NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles
NCERT Exemplar class 10 maths solutions chapter 9 extends the learning of circles. In earlier classes, we have learned about circles; here, we will learn about tangents at any point of the circle. This chapter covers the definition of The NCERT exemplar class 10 maths chapter 9 solutions have been developed by our experienced Mathematics division subject matter experts to give students a quick and easy understanding of NCERT class 10 maths. These Class 10 maths NCERT exemplar chapter 9 solutions on the chapter circles are exceptionally well structured and detailed. The CBSE syllabus for class 10 is the building block for these NCERT exemplar class 10 maths solutions chapter 9.
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Question:1
If radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is tangent to the other circle is
(A) 3 cm (B) 6 cm (C) 9 cm (D) 1 cm
Answer: (B) 6 cm
SolutionAccording to question
Here A is the center and AB = 4 cm radius of small circle and AD = 5 cm radius of large circle.
We have to find the length of CD
ABD is a right angle triangle.
Hence use Pythagoras theorem in ABD
Hence length of chord is 6 cm.
Question:2
In Figure, if ∠AOB = 125°, then ∠COD is equal to
(A) 62.5° (B) 45° (C) 35° (D) 55°
Answer: (D)
Solution
Let ∠COD = x°
As we know that the sum of opposite sides angles of a quadrilateral circumscribing a circle is equal to 180°.
Hence
∠AOB + ∠COD = 180°
125° + ∠COD = 180°
x° = 180° – 125° ( ∠COD = x°)
x° = 55°
Hence, ∠COD = 55°
Question:3
In Figure, AB is a chord of the circle and AOC is its diameter such that ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to
(A) 65° (B) 60° (C) 50° (D) 40°
Answer: (C)
Solution
We know that the angle subtended by a diameter is right.
Hence B = 90°
We know that sum of interior angle of a triangle is 180°.
In ABC
That is
Question:4
From a point P which is at a distance of 13 cm from the centre O of a circle of radius 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
(A) 60 cm^{2} (B) 65 cm^{2} (C) 30 cm^{2} (D) 32.5 cm^{2}
Answer:
(A) 60 cm^{2}Solution
According to question
Given OP = 13 cm
OQ = OR = radius = 5 cm
In POQ, ( PQ is tangent)
Using Pythagoras theorem in POQ
( because H^{2} = B^{2} + P^{2})
Area of POQ = × perpendicular × base
Area of quadrilateral = 2 × area of POQ
= 2 × 30 = 60 cm^{2}
Question:5
At one end A of a diameter AB of a circle of radius 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8 cm from A is
(A) 4 cm (B) 5 cm (C) 6 cm (D) 8 cm
Answer:
(D) 8 cmSolution
According to questions
Given AO = OB = 5 cm
Distance between XY and CD = 8 cm
Since D is the center of the circle and CD is chord. If we join OD it become the radius of circle that is OD = 5cm
AZ = 8 cm (given)
AZ = AO + OZ
8 = 5 + OZ
OZ = 3cm
In ODZ, use Pythagoras theorem
Length of chord CD = 8 cm
Question:6
In Figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to
(A) 4 cm (B) 2 cm (C) (D)
Answer: (C)
SolutionGiven ∠OTA = 30° , OT = 4cm
Join OA, since AT is tangent.
Hence it is perpendicular to OA
In OAT
Question:7
In Figure, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50° with PQ, then ∠POQ is equal to
(A) 100° (B) 80° (C) 90° (D) 75°
Answer: (A)
SolutionGiven :
We know that tangent is perpendicular to radius.
