case study of chapter 7 class 10th maths

Class 10 Maths Case Study Questions Chapter 7 Coordinate Geometry

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Case study Questions in the Class 10 Mathematics Chapter 7  are very important to solve for your exam. Class 10 Maths Chapter 7 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  Class 10 Maths Case Study Questions  Chapter 7  Coordinate Geometry

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Coordinate Geometry Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 7 Coordinate Geometry

Case Study/Passage-Based Questions

Question 1:

Answer: (d) none of these

(ii) The distance of the bus stand from the house is

Answer: (b) 10 cm

(iii) If the grocery store and electrician’s shop lie on a line, the ratio of the distance of the house from the grocery store to that from the electrician’s shop, is

Answer: (c) 1.2

(iv) The ratio of distances of the house from the bus stand to food cart is

Answer: (c) 1.1

(v) The coordinates of positions of bus stand, grocery store, food cart, and electrician’s shop form a

Question 2:

The class X student’s school in krishnagar has 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 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.

1. Taking A as origin, find the coordinates of P

Answer: a) (4,6)

2. What will be the coordinates of R, if C is the origin?

Answer: c) (10,3)

3. What will be the coordinates of Q, if C is the origin?

b) b) (-6,13)

Answer: d) (13,6)

4. Calculate the area of the triangles if A is the origin

Answer: a) 4.5

5. Calculate the area of the triangles if C is the origin

Answer: d) 4.5

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Coordinate Geometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Class 10 Maths Questions and Answers (Download PDF)

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study. 

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them. 

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers. 

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

Features Of Class 10 Maths Case Study Questions And Answers Pdf

Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-

• Accurate answers of all the Case-based questions given in the PDF.
• Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
• Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.

Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers

Since, CBSE Class 10 Maths Case Study Questions and Answers are prepared by our maths experts referring to the CBSE Class 10 Syllabus, it provided benefits in various way:-

• Case study class 10 maths helps in exam preparation since, CBSE Class 10 Question Papers contain case-based questions.
• It allows students to utilise their learning to solve real life problems.
• Solving case study questions class 10 maths helps students in developing their observation skills.
• Those students who solve Case Study Class 10 Maths on a regular basis become extremely good at answering normal formula based maths questions.
• By using class 10 Maths Case Study questions and answers pdf, students focus more on Selfstudys instead of wasting their valuable time.
• With the help of given solutions students learn to solve all Case Study questions class 10 Maths CBSE chapter wise pdf regardless of its difficulty level.

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Case Study Class 10 Maths Questions

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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.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

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
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

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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• Competency Based Learning in CBSE Schools

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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.

Download NCERT Solutions for Class 10 Maths Chapterwise here .

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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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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.

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CBSE Class 10 Maths: Case Study Questions of Chapter 7 Coordinate Geometry PDF Download

Case study Questions in the Class 10 Mathematics Chapter 7  are very important to solve for your exam. Class 10 Maths Chapter 7 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based   questions for Class 10 Maths Chapter 7  Coordinate Geometry

In CBSE Class 10 Maths Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Coordinate Geometry Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 7 Coordinate Geometry

Case Study/Passage-Based Questions

Question 1:

Answer: (d) none of these

(ii) The distance of the bus stand from the house is

Answer: (b) 10 cm

(iii) If the grocery store and electrician’s shop lie on a line, the ratio of the distance of the house from the grocery store to that from the electrician’s shop, is

Answer: (c) 1.2

(iv) The ratio of distances of the house from the bus stand to food cart is

Answer: (c) 1.1

(v) The coordinates of positions of bus stand, grocery store, food cart, and electrician’s shop form a

Question 2:

The class X student’s school in krishnagar has 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 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.

1. Taking A as origin, find the coordinates of P

Answer: a) (4,6)

2. What will be the coordinates of R, if C is the origin?

Answer: c) (10,3)

3. What will be the coordinates of Q, if C is the origin?

b) b) (-6,13)

Answer: d) (13,6)

4. Calculate the area of the triangles if A is the origin

Answer: a) 4.5

5. Calculate the area of the triangles if C is the origin

Answer: d) 4.5

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Coordinate Geometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Case Study Questions Class 10 Maths with Solutions PDF Download

 case study questions class 10 maths pdf, case study questions class 10 maths with solutions, case study questions class 10 maths cbse chapter wise pdf download, how to solve case-based question in maths.

