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Sample Paper 2024-25: Here you can get Latest CBSE Sample Papers with Solutions in PDF Format Along with Subject wise such as Maths, Chemistry, Biology, Physics, English & Hindi. You can download model papers class wise
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Much of mathematics is about finding a pattern – a recognisable link between quantities that change. In our daily life, we come across many patterns that characterise relations such as brother and sister, father and son, teacher and student. In mathematics also, we come across many relations such as number m is less than number n, line l is parallel to line m, set A is a subset of set B. In all these, we notice that a relation involves pairs of objects in certain order. In this Chapter, we will learn how to link pairs of objects from two sets and then introduce relations between the two objects in the pair. Finally, we will learn about special relations which will qualify to be functions. The concept of function is very important in mathematics since it captures the idea of a mathematically precise correspondence between one quantity with the other.
The concept of set serves as a fundamental part of the present day mathematics. Today this concept is being used in almost every branch of mathematics. Sets are used to define the concepts of relations and functions. The study of geometry, sequences, probability, etc. requires the knowledge of sets. The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”. In this Chapter, we discuss some basic definitions and operations involving sets.
We shall say that a set is a well-defined collection of objects.
The following points may be noted :
- Objects, elements and members of a set are synonymous terms.
- Sets are usually denoted by capital letters A, B, C, X, Y, Z, etc.
- The elements of a set are represented by small letters a, b, c, x, y, z, etc
The Central Board of Secondary Education (CBSE) is considering introducing two levels of science and social science — standard and advanced — for students in Classes 9 and 10, according to officials.
The board already offers two levels of mathematics for Class 10 students. Its curriculum committee has approved the idea, which is in line with the new National Education Policy (NEP). However, the board's governing body is yet to give its nod.
According to the NEP, all subjects and corresponding assessments, beginning with mathematics, could be offered at two levels with some students studying subjects at the standard level and others at the advanced level.
"The proposal was approved by the curriculum committee. However, the logistics and the framework are yet to be worked out. Currently, we offer only one subject at two levels in Class 10. While the syllabus for students opting for maths (standard) and maths (basic) is the same, the question papers and the difficulty level of the questions in the board exam differ," a senior CBSE official said.
Q1. If a and b are two odd positive integers such that a > b, then prove that one of the two numbers \(\frac{a+b}{2} \) and \(\frac{a-b}{2} \) is odd and the other is even.
Solution: We know that any odd positive integer is of the form 4q+1 or, 4q+3 for some whole number q. Now that it’s given a > b
So, we can choose a= 4q+3 and b= 4q+1.
∴ \( \frac{a+b}{2}=\frac{\left[ (4q+3)+(4q+1) \right]}{2} \)
⇒ \(\Rightarrow \frac{a+b}{2}=\frac{8q+4}{2} \)(a+b)/2 = (8q+4)/2 ⇒ (a+b)/2 = 4q+2 = 2(2q+1) which is clearly an even number.
Now, doing (a-b)/2
⇒ \(\Rightarrow \frac{a-b}{2}=\frac{\left[ (4q+3)-(4q+1) \right]}{2} \)
⇒ (a-b)/2 = (4q+3-4q-1)/2
⇒ (a-b)/2 = (2)/2
⇒ (a-b)/2 = 1 which is an odd number.
Hence, one of the two numbers (a+b)/2 and (a-b)/2 is odd and the other is even.
2
Check out the schedule for all subjects below.
| Date and Time |
Subject Code |
Subject Name |
| Saturday, 15th February, 2025 |
||
| 10:30 AM - 01:30 PM |
101 |
English (Communicative) |
| 184 |
English (Language and Literature) |
|
| Monday, 17th February, 2025 |
||
| 10:30 AM - 12:30 PM |
036 |
Hindustani Music (Per Ins) |
| 131 |
Rai |
|
| 132 |
Gurung |
|
| 133 |
Tamang |
|
| 134 |
Sherpa |
|
| 254 |
Elements of Book Keeping & Accountancy |
|
| 418 |