Hence
OPR =
OPR = OPQ + QPR
= OPQ +
OPQ =
OPQ = OQD
Hence , OPQ =
We know that the sum of interior angles of a triangle is
In
O+Q +P =
O +40+40 = 180
o =
Hence OPQ =
Question:8
In Figure, if PA and PB are tangents to the circle with centre O such that APB = 50°, then OAB is equal to
(A) 25° (B) 30° (C) 40° (D) 50°
Answer: (A) 25°
SolutionGiven : APB = 50°
We know that length of tangents drawn from an external point is equal
Hence, PA = PB
Since, PA = PB
Let PAB = PBA = x^{0}
In PAB
p+A+B = ( Sum of interior angles of a tangent is )
PAB = PBA =
PAO = ( tangent is perpendicular to radius)
PAO = PAB +OAB
Question:9
If two tangents inclined at an angle 60° are drawn to a circle of radius 3 cm,then length of each tangent is equal to
(A) (B) 6 cm (C) 3 cm (D)
Answer:
(D)Solution
According to question
Given OQ = OR = 3 cm (Radius)
Draw line OP which bisect . That is
In
Here PQ = PR
Hence , PQ = PR=
Question:10
In Figure, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and BQR = 70°,then AQB is equal to
( A) 20° (B) 40° (C) 35° (D) 45°
Answer:
Answer(B)Solution
Given :,
Since DQ perpendicular to PR
Since DQ bisect
Hence ,
Question:1
Answer:False
SolutionHere CA and CB are the two tangents which is drawn on chord AB and also we know that tangent and radius are perpendicular to each other.
i.e.,
In quadrilateral, ABCD
[Sum of interior angles of a quadrilateral is ]
Here we conclude that angle between the tangents at A and B is .
Therefore the given statement is False.
Question:2
Answer:
FalseSolution
CaseI – When external point P is very close to circle
Here C is the center of circle. Let this radius is 5 and distance of Px and Py is 3.Or we can say that CaseI contradict the given statement.
CaseII – When external point P is far the circle.
Here radius of the given circle is 5 cm and let the length of Px and Py is 10 cm.
Hence according to CaseII and given statement is True.
Hence from the above two cases we conclude that the tangent’s length is depend on the distance of external point from the circle.
Therefore given statement is False because length of tangent from an external point of a circle may or may not be greater than the radius of the circle.
Therefore the given statement is False.
Question:3
Answer:
TrueSolution
Here O is the center of given circle and PA is tangent which is drawn from an external point P. OA is radius of circle.We know that tangent and radius is always perpendicular to each other.
is a right angle triangle Use Pythagoras theorem in
…..(i)
From equation (i) we can say that PA is always less than PO.
In other words we can say that the length of hypotenuse is always greater than the length of perpendicular in a right angle triangle.
i.e., OP PA
Hence the given statement is True.
Question:4
Answer:
TrueSolution
It is possible that the angles of two line may be in only two conditions.
1. When lines are parallel
2.When both the lines are coincide.
Hence the given statement is True
This may be possible only when tangents are parallel or when both the tangents are coincide.
Question:5
Answer:
TrueSolution
Given
Draw line OP from point O to P which bisect .
i.e ,
In
Hence the given statement is True.
Question:6
Answer:
FalseSolution
Given
Draw line OP from point O to P which bisect P. Which bisect
i.e,
In
OP = 2a
Hence the value of OP = 2a
Hence the given statement is False.
Question:7
Answer:
Answer: trueSolution
Given ABC is a isosceles triangle and AB = AC.
To Prove :
Proof : AB = AC (Given)
[ angle between chord of a circle and tangent is equal to angle made by chord in alternate segment]
Also
i.e ,
Hence true
Question:8
Answer:
FalseSolution
According to question.
Here C_{1}, C_{2}, C_{3} are the circle with center O_{1}, O_{2}, O_{3} respectively.
C_{1}, C_{2}, C_{3} touches line PQ at point A. Here PQ is the tangent at each circle.
If we join O_{1}, O_{2}, O_{3} to point A then the line is perpendicular to line PQ because if we draw a line from the center of circle at any point then the tangent at that point is perpendicular to the radius.
But here line joining the centers is not bisecting the line PQ because it depend on the length of PQ.