• First of all, a student needs to read the complete passage thoroughly. Then start solving the question
• After reading the question try to understand from which topics the question is asked. and try to remember all the concepts of that topic.
• Sometimes the question is very tricky and you will find it very difficult to understand. In that case, Read the question and passage again and again.
• After solving the answer check your answer with the options given.
• Remember, write only answering your answer book

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CBSE 10th Standard Maths Subject Case Study Questions With Solution 2021 Part - II

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) Proportional expense for each person is

(iii) The fixed (or constant) expense for the party is

(iv) If there would be 15 guests at the lunch party, then what amount Mr Jindal has to pay?

(v) The system of linear equations representing both the situations will have

(ii) Represent the situation faced by Suman, algebraically

(iii) The price of one Physics book is

(iv) The price of one Mathematics book is

(v) The system of linear equations represented by above situation, has

(ii) Represent algebraically the situation of day- II.

(iii) The linear equation represented by day-I, intersect the x axis at

(iv) The linear equation represented by day-II, intersect the y-axis at

(v) Linear equations represented by day-I and day -II situations, are

Amit is preparing for his upcoming semester exam. For this, he has to practice the chapter of Quadratic Equations. So he started with factorization method. Let two linear factors of  \(a x^{2}+b x+c \text { be }(p x+q) \text { and }(r x+s)\) \(\therefore a x^{2}+b x+c=(p x+q)(r x+s)=p r x^{2}+(p s+q r) x+q s .\) Now, factorize each of the following quadratic equations and find the roots. (i) 6x 2 + x - 2 = 0

(ii) 2x 2 -+ x - 300 = 0

(iii) x 2 -  8x + 16 = 0

(iv) 6x 2 -  13x + 5 = 0

(v) 100x 2 - 20x + 1 = 0

(ii) Difference of pairs of shoes in 17 th  row and 10 th row is

(iii) On next day, she arranges x pairs of shoes in 15 rows, then x =

(iv) Find the pairs of shoes in 30 th row.

(v) The total number of pairs of shoes in 5 th and 8 th row is

(ii) The number on first card is

(iii) What is the number on the 19 th card?

(iv) What is the number on the 23 rd card?

(v) The sum of numbers on the first 15 cards is 

A sequence is an ordered list of numbers. A sequence of numbers such that the difference between the consecutive terms is constant is said to be an arithmetic progression (A.P.). On the basis of above information, answer the following questions. (i) Which of the following sequence is an A.P.?

(ii) If x, y and z are in A.P., then

(iii) If a 1  a 2 , a 3  ..... , a n are in A.P., then which of the following is true?

(iv) If the n th term (n > 1) of an A.P. is smaller than the first term, then nature of its common difference (d) is

(v) Which of the following is incorrect about A.P.?

(ii) Find the radius of the core.

(iii) S 2 =

(iv) What is the diameter of roll when one tissue sheet is rolled over it?

(v) Find the thickness of each tissue sheet

(ii) Distance travelled by aeroplane towards west after   \(1 \frac{1}{2}\)   hr is

(iii) In the given figure, \(\angle\) POQ is 

(iv) Distance between aeroplanes after  \(1 \frac{1}{2}\)   hr is

(v) Area of \(\Delta\) POQ is

(ii) The value of x is

(iii) The value of PR is 

(iv) The value of RQ is 

(v) How much distance will be saved in reaching city Q after the construction of highway? 

(ii) Length of BC =

(iii) Length of AD =

(iv) Length of ED = 

(v) Length of AE = 

(ii) The value of x + y is 

(iii) Which of the following is true?

(iv) The ratio in which B divides AC is

(v) Which of the following equations is satisfied by the given points?

(ii) The value of x is equal to

(iii) If M is any point exactly in between city A and city B, then coordinates of M are

(iv) The ratio in which A divides the line segment joining the points O and M is

(v) If the person analyse the petrol at the point M(the mid point of AB), then what should be his decision?

