be grazed by the horses.

15 Arun, a 10th standard student, makes a project on coronavirus in science for an exhibition in his school. In this project, he picks a sphere which has volume 38808 cm3 and 11 cylindrical shapes, each of volume 1540 cm3 with length 10 cm.

Based on the above information, answer the following questions.

(i) Diameter of the base of the cylinder is

(ii) Diameter of the sphere is

(iii) Total volume of the shape formed is

(iv) Curved surface area of the one cylindrical shape is

(v) Total area covered by cylindrical shapes on the surface of sphere is

• In which age group, will the maximum number of children belong?
• Find the mode of the ages of children playing in the park?

17 Piggy bank or Money box( a coin container) is normally used by children. Piggy bank serves as a pedagogical device to teach about saving money to children. Generally, piggy banks have openings besides the slot for inserting coins but some do not have openings. We have to smash the piggy bank with a hammer or by other means, to get the money inside it. A child Shreya has a Piggybank. She saves her money in her Piggybank. One day she found that her Piggybank contains hundred 50 paisa coins, fifty 1 rupees coin, twenty 2 rupees coin, and ten 5 rupees coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down.

(a) The probability that the fallen coin will be 50 paisa coin, is -------

(b) The probability that the fallen coin will be 5 rupees coin, is---------

(c) The probability that the fallen coin will be 2 rupees coin, is---------

(d) The probability that the fallen coin will be 2 rupees coin or 5 rupees coin, is--------

• Find the probability of getting no heads
• Find the probability of getting one tail

Download answers to CBSE Class 10 Maths Important Case Study Questions

Chapter-wise important case study questions cbse class 10 maths.

• CBSE class 10 Maths syllabus 2024
• NCERT Book for Class 10 Maths
• NCERT Solutions for Class 10 Maths
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• CBSE Class 10 Maths Topper Answer Sheet
• CBSE Class 10 Maths Important Formulas for Last Minute Revision
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CBSE 10th Standard Maths Subject Polynomials Case Study Questions With Solution 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) The expression of the polynomial represented by the graph is

(iii) Find the value of the polynomial represented by the graph when x = 6.

(iv) The sum of zeroes of the polynomial x 2  + 2x - 3 is

(v) If the sum of zeroes of polynomial at 2  + 5t + 3a is equal to their product, then find the value of a.

(ii) Find the value of \(\alpha\) + \(\beta\) +  \(\alpha\) \(\beta\) .

(iii) The value of p(2) is

(iv) If \(\alpha\) and \(\beta\) are zeroes of  \(x^{2}+x-2, \text { then } \frac{1}{\alpha}+\frac{1}{\beta}=\)

(v) If sum of zeroes of  \(q(x)=k x^{2}+2 x+3 k\)  is equal to their product, then k =

(ii) The axis of symmetry of the given parabola is

(iii) The zeroes of the polynomial, represented in the given graph, are

(iv) Which of the following polynomial has -2 and -3 as its zeroes?

(v) For what value of 'x', the value of the polynomial  \(f(x)=(x-3)^{2}+9 \text { is } 9 ?\)

(ii) What will be the expression of the polynomial given in diagram?

(iii) What is the value of the polynomial, represented by the graph, when x = 4?

(iv) If the tunnel is represented by x 2  + 3x - 2, then its zeroes are

(v) If one zero is 4 and sum of zeroes is -3, then representation of tunnel as a polynomial is

(ii) The sum of product of zeroes taken two at a time is

(iii) Product of zeroes of polynomial p(x) is

(iv) The value of the polynomial p(x), when x = 4 is

(v) If  \(\alpha,\beta,\gamma\)  are the zeroes of a polynomial g(x) such that  \(\alpha+\beta+\gamma=3, \alpha \beta+\beta \gamma+\gamma \alpha=-16\)   and  \(\alpha \beta \gamma=-48\)   then, g(x) =

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Cbse 10th standard maths subject polynomials case study questions with solution 2021 answer keys.