Physical Activity Trainer |
|
| Tuesday, 18th February, 2025 |
||
| 403 |
Security |
|
| 404 |
Automotive |
|
| 405 |
Introduction to Financial Markets |
|
| 406 |
Introduction to Tourism |
|
| 407 |
Beauty & Wellness |
|
| 408 |
Agriculture |
|
| 409 |
Food Production |
|
| 410 |
Front Office Operations |
|
| 411 |
Banking & Insurance |
|
| 412 |
Marketing & Sales |
|
| 414 |
Apparel |
|
| 415 |
Multi-Media |
|
| 416 |
Multi Skill Foundation Course |
|
| 419 |
Data Science |
|
| 420 |
Electronics & Hardware |
|
| 421 |
Foundation Skills for Sciences |
|
| 422 |
Design Thinking & Innovation |
|
| Thursday, 20th February, 2025 |
||
| 10:30 AM - 01:30 PM |
086 |
Science |
| Saturday, 22nd February, 2025 |
||
| 10:30 AM - 01:30 PM |
018 |
French |
| 119 |
Sanskrit (Communicative) |
|
| 122 |
Sanskrit |
|
| Tuesday, 25th February, 2025 |
||
| 10:30 AM - 01:30 PM |
087 |
Social Science |
| Thursday, 27th February, 2025 |
||
| 10:30 AM - 01:30 PM |
003 |
Urdu Course-A |
| 005 |
Bengali |
|
| 006 |
Tamil |
|
| 009 |
Marathi |
|
| 010 |
Gujarati |
|
| 011 |
Manipuri |
|
| 308 |
Urdu Course-B |
|
| Friday, 28th February, 2025 |
||
| 10:30 AM - 01:30 PM |
002 |
Hindi Course-A |
| 008 |
Hindi Course-B |
|
| Saturday, 1st March, 2025 |
||
| 10:30 AM - 12:30 PM |
049 |
Painting |
| Monday, 3rd March, 2025 |
||
| 10:30 AM - 12:30 PM |
413 |
Health Care |
| Wednesday, 5th March, 2025 |
||
| 10:30 AM - 01:30 PM |
154 |
Elements of Business |
| 10:30 AM - 12:30 PM |
401 |
Retail |
| Thursday, 6th March, 2025 |
||
| 10:30 AM - 01:30 PM |
017 |
Tibetan |
| 020 |
German |
|
| 076 |
National Cadet Corps |
|
| 088 |
Bhoti |
|
| 089 |
Telugu - Telangana |
|
| 092 |
Bodo |
|
| 093 |
Tangkhul |
|
| 094 |
Japanese |
|
| 095 |
Bhutia |
|
| 096 |
Spanish |
|
| 097 |
Kashmiri |
|
| 098 |
Mizo |
|
| 099 |
Bahasa Melayu |
|
| Monday, 10th March, 2025 |
||
| 10:30 AM - 01:30 PM |
041 |
Mathematics Standard |
| 241 |
Mathematics Basic |
|
| Wednesday, 12th March, 2025 |
||
| 10:30 AM - 01:30 PM |
007 |
Telugu |
| 016 |
Arabic |
|
| 021 |
Russian |
|
| 023 |
Persian |
|
| 024 |
Nepali |
|
| 025 |
Limboo |
|
| 026 |
Lepcha |
|
| 031 |
Carnatic Music (Vocal) |
|
| 032 |
Carnatic Music (Melodic Instruments) |
|
| 10:30 AM - 12:30 PM |
033 |
Carnatic Music (Percussion Instruments) |
| 034 |
Hindustani Music (Vocal) |
|
| 035 |
Hindustani Music (Melodic Instruments) |
|
| 136 |
Thai |
|
| Thursday, 13th March, 2025 |
||
| 10:30 AM - 01:30 PM |
064 |
Home Science |
| Monday, 17th March, 2025 |
||
| 10:30 AM - 01:30 PM |
012 |
Punjabi |
| 013 |
Sindhi |
|
| 017 |
Malayalam |
|
| 018 |
Odia |
|
| 014 |
Assamese |
|
| 015 |
Kannada |
|
| 091 |
Kokborok |
|
| Tuesday, 18th March, 2025 |
||
| 10:30 AM - 12:30 PM |
165 |
Computer Applications |
| 402 |
Information Technology |
|
| 417 |
Artificial Intelligence | |
Also Read: CBSE Date Sheet 2025: Class XII Time Table PDF Out at cbse.gov.in
Have a look at the following table for the schedule of all subjects:
| Date and Time |
Subject Code |
Subject Name |
| Saturday, 15th February, 2025 |
||
| 10:30 AM - 01:30 PM |
66 |
Entrepreneurship |
| Monday, 17th February, 2025 |
||
| 10:30 AM - 01:30 PM |
48 |
Physical Education |
| Tuesday, 18th February, 2025 |
||
| 10:30 AM - 12:30 PM |
35 |
Hindustani Music (Melodic Instruments) |
| 36 |
Hindustani Music (Percussion Instruments) |
|
| 821 |
Multi-Media |
|
| 804 |
Automotive |
|
| 813 |
Health Care |
|
| 844 |
Data Science |
|
| 847 |
Electronics & Hardware |
|
| Wednesday, 19th February, 2025 |
||
| 10:30 AM - 01:30 PM |
809 |
Food Production |
| 824 |
Office Procedures & Practices |
|
| 830 |
Design |
|
| 10:30 AM - 12:30 PM |
342 |
Early Childhood Care & Education |
| Thursday, 20th February, 2025 |
||
| 10:30 AM - 01:30 PM |
817 |
Typography & Computer Applications |
| Friday, 21st February, 2025 |
||
| 10:30 AM - 01:30 PM |
42 |
Physics |
| Saturday, 22nd February, 2025 |
||
| 10:30 AM - 01:30 PM |
54 |
Business Studies |
| 833 |
Business Administration |
|
| Monday, 24th February, 2025 |
||
| 10:30 AM - 01:30 PM |
29 |
Geography |
| Tuesday, 25th February, 2025 |
||
| 10:30 AM - 01:30 PM |
118 |
French |
| 822 |
Taxation |
|
| 829 |
Textile Design |
|
| 10:30 AM - 12:30 PM |
843 |
Artificial Intelligence |
| Thursday, 27th February, 2025 |