(ii) The centre of circle is the

(iii) The radius of the circle is

(iv) The area of the circle is

(v) If  \(\left(1, \frac{\sqrt{7}}{3}\right)\)   is one of the ends of a diameter, then its other end is

(ii) The distance between A and Cis

(iii) If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B?

(iv) If the fare for first bus is Rs10/km, then what will be the fare for total journey by that bus?

(v) If the fare for second bus is Rs 15/km, then what will be the fare to reach to the destination by this bus?

*****************************************

Cbse 10th standard maths subject case study questions with solution 2021 part - ii answer keys.

(i) (a): 1 st situation can be represented as x + 7y = 650 ...(i) and 2 nd situation can be represented as x + 11y = 970 ...(ii) (ii) (b): Subtracting equations (i) from (ii), we get  \(4 y=320 \Rightarrow y=80\) \(\therefore\)  Proportional expense for each person is Rs 80. (iii) (c): Puttingy = 80 in equation (i), we get x + 7 x 80 = 650 \(\Rightarrow\) x = 650 - 560 = 90 \(\therefore\)  Fixed expense for the party is Rs 90 (iv) (d): If there will be 15 guests, then amount that Mr Jindal has to pay = Rs (90 + 15 x 80) = Rs 1290 (v) (a): We have a 1  = 1, b 1  = 7, c 1  = -650 and  \(a_{2}=1, b_{2}=11, c_{2}=-970 \) \(\therefore \frac{a_{1}}{a_{2}}=1, \frac{b_{1}}{b_{2}}=\frac{7}{11}, \frac{c_{1}}{c_{2}}=\frac{-650}{-970}=\frac{65}{97}\) \(\text { Here, } \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus, system of linear equations has unique solution.

(i) (a): Situation faced by Sudhir can be represented algebraically as 2x + 3y = 850 (ii) (b): Situation faced by Suman can be represented algebraically as 3x + 2y = 900 (iii) (c) : We have 2x + 3y = 850 .........(i) and 3x + 2y = 900 .........(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get 5y = 750 \(\Rightarrow\)   Y = 150 Thus, price of one Physics book is Rs 150. (iv) (d): From equation (i) we have, 2x + 3 x 150 = 850 \(\Rightarrow\) 2x = 850 - 450 = 400 \(\Rightarrow\) x = 200 Hence, cost of one Mathematics book = Rs 200 (v) (a): From above, we have \(a_{1} =2, b_{1}=3, c_{1}=-850 \) \(\text { and } a_{2} =3, b_{2}=2, c_{2}=-900\) \(\therefore \quad \frac{a_{1}}{a_{2}}=\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{3}{2}, \frac{c_{1}}{c_{2}}=\frac{-850}{-900}=\frac{17}{18} \Rightarrow \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\) Thus system of linear equations has unique solution.

(i) (b): Algebraic representation of situation of day-I is 2x + y = 1600. (ii) (a): Algebraic representation of situation of day- II is 4x + 2y = 3000 \(\Rightarrow\) 2x + y = 1500. (iii) (c) : At x-axis, y = 0 \(\therefore\)   At y = 0, 2x + y = 1600 becomes 2x = 1600 \(\Rightarrow\) x = 800 \(\therefore\) Linear equation represented by day- I intersect the x-axis at (800, 0). (iv) (d) : At y-axis, x = 0 \(\therefore\) 2x + Y = 1500 \(\Rightarrow\)  y = 1500 \(\therefore\) Linear equation represented by day-II intersect the y-axis at (0, 1500). (v) (b): We have, 2x + y = 1600 and 2x + y = 1500 Since  \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \text { i.e., } \frac{1}{1}=\frac{1}{1} \neq \frac{16}{15}\) \(\therefore\) System of equations have no solution. \(\therefore\) Lines are parallel.