(i) (b): Graph of a quadratic polynomial is a parabolic in shape. (ii) (c): Since the graph of the polynomial cuts the x-axis at (-6,0) and (6, 0). So, the zeroes of polynomial are -6 and 6. \(\therefore\) Required polynomial is p(x) = x 2 - (-6 + 6)x + (-6)(6) = x 2 - 36 (iii) (c) : We have, p(x) = x 2 - 36 Now, p( 6) = 62 - 36 = 36 - 36 = 0 (iv) (b): Letf (x) = x 2 + 2x - 3. Then, \(\text { Sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}=-\frac{(2)}{1}=-2\) (v) (d): The given polynomial is at 2 + 5t + 3a Given, sum of zeroes = product of zeroes. \(\Rightarrow \quad \frac{-5}{a}=\frac{3 a}{a} \Rightarrow a=\frac{-5}{3}\)

(i) (b): Given, a and \(\beta\) are the zeroes of  \(p(x)=x^{2}-24 x+128\) \(\text { Putting } p(x)=0 \text { , we get }\) \( x^{2}-8 x-16 x+128=0 \) \(\Rightarrow x(x-8)-16(x-8)=0 \) \(\Rightarrow (x-8)(x-16)=0 \Rightarrow x=8 \text { or } x=16 \) \(\therefore \alpha=8, \beta=16\) (ii) (c) :  \(\alpha+\beta+\alpha \beta =8+16+(8)(16) =24+128=152 \) (iii) (d) :  \(p(2)=2^{2}-2 4(2)+128=4-48+128=84\) (iv) (a): Since a and \(\beta\) are zeroes of  \(x^{2}+x-2\) \(\therefore \quad \alpha+\beta=-1 \text { and } \alpha \beta=-2 \) \(\text { Now, } \frac{1}{\alpha}+\frac{1}{\beta}=\frac{\beta+\alpha}{\alpha \beta}=\frac{-1}{-2}=\frac{1}{2}\) (v) (c): Sum of zeroes  \(=\frac{-2}{k}\) Product of zeroes  \(=\frac{3 k}{k}=3\) According to question, we have  \(\frac{-2}{k}=3\) \(\Rightarrow \quad k=\frac{-2}{3}\)

(i) (b): The shape of the path of the soccer ball is a parabola. (ii) (c): The axis of symmetry of the given curve is a line parallel to y-axis. (iii) (a): The zeroes of the polynomial, represented in the given graph, are -2 and 7, since the curve cuts the x-axis at these points. (iv) (d): A polynomial having zeroes -2 and -3 is  \(p(x)=x^{2}-(-2-3) x+(-2)(-3)=x^{2}+5 x+6\) (v) (c): We have  \(f(x)=(x-3)^{2}+9\) \(\text { Now, } 9=(x-3)^{2}+9 \) \(\Rightarrow(x-3)^{2}=0 \Rightarrow x-3=0 \Rightarrow x=3\)

(i) (a): Since, the graph intersects the x-axis at two points, namely x = 8, -2. So, 8, - 2 are the zeroes of the given polynomial. (ii) (b): The expression of the polynomial given in diagram is  \(-x^{2}+6 x+16\) (iii) (c) : Let  \(p(x)=-x^{2}+6 x+16\) \(\text { When } x=4, p(4)=-4^{2}+6 \times 4+16=24\) (iv) (d): Let  \(f(x)=-x^{2}+3 x-2\) Now, consider  \(f(x)=0 \Rightarrow-x^{2}+3 x-2=0\) \(\begin{aligned} &\Rightarrow x^{2}-3 x+2=0 \Rightarrow(x-2)(x-1)=0\\ &\Rightarrow x=1,2 \text { are its zeroes. } \end{aligned}\) (v) (b): Let a and \(\beta\)  are the zeroes of the required polynomial. Given  \(\alpha + \beta = - 3\) If \(\alpha\) = 4, then  \(\beta\) = -7 \(\therefore \quad \text { Representation of tunnel is }-x^{2}-3 x+28 \text { . }\)