||
| 10:30 AM - 01:30 PM |
43 |
Chemistry |
| Friday, 28th February, 2025 |
||
| 10:30 AM - 01:30 PM |
805 |
Financial Markets Management |
| 807 |
Beauty & Wellness |
|
| 828 |
Medical Diagnostics |
|
| Saturday, 1st March, 2025 |
||
| 10:30 AM - 01:30 PM |
46 |
Engineering Graphics |
| 10:30 AM - 12:30 PM |
57 |
Bharatanatyam - Dance |
| 58 |
Kuchipudi - Dance |
|
| 59 |
Odissi - Dance |
|
| 60 |
Manipuri - Dance |
|
| 61 |
Kathakali - Dance |
|
| 10:30 AM - 01:30 PM |
816 |
Horticulture |
| 823 |
Cost Accounting |
|
| 836 |
Library & Information Science |
|
| Monday, 3rd March, 2025 |
||
| 10:30 AM - 01:30 PM |
74, 801 |
Legal Studies; Retail |
| Tuesday, 4th March, 2025 |
||
| 10:30 AM - 01:30 PM |
76, 811 |
National Cadet Corps; Banking |
| Wednesday, 5th March, 2025 |
||
| 10:30 AM - 01:30 PM |
808, 812 |
Agriculture; Marketing |
| Thursday, 6th March, 2025 |
||
| 10:30 AM - 01:30 PM |
837 |
Fashion Studies |
| Friday, 7th March, 2025 |
||
| 10:30 AM - 01:30 PM |
835 |
Mass Media Studies |
| 848 |
Design Thinking & Innovation |
|
| Saturday, 8th March, 2025 |
||
| 10:30 AM - 01:30 PM |
41 |
Mathematics |
| 241 |
Applied Mathematics |
|
| Monday, 10th March, 2025 |
||
| 10:30 AM - 12:30 PM |
806 |
Tourism |
| 10:30 AM - 01:30 PM |
827 |
Air-Conditioning & Refrigeration |
| 831 |
Salesmanship |
|
| Tuesday, 11th March, 2025 |
||
| 10:30 AM - 01:30 PM |
1 |
English Elective |
| 301 |
English Core |
|
| Wednesday, 12th March, 2025 |
||
| 10:30 AM - 01:30 PM |
841 |
Yoga |
| 845 |
Physical Activity Trainer |
|
| Thursday, 13th March, 2025 |
||
| 10:30 AM - 01:30 PM |
803 |
Web Application |
| Saturday, 15th March, 2025 |
||
| 10:30 AM - 01:30 PM |
002 |
Hindi Elective |
| 302 |
Hindi Core |
|
| Monday, 17th March, 2025 |
||
| 10:30 AM - 01:30 PM |
003 |
Urdu Elective |
| 022 |
Sanskrit Elective |
|
| 10:30 AM - 12:30 PM |
031 |
Carnatic Music Vocal |
| 032 |
Carnatic Music Melodic Instruments |
|
| 033 |
Carnatic Music Percussion Instruments |
|
| 056 |
Kathak - Dance |
|
| 10:30 AM - 01:30 PM |
303 |
Urdu Core |
| 810 |
Front Office Operations |
|
| 814 |
Insurance |
|
| 818 |
Geospatial Technology |
|
| 819 |
Electrical Technology |
|
| Tuesday, 18th March, 2025 |
||
| 10:30 AM - 12:30 PM |
049 |
Painting |
| 050 |
Graphics |
|
| 051 |
Sculpture |
|
| 052 |
Applied Art (Commercial Art) |
|
| Wednesday, 19th March, 2025 |
||
| 10:30 AM - 01:30 PM |
30 |
Economics |
| Thursday, 20th March, 2025 |
||
| 10:30 AM - 12:30 PM |
034 |
Hindustani Music (Vocal) |
| Friday, 21st March, 2025 |
||
| 10:30 AM - 01:30 PM |
045 |
Biotechnology |
| 073 |
Knowledge Tradition & Practices of India |
|
| 188 |
Bhoti |
|
| 191 |
Kokborok |
|
| 820 |
Electronic Technology |
|
| 825 |
Shorthand (English) |
|
| 826 |
Shorthand (Hindi) |
|
| 834 |
Food Nutrition & Dietetics |
|
| Saturday, 22nd March, 2025 |
||
| 10:30 AM - 01:30 PM |
028 |
Political Science |
| Monday, 24th March, 2025 |
||
| 10:30 AM - 01:30 PM |
322 |
Sanskrit Core |
| Tuesday, 25th March, 2025 |
||
| 10:30 AM - 01:30 PM |
44 |
Biology |
| Wednesday, 26th March, 2025 |
||
| 10:30 AM - 01:30 PM |
55 |
Accountancy |
| Thursday, 27th March, 2025 |
||
| 10:30 AM - 01:30 PM |
39 |
Sociology |
| Saturday, 29th March, 2025 |
||
| 10:30 AM - 01:30 PM |
65 |
Informatics Practices |
| 83 |
Computer Science |
|
| 802 |
Information Technology |
|
| Tuesday, 1st April, 2025 |
||
| 10:30 AM - 01:30 PM |
27 |
History |
| Wednesday, 2nd April, 2025 |
||
| 10:30 AM - 01:30 PM |
104 |
Punjabi |
| 105 |
Bengali |
|
| 106 |
Tamil |
|
| 107 |
Telugu |
|
| 108 |
Sindhi |
|
| 109 |
Marathi |
|
| 110 |
Gujarati |
|
| 111 |
Manipuri |
|
| 112 |
Malayalam |
|
| 113 |
Odia |
|
| 114 |
Assamese |
|
| 115 |
Kannada |
|
| 116 |
Arabic |
|
| 117 |
Tibetan |
|
| 120 |
German |
|
| 121 |
Russian |
|
| 123 |
Persian |
|
| 124 |
Nepali |
|
| 125 |
Limboo |
|
| 126 |
Lepcha |
|
| 189 |
Telugu Telangana |
|
| 192 |
Bodo |
|
| 193 |
Tangkhul |
|
| 194 |
Japanese |
|
| 195 |
Bhutia |
|
| 196 |
Spanish |
|
| 197 |
Kashmiri |
|
| 198 |
Mizo |
|
| Thursday, 3rd April, 2025 |
||
| 10:30 AM - 01:30 PM |
064 |
Home Science |
| Friday, 4th April, 2025 |
||
| 10:30 AM - 01:30 PM |
037 |
Psychology |
Keep checking this page for a detailed subject-wise schedule!