Hence the given statement is False.
Question:9
Answer:
TrueSolution
According to question.
Here C_{1}, C_{2} circles pass through the point P and Q.
We know that the perpendicular bisector of chord of a circle is always passes through the center of the circle. Hence the perpendicular bisector of line PQ passes through the center of circles of C_{1}, C_{2}.
Hence the given statement is True.
Question:10
Answer:
TrueSolution: First of all we solve the question according to give conditions. If we able to prove it then it will be true otherwise it will be false.
Given :BAC =
Diagram : Construct figure according to given conditions then join BC and OC.
To Prove : BC = BD
Proof :BAC = (Given)
[ angle between chord and tangent id equal to the angle made by chord in alternate segment]
[ Radius and tangent’s angle is always ]
In OAC
OA = OC (both are radius of circle)
[opposite angles of an isosceles triangle is equal]
In
[ sum of interior angle of a trianglE ]
In BCD we conclude that
and
[sides which is opposite to equal angles is always equal]
Hence Proved.
Hence the given statement is true.
Question:1
Answer:
Radius = 3 cmSolution
According to question
Given : AC = 8 CM
OC = 5 CM
AC = AB+BC
8 = BC+BC ( AB =BC)
BC = 4CM
In using Pythagoras theorem
Hence radius of inner circle is 3 cm.
Question:2
Answer:
SolutionAccording to question
To Prove : QORP is a cyclic quadrilateral.
OQ PQ, OR PR ( PQ, PR are tangents)
Hence,
We know that sum of interior angles of quadrilateral is
[Given (i)]
Here we found that sum of opposite angles of quadrilateral is
Hence QORP is a cyclic quadrilateral.
Hence proved
Question:3
Answer:
SolutionAccording to question
Given :
To Prove : BC + BD = BO, i.e., BO = 2BC
Here OC BC, OD BD ( BC, BD are tangents)
Join OB which bisect
In
Hence Proved
Question:4
Answer:
SolutionHere PQ and PR are tangents and O is the center of circle.
Let us join OQ and OR.
Here
( tangent from exterior point is perpendicular to the radius through the point of contact)
In PQO andPRO
(Radius of circle)
(Common side)
Hence, [RHS interior]
Hence, [By CPCT]
Hence O lie on angle bisector of
Hence Proved.
Question:5
In Figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB = CD.
Answer:
SolutionTo Prove : AB = CD
Extend AB and CD then they meet at point E.
Here EA = EC ( length of tangent drawn from same point is equal)
Also EB = ED ( length of tangent drawn from some point is equal)
AB = AE  BE
CD = CE DF
From equal axiom if equal are subtracted from equal then the result is equal.
Hence Proved
Question:6
In Figure, AB and CD are common tangents to two circles of unequal radii.
In above question, if radii of the two circles are equal, prove that AB = CD.
Answer:
SolutionTo Prove AB = CD
According to question
It is given that radius of both circles are equal
Hence, OA = OC = PB = PD
Here,
( tangent at any point is perpendicular to the radius at the point of contact)
Hence ABCD is a rectangle.
Opposite side of a rectangle are equal
Hence Proved.
Question:7
In Figure, common tangents AB and CD to two circles intersect at E. Prove that AB = CD.
Answer:
SolutionTo Prove : AB = CD
We know that the length of tangents drawn from some point to a circle is equal.
So, EB = ED
AE = CE
Here AB = AE + EB
CD = CE + ED
From Euclid axiom is equals are added in equals then the result is also equal.
AB = CD
Hence Proved
Question:8
Answer:
SolutionAccording to question
To Prove : R bisects the arc PRQ
Here …..(i) ( Alternate interior angles)
We know that angle between tangent and chord is equal to angle made by chord in alternate segment.
…..(ii)
From equation (i) and (ii)
We know that sides opposite to equal angles and equal.
