Subject-wise New CBSE Syllabus 2021-22 (PDF) for 9th, 10th, 11th & 12th has been officially released online at cbseacademic.nic.in. The new CBSE Syllabus is applicable for CBSE Academic Session 2021-22. The board has already instructed CBSE Schools to start new CBSE Academic Session 2021-22 from April and the board has released the new syllabus before April 2021.
We are providing here the CBSE Class 10 Syllabus 2021-22. Students of class 10 can download from here the new syllabus of all subjects to know the course structure and assessment scheme for the new academic year and plan their studies accordingly.
We are providing here the CBSE Class 10 Syllabus 2021-22. Students of class 10 can download from here the new syllabus of all subjects to know the course structure and assessment scheme for the new academic year and plan their studies accordingly.
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.
CBSE Class 10 Syllabus
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
CBSE 10th Syllabus & Exam Pattern 2024
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.
CBSE Class 10 Syllabus 2024 : Study Scheme
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
CBSE 10 Class Syllabus 2024 : All Subject Wise
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.
CBSE 10th Syllabus 2024 : Compulsory 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 Information Technology Syllabus 2024
CBSE Class 10 Syllabus 2024: Subjects of Internal Assessment
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:
How to Download CBSE Class 10th Syllabus 2024?
Follow the below step-by-step process to download CBSE Class 10th Board Exam Syllabus 2024.
Students first search on the google cbseacademic.nic.in.
Now click on CBSE Academic Website shown in the First Position on Google.
After clicking the Academic Website, a new page will open.
Click on the “Curriculum” Tab and Select the Curriculum 2024 tab shown on the top left side of the homepage.
Now click the “Secondary Curriculum (IX – X) option.
Your Class X All Subjects Syllabus will display on the screen.
Download PDF of CBSE Class 10 Syllabus 2024 for further use.
10th Class CBSE Syllabus: Important Links
CBSE Class 10 Syllabus 2024 : Important Points
Kindly read the following important points regarding the CBSE class 10 syllabus 2024:
Candidates are required to go through the latest CBSE class 10th board examination curriculum.
Analysing the syllabus for the CBSE class 10th board will give candidates a blueprint for the exam.
Familiarising with the CBSE class 10th syllabus 2024 will help candidates to become efficient and effective in their studies.
The syllabus for CBSE class 10th includes some major subjects like Science, Math, Social Science, Hindi, English, Sanskrit, and Computer Science.
How to Prepare for the CBSE 10th Board Exams?
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.”
The board offers a 25 per cent relaxation in attendance requirements for students facing specific circumstances—such as medical emergencies, participation in National or International sports tournaments, or other serious reasons, provided they submit the necessary documentation. However, students with attendance below 75 per cent will not be permitted to sit for the CBSE Class 10 and Class 12 board examinations.
The locus of a point which moves in a plane in such a manner that its distance from a given fixed point is always constant, is called a circle.
The fixed point is called the centre and constant distance is called the radius of the circle. In the figure, ‘O’ is centre and OP = r is a radius. We denote it by C(O, r).
A line segment, terminating (or having its end points) on the circle is called a chord. A chord, passing through the centre is called a diameter of the circle.
A line which intersects a circle in two distinct points is called a secant of the circle.
A line intersecting the circle in exactly one point is called a tangent to the circle.
In the figure, PQ is a chord, AB is a diameter, XY is a secant and ST is a tangent to the circle at C.
Note: (i) Diameter is the longest chord in a circle. (ii) Diameter = 2 × Radius
The length of the complete circle is called its circumference, whereas a piece of a circle between two points is called an arc.
Note: (i) A diameter of a circle divides it into two equal arcs, each of which is called a semicircle. (ii) If the length of an arc is less than the semicircle, then it is a minor arc, otherwise, it is a major arc.
The region consisting of all points lying on the circumference of a circle and inside it is called the interior of the circle.
The region consisting of all points lying outside a circle is called the exterior of the circle.
The region consisting of all points which are either on the circle or lie inside the circle is called the circular region.
A chord of a circle divides it into two parts. Each part is called a segment.
The part containing the minor arc is called the minor segment, and the part containing the major arc is called the major segment.
A quadrilateral of which all the four vertices lie on a circle is called a cyclic quadrilateral. The four vertices A, B, C and D are said to be concyclic points.
Fundamentals of Circles
Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.
If two chords of a circle subtend equal angles at the centre, then the chords are equal.
The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
The perpendicular from the centre of a circle to a chord bisects the chord.
Equal chords of a circle are equidistant from the centre whereas the equidistant chords from the centre are equal.
Chords corresponding to equal arcs are equal.
Congruent arcs of a circle subtend equal angles at the centre.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part.
Angles in the same segment are equal, whereas the angle in a semicircle is a right angle.
The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
If the opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, then the four points are cyclic.
Theorems 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′Sarc(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.
Q9. For what value of a is 2x3 + ax2 + 11x + a + 3 exactly divisible by (2x - 1).
Q10. If x - 2 is a factor of a polynomial f(x) = x5 - 3x4 - ax3 + 2ax + 4, then find the value of a.
Q11. Find the value of a and b so that x2 - 4 is a factor of ax4 + 2x3 - 3x2 + bx - 4
Q12. If x = 2 and x = 0 are zeroes of the polynomial 2x3 - 3x2 + px + q, then find the value of p and q.
Q13. Find the value of a and b, so that x3 - ax2 - 13x + b is exactly divisible by (x - 1) as well as (x + 3).
Q14. The polynomial x3 - mx2 + 4x + 6 when divided by (x + 2) leaves remainder 14, find m.
Q15. If the polynomial ax3 + 3x2 - 13 and 2x3 - 15x + a, when divided by (x - 2) leave the same remainder. Find the value of a.
Q16. If both (x - 2) and \(\left( {x - \dfrac{1}{2}} \right)\) are the factors of px2 + 5x + r, show that p = r.
Q17. If f(x) = x4 - 2x3 + 3x2 - ax + b is divided by x - 1 and x + 1 the remainders are 5 and 19 respectively, then find a and b.
Q18. If A and B be the remainders when the polynomials x3 + 2x2 - 5ax - 7 and x3 + ax2 - 12x + 6 are divided by (x + 1) and (x - 2) respectively and 2A + B = 6, find the value of a.
Q19. Show that x + 1 and 2x - 3 are factors of 2x3 - 9x2 + x + 12.
Q20. If sum of remainders obtained by dividing ax3 - 3ax2 + 7x + 5 by (x + 1) is -36. find a.
Proof: In triangles AOB and COD, we have
Proof: Since the perpendicular from the centre of a circle to a chord bisects the chord. Therefore,
Proof: Since the perpendicular From the Centre of a circle to a chord bisects the chord.
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
Proof: If possible, Let ABCD be not cyclic quadrilateral. Draw a circle passing through three non-collinear points 