CBSE Date Sheet 2025 Live: The Central Board of Secondary Education, CBSE, on November 20 released the timetable for classes 10 and 12 board exams. Students appearing for the examinations can download the subject-wise date and time for both classes on the official website at cbse.gov.in. Direct link to check CBSE Class 10th, 12th exam 2025 date sheet
The CBSE 10th and 12th exams are scheduled to begin on February 15. The class 10th exam will begin with Communicative English/English Language and Literature. The exam will be held between 10.30 am to 1.30pm. The Class 12th exams will begin with Entrepreneurship between 10.30 am to 1.30pm.
Furthermore, in order to be eligible for the board examinations 2025, it is essential for students to maintain an attendance of 75 per cent or more.
Around 44 lakh students from 8,000 schools in India and abroad are eligible to appear for the Class 10 and 12 board exams this year.
Meanwhile, the Uttar Pradesh Madhyamik Shiksha Parishad (UPMSP) announced the board exam dates 2025.
Both examinations will be held from February 24 to March 12.
CBSE date sheet 2025: Here's how to download
Follow this blog for the live updates on CBSE board exam date sheet.
NCERT Solutions for Class 6 Civics. Chapter 8 Rural Livelihoods. PDF Download Free.
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NCERT Solutions for Class 6 Civics PDF Download Free. Social and Political Life. CBSE Book Question Answers.
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NCERT Solutions for Class 6 Civics. Chapter 2 Diversity and Discrimination. PDF Download Free.
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CBSE Class 10 Syllabus 2024 PDF Download | New | CBSE 10th Syllabus Subject Wise | CBSE 10th Board English Syllabus | Class 10 CBSE Syllabus Maths | Science | Hindi | Social Science.
CBSE Class 10 Syllabus 2024: The CBSE Class 10 Syllabus is out on the cbse board’s official website, cbse.gov.in. Download the all subject-wise PDF of latest syllabus from this article.
Students preparing for the board exam should know the class 10 syllabus. Students should be aware of all the topics, the examination pattern, important details regarding the syllabus, etc. There are no changes in terms of class 10 reduced syllabus or increase in the syllabus. Like every year, the new session for 2023-24 starts in April or May. Below is a table on the class 10 syllabus overview:
| CBSE Class 10 Syllabus | Details |
|---|---|
| Conducting Body | Central Board of Secondary Education (CBSE) |
| Category | CBSE Syllabus |
| Academic Session | 2023-24 |
| Class | 10th |
| Frequency of Conduction | Once in an Academic year |
| Mode of Exam | Offline |
| Exam Duration | 3 Hours |
| Question Paper Marks | 100 marks (Theory marks + Internal Assessments) |
| All Subjects | Maths, Science, Social Science, English Language & Literature, Communicative English, Sanskrit, Hindi Course A, Hindi Course B, Computer Application, Information Technology, Home Science, Painting, Elements of Business. |
| Official Website | cbse.nic.in |
As per the new Class 10 exam pattern the CBSE Board will conduct class 10th board exam once. Therefore, the CBSE Class 10 Syllabus 2024 has been released and advised the students to use.
The new cbse syllabus of Class 10th enables the students to understand the study scheme, types of questions, exam pattern and many more exam related things. Using the latest CBSE Syllabus, students will be able to get an overview of their course structure so that they can begin their board exam preparation from early on.
In CBSE Class 10th there are lots of subjects but all students study 5 compulsory subjects, 2 optional subjects and 2 subjects for internal assessment. Below given table contains the scheme of studies for class 10 CBSE students.
| Type | Subject Name |
|---|---|
| Name of Compulsory Subjects | Science |
| Maths | |
| Social Science | |
| Language A | |
| Language B | |
| Optional Subjects | Skill Subject |
| Language 3 / Any Academic subject apart from above selected subject | |
| Subjects of Internal Assessment | Art Education |
| Health and Physical Education |
The class 10 syllabus is released for subjects including Mathematics, Science, English Grammar, Social Science, Hindi, and English. The board has provided the CBSE class 10 syllabus 2024 on its official portal. Refer to the tables below to learn more about the CBSE class 10 syllabus for various subjects.
The board released the CBSE class 10 syllabus 2024 on its official portal. Refer to the table below to get the direct download link.
| CBSE Class 10 Computer Applications Syllabus 2023-24 |
| CBSE Class 10 Information Technology Syllabus 2024 |
There are a total of 12 Subjects in the Academic Electives Syllabus 2024 such as Carnatic Music (Vocal), Carnatic Music (Melodic Instruments), Carnatic Music (Percussion Instruments), Hindustani Music (Vocal), Hindustani Music (Melodic Instruments), Hindustani Music (Percussion Instruments), Painting, Home Science, National Cadet Corps (NCC), Computer Applications, Elements of Business, Elements of Book Keeping and Accountancy.
The given table has PDF links for the CBSE class 10th syllabus for the academic year 2024. Kindly go through all the subjects and download all the PDFs as mentioned in the above table:
Follow the below step-by-step process to download CBSE Class 10th Board Exam Syllabus 2024.
Kindly read the following important points regarding the CBSE class 10 syllabus 2024:
CBSE Board exams decide the future education journey of the students. The 10th board exam is the path to determine whether students want to opt for Medical, Engineering, Law, or higher studies. Today students have a variety of choices, but for this, it is important to clear their board exams.
It is not only important to study hard, but students should know how to work smart which would boost their performance. Below are a few best tips to prepare for the CBSE 10th board exams:
List down all the important topics you want to study in each subject:-
1. Ensure that you are prepared once with all the topics
2. Prepare a schedule to cover all the important subjects like Math and Science every day within a time frame
3. Do not burden your mind with too many things. Plan the number of days with the subjects and the topics to be covered
4. Take short breaks in between and do something you like
5. Solving sample papers is very important
6. Get your doubts clarified and don’t wait till the last moment
7. Group study is a good way of revising and preparing
8. Try taking a few mock tests timing yourself to get a better hold on writing the exam within the stipulated time
The Central Board of Secondary Education (CBSE) has recently announced guidelines concerning attendance requirements for students preparing for the Class 10 and Class 12 board exams in 2025. According to the notification, all students must maintain a minimum attendance of 75 per cent to qualify for the board exams scheduled for February 2025.