(i) (c): For finding  \(\alpha,\beta,\)   \(\gamma\)  consider p(x) = 0 \(\Rightarrow \quad x^{3}-18 x^{2}+95 x-150=0 \) \(\Rightarrow \quad(x-3)\left(x^{2}-15 x+50\right)=0 \) \(\Rightarrow \quad(x-3)(x-5)(x-10)=0 \Rightarrow x=10 \text { or } x=5 \text { or } x=3 \) \(\text { Thus } \alpha=10, \beta=5 \text { and } \gamma=3\) (ii) (d): Here  \(\alpha=10, \beta=5 \text { and } \gamma=3\) \(\therefore\)   Sum of product of zeroes taken two at a time \(\begin{array}{l} =\alpha \beta+\beta \gamma+\gamma \alpha=(10)(5)+(5)(3)+(3)(10) \\ =50+15+30=95 \end{array}\) (iii) (a): Product of zeroes of polynomial p(x) =  \(\alpha\beta\gamma\) = (10) (5) (3) = 150 (iv) (b): We have  \(p(x)=x^{3}-18 x^{2}+95 x-150\) \(\begin{array}{l} \text { Now, } p(4)=4^{3}-18(4)^{2}+95(4)-150 \\ =64-288+380-150=6 \end{array}\) (v) (d):   \(g(x)=x^{3}-(\alpha+\beta+\gamma) x^{2} +(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma \) \(\Rightarrow g(x)=x^{3}-3 x^{2}-16 x-(-48)=x^{3}-3 x^{2}-16 x+48\)

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NCERT Solutions for Class 10 Maths Chapter 7 Free PDF Download

Ncert solutions for class 10 maths chapter 7 – coordinate geometry.

NCERT Solutions for class 10 maths chapter 7 – Coordinate geometry will help you to make your foundation strong on the concepts of Coordinate geometry class 10. The study of Coordinate geometry class 10  and solving the problems will help you to solve complex problems easily. Coordinate geometry class 10 covers all the exercises provided in the NCERT textbook.

CBSE Class 10 Maths Chapter 7 NCERT Solutions are prepared by our expert at Toppr to help you to prepare for your exams in a better way and enhance your score. Coordinate geometry class 10  provide step by step solutions for the questions given in class 10 maths NCERT textbook as per CBSE Board guidelines and are also prepared according to the exam pattern. With the Toppr app, you can download NCERT Solutions for class 10 maths chapter 7 for free. In case you have a doubt while you are studying, Coordinate geometry class 10, for this we have a team of teachers who prove live doubt solving sessions only for you.

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CBSE Class 10 Maths Chapter 7 NCERT Solutions 

Coordinate geometry class 10 explains the distance between the two points whose coordinates, area of the triangle formed by three given points, coordinates of the point which divides a line segment joining two points in a given ratio, Distance formula, Section formula, Area of a Triangle. Also, the questions are solved with alternative solutions and diagrammatic representation.

Coordinate Geometry Sub-topics

• 7.1 – Introduction to Coordinate Geometry
• 7.2 – Distance Formula
• 7.3 – Section Formula
• 7.4 – Area of a Triangle
• 7.5 – Summary

You can download NCERT Solutions for Class 10 Maths Chapter 7 PDF for free by clicking on the button below.

NCERT Solutions for Class 10 Maths Chapter 7

Q.1 Find the distance between the following pairs of points: 

(2, 3), (4, 1) 

(−5, 7), (−1, 3) 

( a , b ), (− a , − b ) 

Distance between two points ( x 1 , y 1 ) and ( x 2 , y 2  ) is 

√ − ( x −−−−−−−−−−−−−−−− 1 − x 2 ) 2 + ( y 1 − y 2 ) − 2 (i) Distance between (2, 3) and (4, 1) 

= √ − (2 −−−−−−−−−−−−− − 4 ) 2 + (3 − 1 ) − 2 = √ − (−2 −−−−−−−− (2) 2 + ) − 2 = √ − 4+4 −−− = 8√ = 2 2√ 

(ii) Distance between (−5, 7) and (−1, 3) 

= √ − (−5 −−−−−−−−−−−−−−−−− − (−1) ) 2 + (7 − 3 ) − 2 = √ − (−4 −−−−−−−− (4) 2 + ) − 2 = √ − 16 −−−−− + 16 = √ −− 32= 4 2√ 

(iii )Distance between ( a , b ) and (− a , − b ) 

= √ − ( a − −−−−−−−−−−−−−−−−−− (− a ) ) 2 + ( b − (− b ) ) − 2 = √ − (2 a −−−−−−−− ) 2 + (2 b ) − 2 = √ − 4 −−−−−−− a 2 + 4 b 2 = 2 √ − a −−−−− 2 + b 2 #465323 Topic: Distance Between Two Points 

Q.2 Find the distance between the points (0, 0) and (36, 15) . 