(i) (b): We have  \(6 x^{2}+x-2=0\) \(\Rightarrow \quad 6 x^{2}-3 x+4 x-2=0 \) \(\Rightarrow \quad(3 x+2)(2 x-1)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{-2}{3}\) (ii) (c):  \(2 x^{2}+x-300=0\) \(\Rightarrow \quad 2 x^{2}-24 x+25 x-300=0 \) \(\Rightarrow \quad(x-12)(2 x+25)=0 \) \(\Rightarrow \quad x=12, \frac{-25}{2}\) (iii) (d):   \(x^{2}-8 x+16=0\) \(\Rightarrow(x-4)^{2}=0 \Rightarrow(x-4)(x-4)=0 \Rightarrow x=4,4\) (iv) (d):   \(6 x^{2}-13 x+5=0\) \(\Rightarrow \quad 6 x^{2}-3 x-10 x+5=0 \) \(\Rightarrow \quad(2 x-1)(3 x-5)=0 \) \(\Rightarrow \quad x=\frac{1}{2}, \frac{5}{3}\) (v) (a):  \(100 x^{2}-20 x+1=0\) \(\Rightarrow(10 x-1)^{2}=0 \Rightarrow x=\frac{1}{10}, \frac{1}{10}\)  

Number of pairs of shoes in 1 st , 2 nd , 3 rd row, ... are 3,5,7, ... So, it forms an A.P. with first term a = 3, d = 5 - 3 = 2 (i) (d): Let n be the number of rows required. \(\therefore S_{n}=120 \) \(\Rightarrow \quad \frac{n}{2}[2(3)+(n-1) 2]=120 \) \(\Rightarrow \quad n^{2}+2 n-120=0 \Rightarrow n^{2}+12 n-10 n-120=0\) \(\Rightarrow \quad(n+12)(n-10)=0 \Rightarrow n=10\) So, 10 rows required to put 120 pairs. (ii) (b): No. of pairs in 1ih row = t 17 = 3 + 16(2) = 35 No. of pairs in 10th row = t 10  = 3 + 9(2) = 21 \(\therefore\) Required difference = 35 - 21 = 14 (iii) (c) : Here n = 15 \(\therefore\) t 15  = 3 + 14(2) = 3 + 28 = 31 (iv) (a): No. of pairs in 30 th row = t 30 = 3 +29(2) = 61 (v) (c): No. of pairs in 5 th row = t 5  = 3 + 4(2) = 11 No. of pairs in 8 th row = t 8  = 3 + 7(2) = 17 \(\therefore\) Required sum = 11 + 17 = 28

Let the numbers on the cards be a, a + d, a + Zd, ... According to question, We have (a + 5d) + (a + 13d) = -76 \(\Rightarrow\) 2a+18d = -76 \(\Rightarrow\) a + 9d= -38 ... (1) And (a + 7d) + (a + 15d) = -96 \(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2) From (1) and (2), we get 2d= -10 \(\Rightarrow\) d= -5 From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7 (i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5 (ii) (d): Number on first card = a = 7 (iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83 (iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103 (v) (d):  \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)

(i) (c) (ii) (c) (iii) (d) (iv) (b) (v) (c)

Here S n = 0.1n 2 + 7.9n (i) (c): S n -1 = 0.1(n - 1) 2 + 7.9(n - 1) = 0.1n 2 + 7.7n - 7.8 (ii) (b): S 1 = t 1  = a = 0.1(1) 2 + 7.9(1) = 8 cm = Diameter of core So, radius of the core = 4 cm (iii) (a): S 2 = 0.1(2) 2 + 7.9(2) = 16.2 (iv) (d): Required diameter = t 2 = S 2 - S 1 = 16.2 - 8 = 8.2 cm (v) (c): As d = t 2 - t 1  = 8.2 - 8 = 0.2 cm So, thickness of tissue = 0.2 \(\div\)   2 = 0.1 cm = 1 mm