The official notification read, “Schools are more than just places for academic learning; they also play an important role in students’ overall development. In addition to providing academic information, schools facilitate extracurricular activities, peer learning, character development, values inculcation, cooperation, teamwork, diversity and inclusion and many other things. As a result, students’ consistent attendance at school is critical to ensure their overall development.”
Terms Related to Circles
(1) Prove that If two arcs of circle are congruent, then corresponding chords are equal.
Given: Arc PQ of a Circle C(O,r) and arc RS of another circle C(O′,r) such that PQ≅RS
To Prove:PQ = RS
Construction: Draw Line segment OP, OQ, O′R and O′S.
Proof:
Case-I When arc(PQ) and arc(RS) are minor Arcs
In triangle OPQ and O′RS, We have
OP = OQ = O′R = O′S = r [Equal radii of two circles]
∠POQ=∠RO′S arc(PQ)≅arc(RS)⇒m(arc(PQ))≅m(arc(RS))⇒∠POQ=∠RO′S
So by SAS Criterion of congruence, we have
ΔPOQ≅ΔRO′S
⇒PQ=RS
Case-II When arc(PQ) and arc(RS) are major arcs.
If arc(PQ), arc(RS) are major arcs, then arc(QP) and arc(SR) are Minor arcs.
So arc(PQ)≅arc(RS)
⇒ arc(QP)≅arc(SR)
⇒QP=SR
⇒PQ=RS
Hence, PQ≅RS⇒PQ=RS
(2) Prove that If two chords of a circle are equal, then their corresponding arcs are congruent.
Given: Equal chords, PQ of a circle C(O,r) and RS of congruent circle C(O′,r)
To Prove:arc(PQ)≅arc(RS), where both arc(PQ) and arc(RS) are minor, major or semi-circular arcs.
Construction: If PQ,RS are not diameters, draw line segments OP, OQ, O′R and O′S.
Proof:
Case I: when arc(PQ) and arc(RS) are diameters
In this case, PQ and RS are semi circle of equal radii, hence they are congruent.
Case II: When arc(PQ) and arc(RS) are Minor arcs.
In triangles POQ and RO′S, we have
PQ=RS
OP=O′R=r and OQ=O′S=r
So by SSS-criterion of congruence, we have
ΔPOQ≅ΔRO′S
⇒ ∠POQ=∠RO′S
⇒ m(arc(PQ))=m(arc(RS))
⇒ arc(PQ)≅arc(RS)
Case III: When arc(PQ) and arc(RS) are major arcs
In this case, arc(QP) And arc(SR)will be minor arcs.
PQ=RS
⇒ QP=SR
⇒ m(arc(QP))=m(arc(SR))
⇒ 360∘−m(arc(PQ))−360∘−m(arc(RS))
⇒ m(arc(PQ))−m(arc(RS))
⇒ arc(PQ)≅arc(RS)
Hence, in all the three cases, we have arc(PQ)≅arc(RS)
(3) Prove that The perpendicular from the centre of a circle to a chord bisects the chord.
Given: A Chord PQ of a circle C(O,r) and perpendicular OL to the chord PQ.
To Prove:LP=LQ
Construction: Join OP and OQ
Proof: In Triangles PLO and QLO, we have
OP=OQ=r [Radii of the same circle]
OL=OL [Common]
And, ∠OLP=∠OLQ [Each equal to 90∘]
So, by RHS-criterion of congruence, we have
ΔPLO≅ΔQLO
⇒ PL=LQ
(4) Prove that The line segment joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
Given: A Chord PQ OF a circle C(O,r) with mid-point M.
To Prove:OM⊥PQ
Construction: Join OP and OQ
Proof: In triangles OPM and OQM, we have
OP=OQ [Radii of the same circle]
PM=MQ [M is mid-point of PQ]
OM=OM
So, by SSC- criterion of congruence, we have
ΔOPM≅ΔOQM
⇒ ∠OMP=∠OMQ
But , ∠OMP+∠OMQ=180∘ [Linear pair]
⇒ ∠OMP+∠OMP=180∘ [∠OMP=∠OMQ]
⇒ 2∠OMP=180∘
⇒ ∠OMP=90∘
(5) Prove that There is one and only circle passing through three given points.
Given: Three non-collinear points P,Q and R.
To Prove: There is one and only one circle passing through P,Q and R.
Construction: Join PQ and QR. Draw perpendicular bisectors AL and BM of PQ and RQ respectively. Since P,Q and R. are not collinear. Therefore, the perpendicular bisectors AL and BM are not parallel.
Let AL And BM intersect at O. Join OP,OQ and OR.
Proof: Since O lies on the perpendicular bisector of PQ.
Therefore,
OP=OQ
Again, O Lies on the perpendicular bisector of QR.
Therefore,
OQ=OR
Thus, OP=OQ=OR=r (say)
Taking O as the centre draw a circle of radius s. Clearly, C(O,s) passes through P, Q and R. This proves that there is a circle passing the points P,Q and R.
We shall now prove that this is the only circle passing through P,Q and R.
If possible, let there be another circle with centre O′ and radius r, passing through the points P,Q and R. Then, O′ will lie on the perpendicular bisectors AL of PQ and BM of QR.