Distance Between two given point= 

√ − ( x −−−−−−−−−−−−−−−− 2 − x 1 ) 2 + ( y 2 − y 1 ) − 2 Here 

x 1 = 0, x 2 = 26 and y 1 = 0, y 2 = 15 ∴ Distance between the points (0, 0) and (36, 15) = 

√ − (36 −−−−−−−−−−−−−− − 0 ) 2 + (15 − 0 ) − 2 = √ − +36 −−−−−−− 2 15 2 = √ − 1296 −−−−−−− + 225 − = √ − 1521 −−− = 39 

Q.3 Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear. 

Three points A , B and C 

are collinear if 

AB + BC = AC 

Here, point A (1, 5), B (2, 3) and 

C (−2, −11). ∴ AB = √ − (2 −−−−−−−−−−−−− − 1 ) 2 + (3 − 5 ) − 2 = √ − 1 −−−−−−− 2 +(− 2 2 − ) = √ − 1+4 −−− = 5√ = 2.23 BC = √ − ((−2) −−−−−−−−−−−−−−−−−−−−−− − (2) ) 2 + ((−11) − (3) ) − 2 = √ − (−4 −−−−−−−−−− ) 2 + (−14 ) − 2 = √ − 16 −−−−−− + 196 = √ −−− 212 = 14.56 AC = √ − ((−2) −−−−−−−−−−−−−−−−−−−−−− − (1) ) 2 + ((−11) + (5) ) − 2 = ( √ − 3 ) 2 + (−16 ) 2 = √ − 9 −−−−− + 256 = √ −−− 265 = 16.27 AB + BC = 2.23 + 14.56 = 16.79 

AB + BC ≃ AC 

Hence the given points are collinear. 

Q. 4 Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6) 

Let the required ratio be 

Take ( x 1 , y 1 ) = (−3, 10); ( x 2 , y 2 ) = (6, −8) and 

( x , y ) = (−1, 6) ∴ ⇒ x −1 = = m 1 xm m 2 1 1 + m 2 x 1 

m 2 ×−3 m 1 + m 2 ⇒− m 1 − m 2 =6 m − 3 m 2 

⇒− m 1 −6 m 1 =−3 m 2 + m 2 

⇒ −7 m 1 = −2 m 2 ⇒ m 1 m 2 

= 2 7 ∴ The required ratio is 2:7 

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NCERT Solutions for Class 10

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NCERT Solutions for Class 10 Maths Ch 7 Coordinate Geometry

NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

• Exercise 7.1
• Exercise 7.2
• Exercise 7.3
• Exercise 7.4

How many exercises in Chapter 7 Coordinate Geometry

What is coordinate geometry, find the point on y-axis which is equidistant from the points (5, − 2) and (− 3, 2)., if the points a (4, 3) and b (x, 5) are on the circle with the centre o (2, 3), find the value of x., contact form.

• Exponents and Powers Class 8 Case Study Questions Maths Chapter 10

Last Updated on September 8, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 8 maths. In this article, you will find case study questions for CBSE Class 8 Maths Chapter 10 Exponents and Powers. It is a part of Case Study Questions for CBSE Class 8 Maths Series.

Table of Contents

Case Study Questions on Exponents and Powers

In a class science teacher give some information to all students about Solar system in following manner. Distance of earth from sun = 149600000 km Mass of earth = 5970000000000000000000000 kg Mass of Mars = 642000000000000000000000000000 kg Mass of sun = 1990000000000000000000000000000 kg Mass of moon = 73500000000000000000000 kg

Now asked them to answer some question.

Q. 1. Write distance of earth from sun in standard form? (a) 1.496×10 8 km (b) 14.96×10 8 km (c) 1.496×10 9 km (d) 14.96×10 9 km

Difficulty Level: Medium

Ans. Option (a) is correct. Explanation: In standard notation we write by using power of 10 so 149600000 km = 1.496 × 10 8 km

Q. 2. Write mass of sun in standard notation? (a) 19.9 × 10 28 kg (b) 1.99 × 10 28 kg (c) 1.99 × 10 30 kg (d) 19.9 × 10 30 kg

Ans. Option (c) is correct.

Q. 3. Mass of earth in standard notation: (a) 5.97 × 10 24 kg (b) 59.7 × 10 24 kg (c) 5.97 × 10 26 kg (d) 59.7 × 10 26 kg

Ans. Option (a) is correct.