(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\)  Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time =  \(\frac{3}{2}\)  hr. \(\therefore\)  Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after  \(1\frac{1}{2}\)   hour  \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of  \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\)   [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (b): If \(\Delta\) AED and \(\Delta\) BEC, are similar by SAS similarity rule, then their corresponding proportional sides are  \(\frac{B E}{A E}=\frac{C E}{D E}\) (ii) (c): By Pythagoras theorem, we have \(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\) (iii) (a): Since \(\Delta\) ADE and \(\Delta\) BCE are similar. \(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \) \(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\) (iv) (b): \(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \) \(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\) (v) (d) :   \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\) \(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \) \(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (a): We have, OA = 2 \(\sqrt{2}\) km \(\Rightarrow \sqrt{2^{2}+y^{2}}=2 \sqrt{2} \) \(\Rightarrow 4+y^{2}=8 \Rightarrow y^{2}=4 \) \(\Rightarrow y=2 \quad(\because y=-2 \text { is not possible })\) (ii) (c): We have OB = 8 \(\sqrt{2}\) \(\Rightarrow \sqrt{x^{2}+8^{2}}=8 \sqrt{2} \) \(\Rightarrow x^{2}+64=128 \Rightarrow x^{2}=64 \) \(\Rightarrow x=8 \quad(\because x=-8 \text { is not possible })\) (iii) (c) : Coordinates of A and Bare (2, 2) and (8, 8) respectively, therefore coordinates of point M are \(\left(\frac{2+8}{2}, \frac{2+8}{2}\right)\) i.e .,(5.5) (iv) (d): Let A divides OM in the ratio k: 1.Then \(2=\frac{5 k+0}{k+1} \Rightarrow 2 \mathrm{k}+2=5 k \Rightarrow 3 k=2 \Rightarrow k=\frac{2}{3}\) \(\therefore\) Required ratio = 2 : 3 (v) (b): Since M is the mid-point of A and B therefore AM = MB. Hence, he should try his luck moving towards B.

(i) (c): Required coordinates are  \(\left(0, \frac{4}{3}\right)\) (ii) (c) (iii) (a): Radius = Distance between (0,0) and  \(\left(\frac{4}{3}, 0\right)\) \(=\sqrt{\left(\frac{4}{3}\right)^{2}+0^{2}}=\frac{4}{3} \text { units }\) (iv) (b): Area of circle = \(\pi\) (radius) 2 \(=\pi\left(\frac{4}{3}\right)^{2}=\frac{16}{9} \pi \text { sq. units }\) (v) (d): Let the coordinates of the other end be (x,y). Then (0,0) will bethe mid-point of  \(\left(1, \frac{\sqrt{7}}{3}\right)\)  and (x, y). \(\therefore\left(\frac{1+x}{2}, \frac{\frac{\sqrt{7}}{3}+y}{2}\right)=(0,0) \) \(\Rightarrow \frac{1+x}{2}=0 \text { and } \frac{\frac{\sqrt{7}}{3}+y}{2}=0 \) \(\Rightarrow x=-1 \text { and } y=-\frac{\sqrt{7}}{3}\) Thus, the coordinates of other end be  \(\left(-1, \frac{-\sqrt{7}}{3}\right)\)

Coordinates of A, Band Care (-2, -3), (2, 3) and (3,2). (i) (d): Required distance  \(=\sqrt{(2+2)^{2}+(3+3)^{2}}\) \(=\sqrt{4^{2}+6^{2}}=\sqrt{16+36}=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) (ii) (d): Required distance  \(=\sqrt{(3+2)^{2}+(2+3)^{2}}\) \(=\sqrt{5^{2}+5^{2}}=5 \sqrt{2} \mathrm{~km}\) (iii) (b): Distance between Band C \(=\sqrt{(3-2)^{2}+(2-3)^{2}}=\sqrt{1+1}=\sqrt{2} \mathrm{~km}\) Thus, distance travelled by first bus to reach to B \(=A C+C B=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2} \mathrm{~km} \approx 8.48 \mathrm{~km}\) and distance travelled by second bus to reach to B \(=A B=2 \sqrt{13} \mathrm{~km} \approx 7.2 \mathrm{~km}\) \(\therefore\)  Distance of first bus is greater than distance of the second bus, therefore second bus should be chosen. (iv) (d): Distance travelled by first bus = 8.48 km \(\therefore\) Total fare = 8.48 x 10 = Rs 84.80 (v) (b): Distance travelled by second bus = 7. 2 km \(\therefore\) Total fare = 7.2 x 15 = Rs 108  

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• 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.

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• 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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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)

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

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

= 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

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

= 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 a triangle = 1/2 × [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 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.

Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]

= 1/2 (12 + 0 + 9)

= 21/2 square units

Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]

= 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)

Formula, to find area of a triangle = 1/2 × [x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )]

Now, a rea of ΔABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

= 1/2 (-8 + 18 – 16)

= -3 square units

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

Area of ΔACD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]

= 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

OA = √[(x – 6) 2  + (y + 6) 2 ]

OB = √[(x – 3) 2  + (y + 7) 2 ]

OC = √[(x – 3) 2  + (y – 3) 2 ]

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 = √[(x 1 + 1) 2  + (y 1 – 2) 2 ]

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:

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

= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ (- 12 – 3 + 24 )

= 9/2 sq unit

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

= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= 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 [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

The area of ∆ ADE can be calculated as follows:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= 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.

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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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Case Study Questions for Class 7 Maths

• Last modified on: 7 months ago
• Reading Time: 7 Minutes

Table of Contents

Here in this article, we are providing case study questions for class 7 maths.

Maths Class 7 Chapter List

Latest chapter list (2023-24).

There is total 13 chapters.

Chapter 1 Integers Case Study Questions Chapter 2 Fractions and Decimals Case Study Questions Chapter 3 Data Handling Case Study Questions Chapter 4 Simple Equations Case Study Questions Chapter 5 Lines and Angles Case Study Questions Chapter 6 The Triangles and its Properties Case Study Questions Chapter 7 Comparing Quantities Case Study Questions Chapter 8 Rational Numbers Case Study Questions Chapter 9 Perimeter and Area Case Study Questions Chapter 10 Algebraic Expressions Case Study Questions Chapter 11 Exponents and Powers Case Study Questions Chapter 12 Symmetry Case Study Questions Chapter 13 Visualising Solid Shapes Case Study Questions

Old Chapter List

Chapter 1 Integers Chapter 2 Fractions and Decimals Chapter 3 Data Handling Chapter 4 Simple Equations Chapter 5 Lines and Angles Chapter 6 The Triangles and its Properties Chapter 7 Congruence of Triangles Chapter 8 Comparing Quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

Deleted Chapter:

• Chapter 7 Congruence of Triangles
• Chapter 10 Practical Geometry

Tips for Answering Case Study Questions for Class 7 Maths in Exam

1. Comprehensive Reading for Context: Prioritize a thorough understanding of the provided case study. Absorb the contextual details and data meticulously to establish a strong foundation for your solution.

2. Relevance Identification: Pinpoint pertinent mathematical concepts applicable to the case study. By doing so, you can streamline your thinking process and apply appropriate methods with precision.

3. Deconstruction of the Problem: Break down the complex problem into manageable components or steps. This approach enhances clarity and facilitates organized problem-solving.

4. Highlighting Key Data: Emphasize critical information and data supplied within the case study. This practice aids quick referencing during the problem-solving process.

5. Application of Formulas: Leverage pertinent mathematical formulas, theorems, and principles to solve the case study. Accuracy in formula selection and unit usage is paramount.

6. Transparent Workflow Display: Document your solution with transparency, showcasing intermediate calculations and steps taken. This not only helps track progress but also offers insight into your analytical process.

7. Variable Labeling and Definition: For introduced variables or unknowns, offer clear labels and definitions. This eliminates ambiguity and reinforces a structured solution approach.

8. Step Explanation: Accompany each step with an explanatory note. This reinforces your grasp of concepts and demonstrates effective application.

9. Realistic Application: When the case study pertains to real-world scenarios, infuse practical reasoning and logic into your solution. This ensures alignment with real-life implications.

10. Thorough Answer Review: Post-solving, meticulously review your answer for accuracy and coherence. Assess its compatibility with the case study’s context.

11. Solution Recap: Before submission, revisit your solution to guarantee comprehensive coverage of the problem and a well-organized response.

12. Previous Case Study Practice: Boost your confidence by practicing with past case study questions from exams or textbooks. This familiarity enhances your readiness for the question format.

13. Efficient Time Management: Strategically allocate time for each case study question based on its complexity and the overall exam duration.

14. Maintain Composure and Confidence: Approach questions with poise and self-assurance. Your preparation equips you to conquer the challenges presented.

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