Since two lines cannot intersect at more than one point, so O′ must coincide with O. Since OP=r, O′P=s and O and O′ coincide, we must have
r=s
⇒ C(O,r)=C(O′,s)
Hence, there is one and only one circle passing through three non-collinear points P,Q and R.
(6) Prove that If two circles intersect in two points, then the through the centre is perpendicular to the common chord.
Given: Two circles C(O,r) and C(O′,s) intersecting at points A and B.
To Prove:OO′ is perpendicular bisector of AB.
Construction: Draw line segments
OA,OB,O′A and O′B
Proof: In triangles OAO′ and OBO′, we have
OA=OB=r
O′A=O′B=s
And, OO′=OO′
So, by SSS-criterion of congruence, we have
ΔOAO≅ΔOBO′
⇒ ∠AOO′=∠BOO′
⇒ ∠AOM=∠BOM [∠AOO′=∠AOM and ∠BOM=∠BOO′]
Let M be the point of intersection of AB and OO′
In triangles AOM and BOM, we have
OA=OB=r
⇒ ∠AOO′=∠BOO′
⇒ ∠AOM=∠BOM [∠AOO′=∠AOM and ∠BOM=∠BOO′]
Let M be the point of intersection of AB and OO′
In triangles AOM and BOM, we have
OA=OB=r
∠AOM=∠BOM
And OM=OM
So, by SAS-criterion of congruence, we have
ΔAOM≅ΔBOM
⇒ AM=BM and ∠AMO=∠BOM
But, ∠AOM+∠BMO=180∘
2∠AOM=180∘
⇒ ∠AOM=90∘
Thus, AM=BM and ∠AOM=∠BMO=90∘
Hence, OO′ is the perpendicular bisector of AB.
(7) Prove that Equal chords of a circle subtend equal angle at the centre.
Given: Two Chord AB and CD of circle C(O,r) such that AB=CDand OL⊥AB and OM⊥CD
To Prove: Chord AB and CD are equidistant from the centre O i.e OL=OM.
Construction: Join OA and OC.
Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
OL⊥AB⇒AL=12AB..........(i)
And, OM⊥CD⇒CM=12CD..........(ii)
But, AB=CD
⇒ 12AB=12CD
⇒ AL=CM [Using (i) and (ii) ]….......(iii)
Now, in right triangles OAL and OCM, we have
OA=OC [Equal to radius of the circle]
AL=CM [From equation (iii)]
And, ∠ALO=∠CMO [Each equal to 90∘]
So by RHS criterion of convergence, we have
ΔOAL≅ΔOCM
⇒ OL=OM
Hence, equal chord of a circle are equidistant from the centre.
(8) Prove that Chords of a circle which are equidistant from the centre are equal.
Given: Two Chords AB and CD of a circle C(O,r) which are equidistant from its centre i.e. OL=OM, where OL⊥AB and OM⊥CD.
To Prove: Chords are Equal i.e. AB=CD
Construction: Join OA and OC
Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord.
Therefore,
OL⊥AB
⇒ AL=BL
⇒ AL=12AB
And, OM⊥CD
⇒ CM=DM
⇒ CM=12CD
In triangles OAL and OCM, we have
OA=OC [Each equal to radius of the given Circle]
∠OLA=∠OMC [Each equal to 90∘]
And, OL=OM [Given]
So, by RHS, criterion of convergence, we have
ΔOAL≅ΔOCM
⇒ AL=CM
⇒ 12AL=12AB
⇒ AB=CD
Hence, the chords of a circle which are equidistant from the centre are equal.
(9) Prove that Equal chords of a circle subtend equal angle at the centre.
Given: A circle C(O,r) and its two equal chords AB and CD.
To Prove:∠AOB=∠COD
Proof:In triangles AOB and COD, we have
AB=CD [Given]
OA=OC [Each equal to r]
OB=OD [Each equal to r]
So, by SSC-criterion of Congruence, we have
ΔAOB≅ΔCOD
⇒ ∠AOB=∠COD
(15) Prove that If the angles subtended by two chords of a circle at the centre are equal, the chords are equal.
Given: Two Chord AB and CD of a circle C(O,r) such that ∠AOB=∠COD
To Prove: AB=CD
Proof: In triangles AOB and COD, we have
OA=OC [Each equal to r]
∠AOB=∠COD [Given]
OB=OD [Each equal to r]
So, by SAS-criterion of congruence, we have
ΔAOB≅ΔCOD
⇒ AB=CD
(11) Prove that Of any two chords of a circle, the larger chord is nearer to the centre.
Given: Two Chord AB and CD of a circle with Centre O such that AB>CD
To Prove: Chord AB is nearer to the centre of the circle i.e. OL<OM, where OL and OM are perpendiculars from O to AB and CD respectively
Construction: Join OA and OC.
Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
OL⊥AB⇒AL=12AB
And, OM⊥CD⇒CM=12CD
In right triangles OAL and OCM, we have
OA2=OL2+AL2
And, OC2=OM2+CM2
⇒ OL2+AL2=OM2+CM2.. (i) [OA=OC⇒OA2=OC2]
Now, AB>CD
⇒ 12AB>12CD
⇒ AL>CM
⇒ AL2>CM2
⇒ OL2+AL2>OL2+CM2 [Adding OL2 on both sides]
⇒ OM2+CM2>OL2+CM2 [using equation (i)]
⇒ OM2>OL2
⇒ OM>OL
⇒ OL<OM
Hence, AB is nearer to the centre than CD.
(12) Prove that Of any two chords of a circle, the chord nearer to the centre is larger.
Given: Two Chord AB and CD of a circle C(O,r) such that OL<OM, where OL and OM are perpendiculars From O on AB and CD respectively.