Q. 4. Calculate the total mass of earth and moon.

Difficulty Level: Hard

Sol. Mass of earth = 5.97 × 10 24 kg Mass of moon = 7.35 × 10 22 kg Change mass of earth = 597.0 × 10 22 kg Sum of mass of earth and moon = 597.0 × 10 22 + 7.35 × 10 22 = 604.35 × 10 22 kg

Q. 5. Calculate difference of mass of mars from mass of sun?

Sol. Mass of sun = 1.99 × 10 30 kg Change mass of sun = 19.9 × 10 29 kg Mass of mars = 6.42 × 10 29 kg Difference between mass of sun and mass of mars is 19.9 × 1029 kg – 6.42 × 10 29 kg = 13.52 × 10 29 kg Hence, difference between mass of sun and mars =13.52×10 29 kg.

• Mensuration Class 8 Case Study Questions Maths Chapter 9
• Algebraic Expressions and Identities Class 8 Case Study Questions Maths Chapter 8
• Comparing Quantities Class 8 Case Study Questions Maths Chapter 7
• Cube and Cube Roots Class 8 Case Study Questions Maths Chapter 6
• Square and Square Roots Class 8 Case Study Questions Maths Chapter 5
• Data Handling Class 8 Case Study Questions Maths Chapter 4
• Understanding Quadrilaterals Class 8 Case Study Questions Maths Chapter 3

Linear Equations in One Variable Class 8 Case Study Questions Maths Chapter 2

Rational numbers class 8 case study questions maths chapter 1, download ebooks for cbse class 8 maths.

• Rational Numbers Topicwise Worksheet for CBSE Class 8 Maths
• Linear Equations in One Variable Worksheet for CBSE Class 8 Maths
• Understanding Quadrilaterals Worksheet for CBSE Class 8 Maths
• Data Handling Worksheet for CBSE Class 8 Maths
• Squares and Square Roots Worksheet for CBSE Class 8 Maths
• Cube and Cube Roots Worksheet for CBSE Class 8 Maths
• Comparing Quantities Worksheet for CBSE Class 8 Maths
• Algebraic Expressions and Identities Worksheet for CBSE Class 8 Maths

Topics from which case study questions may be asked

• Use of Exponents to Express Small Numbers in Standard Form
• Comparing very large number and very small numbers
• Powers with negative Exponents
• Laws of Exponents

When any non-zero integer is multiplied repeatedly with itself, it is called exponential form of the given number.

a b Here, a = base and b = exponent and we read it as ‘a’ raised to the power ‘b’.

Case study questions from the above given topic may be asked.

Frequently Asked Questions (FAQs) on Exponents and Powers Case Study

Q1: what are the important topics covered in chapter 10 exponents and powers for class 8.

A1: The key topics include laws of exponents, simplifying expressions using exponents, powers with negative exponents, and expressing numbers in standard and exponential forms.

Q2: What is an exponent in mathematics?

A2: An exponent refers to the number of times a number, known as the base, is multiplied by itself. For example, in 2 3 , 2 is the base and 3 is the exponent, meaning 2 × 2 × 2 = 8.

Q3: What are the laws of exponents?

A3: – Product law: $a^m \times a^n=a^{m+n}$ – Quotient law: $a^m \div a^n=a^{m-n}$ – Power of a power law: $\left(a^m\right)^n=a^{m \times n}$ – Zero exponent rule: $a^0=1$ (where $a \neq 0$ ) – Negative exponent rule: $a^{-m}=\frac{1}{a^m}$

Q4: How are negative exponents handled?

A4: Negative exponents represent reciprocals. For example, 2 −3 = 1/8​. So, when you have a negative exponent, you convert it into a fraction.

Q5: What is the power of zero?

A5: any non-zero number raised to the power of zero is always equal to 1.

Q6: How do we express large numbers using exponents?

A6: Large numbers are often written using exponents to make them more manageable. For example, 1 billion can be written as 10 9 .

Q7: Why do we use exponents in mathematics?

A7: Exponents simplify the representation and calculation of large or small numbers. They are essential in scientific notation, which is widely used in scientific fields to manage very large or very small values.

Q8: Are there any online resources or tools available for practicing Mensuration case study questions?