PR = QR
Hence R bisects the arc PRQ
Hence Proved
Question:9
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
Answer:
SolutionAccording to question
To Prove :
X_{1}, Y_{1}, X_{2}, Y_{2} are tangents at point A and B respectively.
Take a point C and join AC and AB.
Now …..(i) (Angle in alternate segment)
Similarly,
…..(ii)
From equation (i) and (ii)
…..(iii)
From equation (3)
Hence Proved
Question:10
Answer:
SolutionAccording to question
Let us take a chord EF  XY
Here
( tangent at any point of the circle is perpendicular to the radius through the point of contact)
(Corresponding angles)
Thus AB bisects EF
Hence AB bisects all the chords which are parallel to the tangent at the point A.
Hence Proved
Question:1
If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
Answer:
SolutionGiven ABCDEF hexagon circumscribe a circle.
To Prove : AB + CD + EF = BC + DE + FA
Proof : Here AR = AS [ length of tangents drawn from a point are always equal]
Similarly,
BS = BT
CT = CU
DU = DV
EV = EW
FW = FR
Now AB + CD + EF = (AS + SB) + (CU + UD) + (EW + WF)
= (AR + BT) + (CT + DV) + (EV + FR)
= (AR + FR) + (BT + CT) + (DV + EV)
Now according to Euclid’s axiom when equals are added in equals then the result is also equal
AB + CD + EF = AF + BC + DE
i.e., AB + CD + EF = BC + DE + FA
Hence Proved
Question:2
Answer:
SolutionGiven : BC = a, CA = b, AB = c
Here, AF = AE = Z1 ( tangents drawn from an external point to the circle are equal in length)
Here AB + BC + CA = c + a + b
(AB + FB) + (BC + DC) + (CE + EA) = a + b + c

Hence Proved
Question:3
Answer:
CE = CA [ Tangents from an external point to a circle are equal in length]
Similarly,
DE = DB and PB = PA
Perimeter of DPCD
= PC + CD + PD
= PC + CE + ED + PD ( CD = CE + ED)
= PC + CA + DB + PD
= PA + PB [ PC + CA = PA and DB + PD = PB]
= PA + PA [ PB = PA]
= 2PA
= 2 × 10 [ PA = 10]
= 20 cm
Question:4
Answer:
SolutionHere [ AC is a diameter line, Angle in semi circle formed, is ]
In ABC
[ sum of interior angles of a triangle is ]
......(I)
We know that diameter of a circle is perpendicular to the tangent.
Equate equation (i) and (ii) we get
Hence Proved
Question:5
Answer:
4.8 cmSolution
Given : Radii of two circles are OP = 3 cm and and intersection point of two circles are P and Q. Here two tangents drawn at point P are OP and P
Also apply Pythagoras theorem in we get
Equate equation (i) and (ii) we get
Put x = 1.8 in equation (i) we get
Question:6
Answer:
SolutionLet O be the center of the given circle.
Suppose P meet BC at point R
Construction: Join point P and B
To Prove: BR = RC
Proof: [Given]
In ABC
[ Sum of interior angle of a triangle is 180°]
Also [ tangent and chord made equal angles in alternate segment]
[ angle in semi circle formed is 90°]
Equal equation (i) and (ii) we get
[Side opposite to equal angles are equal]
Also, PR = BR …..(iv) [ tangents drawn to a circle from external point are equal]
From equation (iii) and (iv)
BR = RC
Hence Proved
Question:7
Answer:
Solution
In the given figure PQ and PR are two tangents drawn from an external point P.
PQ = PR [ lengths of tangents drawn from on external point to a circle are equal]
[ angels opposite to equal sides are equal]
In PQR
[ sum of angels of a triangle is 180°]
SROP (Given)
[Alternate interior angles]
Also [Alternate segment angles]
In QRS
Question:8
Answer:
SolutionGiven AB is a diameter and AC is a chord of circle with center O.