To Prove:AB>CD
Construction: Join OA and OC.
Proof: Since the perpendicular From the Centre of a circle to a chord bisects the chord.
AL=12AB and CM=12CD
In right triangles OAL and OCM, we have
OA2=OL2+AL2 and, OC2=OM2+CM2
⇒ AL2=OA2−OL2....... (i)
And, CM2=OC2−OM2.......(ii)
Now, OL<OM
⇒ OL2<OM2
⇒ −OL2>−OM2
⇒ OA2−OL2>OA2−OM2 [adding OA2 on both sides]
⇒ OA2−OL2>OC2−OM2 [OA2=OC2]
⇒ AL2>CM2
⇒ AL>CM
⇒ 2AL>2CM
⇒ AB>CD
(13) Prove that The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Given: An arc PQ of a circle C(O,r) and a point R on the remaining part of the circle i.e. arc QP.
To Prove:∠POQ=2∠PRQ
Construction: join RO and produce it to a point M outside the circle.
Proof: We shall consider the following three different cases:
Case I: when arc(PQ) is a minor arc.
We know that an exterior angle of a triangle is equal to the sum of the interior oppsite angles.
In ΔPOQ, ∠POM is the exterior angle.
∠POM=∠OPR+∠ORP
⇒∠POM=∠ORP+∠ORP [OP=OR=r, ∠OPR=∠ORP]⇒∠POM=2∠ORP.....................(i)
In ΔQOR, ∠QOM is the exterior angle.
∠QOM=∠OQR+∠ORQ
⇒∠QOM=∠OQP+∠ORQ [OQ=OR=r, ∠ORQ=∠OQR]
⇒∠QOM=2∠ORQ................(ii)
Adding equation (i) and (ii), we get
∠POM+∠QOM=2∠ORP+2∠ORP
⇒∠POM+∠QOM=2(∠ORP+∠ORP)
⇒ ∠POM=2∠PRQ
Case II: when arc(PQ) is a semi-circle
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles.
In ΔPOQ, we have
∠POM=∠OPR+∠ORP
⇒∠POM=∠ORP+∠ORP [OP=OR=r, ∠OPR=∠ORP]
⇒∠POM=2∠ORP.....................(iii)
In ΔQOR, We have
∠QOM=∠ORQ+∠OQR
⇒∠QOM=∠ORQ+∠ORQ [OQ=OR=r, ∠ORQ=∠OQR]
⇒∠QOM=2∠ORQ................(iv)
Adding equations (iii) and (iv), we get
∠POM+∠QOM=2(∠ORP+∠ORQ)
∠POQ=2∠PRQ
Case III: When arc(PQ) is a major arc.
We know that an exterior angle of a triangle is equal to the sum of the interior opposite angles
In ΔPOR, we have
∠POM=∠ORP+∠ORP [OP=OR=r, ∠OPR=∠ORP]
⇒∠POM=2∠ORP ...................(v)
In ΔQOR, we have
∠QOM=∠ORQ+∠OQR
⇒∠QOM=2∠ORQ ...................(vi)
Adding equations (v) and (vi), we get
∠POM+∠QOM=2(∠ORP+∠ORP)
⇒ Reflex ∠POQ=2∠PRQ
(14)Prove that Angles in the same segment of a circle are equal.
Given: A circle C(O,r), an arc PQ and two angles ∠PRQ and ∠PSQ in the same segment of the circle.
To Prove:∠PRQ= ∠PSQ
Construction: Join OP and OQ
Proof: we know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point in the remaining part of the circle. So we have
∠POQ=2∠PRQ and ∠POQ=2∠PSQ
⇒2∠PRQ=2∠PSQ
⇒∠PRQ=∠PSQ
We have
Reflex ∠POQ=2∠PRQ and ∠POQ=2∠PSQ
⇒2∠PRQ=2∠PSQ
⇒∠PRQ=∠PSQ
Thus , in both the cases, we have
∠PRQ=∠PSQ
(15) Prove that The angle in a semi-circle is a right angle.
Given:PQ is a diameter of a circle C(O,r) and ∠PRQ is an angle in semi-circle.
To Prove:∠POQ=90∘
Proof: we know that the angle subtended by an arc of a circle at its centre is twice the angle formed by the same arc at a point on the circle. So, we have
∠POQ=∠PRQ
⇒180∘=2∠PRQ [POQ is a straight line]
⇒∠PRQ=90∘
(16) Prove that The opposite angles of a cyclic quadrilateral are supplementary.
Given: A Cyclic quadrilateral ABCD
To Prove:∠A+∠C=180∘ and ∠B+∠D=180∘
Construction: Join AC and BD.
Proof: Consider side AB of quadrilateral ABCD as the Chord of the circle. Clearly, ∠ACB and ∠ADB are angles in the same segment determined by chord AB of the Circle.
∠ACB=∠ADB ………….(i)
Now , consider the side BC of quadrilateral ABCD as the chord of the circle. We find that ∠BAC and ∠BDC are angles in the same segment
∠BAC = ∠BDC [angles in the same segment are equal]..(ii)
Adding equation (i) and (ii), we get
⇒ ∠ACB+∠BAC=∠ADB+∠BDC
⇒ ∠ACB+∠BAC=∠ADC
⇒ ∠ABC+∠ACB+∠BAC=∠ABC+∠ADC
⇒ 180∘=∠ABC+∠ADC [sum of angle of triangle is 180∘ ]
⇒ ∠ABC+∠ADC=180∘
⇒ ∠B+∠D=180∘
But, ∠A+∠B+∠C+∠D=360∘
∠A+∠C=360∘−(∠B+∠D)
⇒∠A+∠C=360∘−180∘=180∘
Hence, ∠A+∠C=180∘ and ∠B+∠D=180∘
The converse of this theorem is also true as given below.