A8: We provide case study questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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Home » 7th Class » Class 7 Maths Notes for Exponents and Powers (PDF) – Study Material

Class 7 Maths Notes for Exponents and Powers (PDF) – Study Material

Class 7 Maths Exponents and Powers – Get here the Notes, Question & Practice Paper of Class 7 Maths for topic Exponents and Powers Notes. Exponents and Powers Notes for Class 7 Maths are here. You can download the Exponents and Powers Notes PDF to study all the topics in this chapter. Moreover the Class 7 Maths notes include chapter summary, definitions, examples, and key pointers for Exponents and Powers . Thus if you are studying class Maths (गणित), then the  Exponents and Powers notes  will help you easily understand the topic and ace it.

Class 7 Maths Notes for Exponents and Powers

Exponents and Powers is a critical part in the study of Maths . In India, it is taught in class. Therefore the Class 7 Notes for Maths topic Exponents and Powers have been compiled by teachers and field experts. They explain the complete chapter of Exponents and Powers in one-shot .

Exponents and Powers Notes Download Link – Click Here to Download PDF

Exponents and Powers Notes for Class 7 Maths

Exponents and Powers Class 7 notes is as follows. You can view the document here and also download it use it anytime for future reference whenever you want to brush up your concepts of Maths.

Chapter 11 – Exponents and Powers

Important Concepts: – The continued product of a number multiplied with itself a number of times can be written as the number raised to the power a natural number, equal to the number of times the number is multiplied with itself.

Ex: 5 × 5 × 5 can be written as 5 3 and it read as 5 raised to the power 3 or third power of 5 Similarly, a × a = a 2 , a × a × a = a 3 , a × a × a × a = a 4 In general, if n is a natural number, then, a × a × a × a × ………… × a = a n (a n is called n th power of a)

Candidates who are ambitious to qualify the Class 7 with good score can check this article for Notes, Study Material, Practice Paper. Above we provided the link to access the Notes , Important Question and Practice Paper of Class 7 Maths for topic Exponents and Powers.

All Topics Class 7 Maths Notes

Chapter wise notes for Maths (गणित) are given below.

• Fractions and Decimals
• Data Handling
• Simple Equations
• Lines & Angles
• Triangles & Its Properties
• Comparing Quantities
• Rational Numbers
• Perimeter and Area
• Algebraic Expressions
• Exponents and Powers
• Visualing Solid Shapes

Class 7 Notes for All Subjects

• Class 7 Maths Notes
• Class 7 Science Notes

NCERT Solutions for Class 7 Maths Exponents and Powers

The Exponents and Powers notes here help you solve the questions and answers . Also, you can complete the Class 7 Exponents and Powers worksheet using the same. In addition you will also tackle CBSE Class 7 Maths Important Questions with these Class 7 notes .

However if you still need help, then you can use the NCERT Solutions for Class 7 Maths Exponents and Powers to get all the answers. Exponents and Powers solutions contain questions, answers, and steps to solve all questions.

Notes for All Classes

• Class 6 Notes
• Class 7 Notes
• Class 8 Notes
• Class 9 Notes
• Class 10 Notes
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Exponents and Powers Notes for Class 7 Maths – An Overview

Class 7 Exponents and Powers Notes for All Boards

You can use the Class 7 Maths notes of Exponents and Powers for all boards.

The education boards in India for which Exponents and Powers notes are relevant are – CBSE, CISCE, AHSEC, CHSE Odisha, CGBSE, HBSE, HPBOSE, PUE Karnataka, MSBSHSE, PSEB, RBSE, TBSE, UPMSP, UBSE, BIEAP, BSEB, GBSHSE, GSEB, JAC, JKBOSE, KBPE, MBOSE, MBSE, MPBSE, NBSE, DGE TN, TSBIE, COHSEM, WBCHSE .

Therefore you can refer to these notes as CBSE, CISCE, AHSEC, CHSE Odisha, CGBSE, HBSE, HPBOSE, PUE Karnataka, MSBSHSE, PSEB, RBSE, TBSE, UPMSP, UBSE, BIEAP, BSEB, GBSHSE, GSEB, JAC, JKBOSE, KBPE, MBOSE, MBSE, MPBSE, NBSE, DGE TN, TSBIE, COHSEM, WBCHSE notes for class Class 7 / Class / Maths for the topic Exponents and Powers.

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Class 7 Maths Notes for Perimeter and Area (PDF) – Study Material

Class 7 maths notes for symmetry (pdf) – study material, related posts.

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