To Prove : BC = BD
Construction : Join B and C
Proof :[Angle is alternate segment]
[Angle in semi circle formed is 90°]
In ABC
[Sum of interior angles of a triangle is 180°]
Also , [Linear pair]
In
From equation (i) and (ii)
[ Sides opposite to equal angles are equal]
Hence Proved
Question:9
Answer:
SolutionLet midpoint of arc is C and DCE be the tangent to the circle.
Construction: Join AB, AC and BC.
Proof: In ABC
AC = BC
[ sides opposite to equal angles are equal]
Here DCF is a tangent line
[ angle in alternate segments are equal]
…..(ii) [From equation (i)]
But Here and are alternate angels.
equation (ii) holds only when ABDCE.
Hence the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining the end points of the arc.
Hence Proved.
Question:10
Answer:
SolutionConstruction : Join AO and OS
and
In and
[Radius are equal]
[Common side]
ED = EB [Tangent drawn from an external point to the line circle are equal to length]
[By SSS congruence criterion]
i.e., is bisector of
Similarly OE is bisector of
In quadrilateral
[ is cyclic quadrilateral]
[ AB is a straight line]
[From equation (ii)]
Similarly
From equation (ii)
Dividing both side by 2
Similarly
Dividing both side by 2
Now
[from equation (iii) and (iv)]
So,OEO’ is straight line
O, E and are collinear.
Hence Proved
Question:11
Answer:
AnswerGiven: Radius = 5 cm, OT = 13 cm
[ PT is tangent]
Using Pythagoras theorem OPT
PT and QT are tangents from same point
PT = QT = 12 cm
AT = PT – PA
AT = 12 – PA ..…(i)
Similarly BT = 12 – QB ..…(ii)
Since PA, PF and BF, BQ are tangents from point A and B respectively.
Hence, PA = AE ..…(iii)
BQ = BE ..…(iv)
AB is tangent at point E
Hence OE AB
ET = OT – OF
ET = 13 – 5
ET = 8 cm
In AET, using Pythagoras theorem
[using (i) and (ii)]
Similiraly
Question:12
Answer:
Given :
Here [ PC is tangent]
Here OC = OA (Radius)
[ Sides opposite to equal angles are equal]
We know that PC is tangent and angles in alternate segment are equal.
Hence
In
[Interior angles sum of triangle is 180°]
[using (i)]
…..(ii)
Here
( using (ii))
…..(iii)
In
[using (iii)]
Hence
Question:13
Answer:
AnswerSolution
According to question
In and
AB = AC [Given]
BO = CO [Radius]
AO = AO [Common side]
[By SSS congruence Criterion]
[CPCT]
In and
AB = AC [given]
AD = AD [common side]
[By SAS congruence Criterion]
…..(i) [CPCT]
…..(ii)
From (i) and (ii)
OA is a perpendicular which bisects chord BC
Let AD = x, then OD = 9 – x
Use Pythagoras in ADC
…..(iii)
In ODC using Pythagoras theorem
…..(iv)
From (iii) and (iv)
i.e., AD = 2 cm, OD = 9 – 2 = 7 cm
Put value of x in (iii)
Areao of
Question:14
Answer:
Answer 24 cmSolution
Let us make figure according to question
Given : OA = 13 cm, Radius = 5 cm
Here AP = AQ (tangent from same point)
OP PA, OQ QA ( AP, AQ are tangents)
In OPA using Pythagoras theorem
........(i)
Perimeter of ABC = AB + BC + CA
= AB + BR + RC + CA
= AB + BD + CQ + CA
[ BP and BR are tangents from point B and CP and CQ are tangents from point C]
= AP + AQ [ AP = AB + BP, AQ = AC + CQ]
= AP + AP [ AP = AQ]
= 2AP
= 2 × 12 [using (i)]
= 24 cm
NCERT Exemplar Solutions Class 10 Maths Chapter 9 Important Topics:
The chapter on Circles through NCERT exemplar class 10 maths solutions chapter 9 covers the belowmentioned topics:
◊ Several theorems about a tangent to the circle and their proofs.