(17) Prove that If the sum of any pair of opposite angles of a quadrilateral is 180∘,then it is cyclic.
Given: A quadrilateral ABCD in which ∠B+∠D=180∘
To Prove: ABCD is acyclic quadrilateral.
Proof: If possible, Let ABCD be not cyclic quadrilateral. Draw a circle passing through three non-collinear points A, B and C. Suppose the circle meets AD or AD produced at D′. Join D′C.
Now, ABCD’ is a cyclic quadrilateral.
∠ABC+∠AD′C=180∘..............(i)
But, ∠B+∠D=180∘
i.e. ∠ABC+∠ADC=180∘..............(ii)
from (i) and (ii), we get
∠ABC+∠AD′C = ∠ABC+∠ADC
⇒ ∠AD′C = ∠ADC
⇒ An exterior angle of ΔCDD′ is equal to interior oppsite angle.
But, this is not possible, unless D′ coincides with D. Thus, the circle passing through A,B,C also passes through D.
Hence, ABCD is a cyclic Quadrilateral.
(18) Prove that If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
Given: A Cyclic quadrilateral ABCD one of whose side AB is produced to E.
To Prove:∠CBE=∠ADC
Proof: Since ABCD is a quadrilateral and the sum of opposite pairs of angles in a cyclic quadrilateral is 180∘
∠ABC+∠ADC=180∘
But, ∠ABC+∠CBE=180∘ [Liner Pairs]
∠ABC+∠ADC=∠ABC+∠CBE
⇒ ∠ADC=∠CBE
Or, ∠CBE=∠ADC
(19) Prove that The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
Given: A Cyclic quadrilateral ABCD in which AP,BP,CR and DR are the bisectors of ∠A, ∠B, ∠C and ∠D respectively such that a quadrilateral PQRS is formed.
To Prove:PQRS is a cyclic quadrilateral.
Proof: In order to prove that PQRS is a cyclic quadrilateral, it is sufficient to show that
∠APB+∠CRD=180∘
Since the sum of the angles of a triangle is 180∘. Therefore, in triangles APB and CRD, we have
∠APB+∠PAB+∠PBA=180∘
And, ∠CRD+∠RCD+∠RDC=180∘
⇒ ∠APB+12∠A+12∠B=180∘
And, ∠CRD+12∠C+12∠D=180∘
⇒ ∠APB+12∠A+12∠B+∠CRD+12∠C+12∠D=180∘+180∘
∠APB+∠CRD+12{∠A+∠B+∠C+∠D}=360∘
∠APB+∠CRD+12{(∠A+∠C)+(∠B+∠D)}=360∘
∠APB+∠CRD+12(180∘+180∘)=360∘
∠APB+∠CRD=180∘
Hence, PQRS is a cyclic Quadrilateral.
(20) Prove that If two sides cyclic quadrilateral are parallel, then the remaining two sides are equal and the diagonals are also equal.
Given: A Cyclic quadrilateral ABCD in which AB∥DC.
To Prove: (i) AD=BC (ii) AC=BD
Proof: In order to prove the desired results, it is sufficient to show that ΔADC≅ΔBCD. Since ABCD is cyclic Quadrilateral and sum of opposite pairs of angles in a cyclic Quadrilateral is 180∘
∠B+∠D=180∘........(i)
Since AB∥DC and BC is a transversal and sum of the interior angles on the same side of a transversal is 180∘
∠ABC+∠BCD=180∘
∠B+∠C=180∘...................(ii)
From (i) and (ii), we get
∠B+∠D=∠B+∠C
⇒ ∠C=∠D.................(iii)
Now, consider triangles ADC and BCD. In ΔADC and ΔBCD, we have
∠ADC=∠BCD [From equation (iii)]
DC=DC [Common]
And, ∠DAC=∠CBD [∠DAC and ∠CBD are angles in the segment of chord CD]
So, by AAS-criterion of congruence, we have
ΔADC≅ΔBCD
⇒ AD=BC and AC=BD
(21) Prove that If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.
Given: A cyclic quadrilateral ABCD such that AD=BC.
To Prove: AB∥CD
Construction: Join BD.
Proof: We have,
AD=BC
⇒ DA⌢≅BC⌢
⇒ m(DA)⌢≅(BC)⌢
⇒ 2∠2=2∠1
⇒ ∠2=∠1
But, these are alternate interior angles. Therefore, AB∥CD.
(22) Prove that An isosceles trapezium is cyclic.
Given: A trapezium ABCD in which AB∥DC and AD=BC
To Prove:ABCD is a cyclic trapezium.
Construction: Draw DE⊥AB and CF⊥AB.
Proof: In order to prove that ABCD is a cyclic trapezium, it is sufficient to show that ∠B+∠D=180∘.
In triangles DEA and CFB, we have
AD=BC [Given]
∠DEA=∠CFB [Each equal to 90∘]
And, DE=CF
So, by RHS-criterion of congruence, we have
ΔDEA≅ΔCFB
⇒ ∠A=∠B and ∠ADE=∠BCF
Now, ∠ADE=∠BCF
⇒ 90∘+∠ADE=90∘+∠BCF
⇒ ∠EDC+∠ADE=∠FCD+∠BCF
⇒ ∠ADC=∠BCD
⇒ ∠D=∠C
Thus, ∠A=∠B and ∠C=∠D.
∠A+∠B+∠C+∠D=360∘
⇒ 2∠B+2∠D=360∘
⇒ ∠B+∠D=180∘
Hence, ABCD is a cyclic quadrilateral.
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