◊ Find the number of tangents that can be drawn from any given point
◊ In this chapter, students will learn that the tangent will be perpendicular to the line joining centre and the point of tangent on the circle.
◊ Class 10 maths NCERT exemplar chapter 9 solutions discusses the method that if we draw two tangents on diametrically opposite points of a circle we will observe that these two tangents are parallel.
NCERT Class 10 Exemplar Solutions for Other Subjects:
NCERT Exemplar Class 10 Maths solutions
NCERT Exemplar Class 10 Science solutions
NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
Chapter 
NCERT exemplar solution for class 10 Maths Chapter 1 Real Numbers 
NCERT exemplar solutions for class 10 Maths Chapter 2 Polynomials 
NCERT exemplar solution for class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables 
NCERT exemplar solution for class 10 Maths Chapter 4 Quadratic Equations 
NCERT exemplar solution for class 10 Maths Chapter 5 Arithmetic Progressions 
NCERT exemplar solution for class 10 Maths Chapter 6 Triangles 
NCERT exemplar solution for class 10 Maths Chapter 7 Coordinate Geometry 
NCERT exemplar solution for class 10 Maths Chapter 8 Introduction to Trigonometry & Its Equations 
NCERT exemplar solution for class 10 Maths Chapter 9 Circles 
NCERT exemplar solution for class 10 Maths Chapter 10 Constructions 
NCERT exemplar solution for class 10 Maths Chapter 11 Areas related to Circles 
NCERT exemplar solution for class 10 Maths Chapter 12 Surface Areas and Volumes 
NCERT exemplar solution for class 10 Maths Chapter 13 Statistics and Probability 
Features of NCERT exemplar class 10 maths solutions chapter 9:
These class 10 maths NCERT exemplar chapter 9 solutions provide an extension to the learning of circles done in class 9. The chapter discusses tangents at any point of the circle along with the theorems such as the perpendicular tangent theorem. The chapter on Circles can be better understood and practiced using these NCERT exemplar class 10 maths chapter 9 solutions Circles and will be enough to solve other books such as A textbook of Mathematics by Monica Kapoor, NCERT class 10 maths, RD Sharma class 10 maths, RS Aggarwal class 10 maths et cetera.
NCERT exemplar class 10 maths solutions chapter 9 pdf download is a free and valuable feature. It provides the students with a pdf download facility to refer to the solutions in an offline environment while studying NCERT exemplar Class 10 maths chapter 9.
Class 10 NCERT Solutions Maths for Other Chapters
Chapter No.  Chapter Name 
Chapter 1  
Chapter 2  
Chapter 3  NCERT solutions for class 10 maths chapter 3 Pair of Linear Equations in Two Variables 
Chapter 4  NCERT solutions for class 10 maths chapter 4 Quadratic Equations 
Chapter 5  NCERT solutions for class 10 chapter 5 Arithmetic Progressions 
Chapter 6  
Chapter 7  NCERT solutions for class 10 maths chapter 7 Coordinate Geometry 
Chapter 8  NCERT solutions for class 10 maths chapter 8 Introduction to Trigonometry 
Chapter 9  NCERT solutions for class 10 maths chapter 9 Some Applications of Trigonometry 
Chapter 10  
Chapter 11  
Chapter 12  NCERT solutions for class 10 chapter maths chapter 12 Areas Related to Circles 
Chapter 13  NCERT solutions class 10 maths chapter 13 Surface Areas and Volumes 
Chapter 14  
Chapter 15 
Frequently Asked Question (FAQs)  NCERT Exemplar Class 10 Maths solutions Chapter 9 Circles
Question: How many tangents can be drawn from a point inside the circle?
Answer:
From any point inside the circle, we can never draw a tangent to the circle.
Therefore, the answer to this question will be zero
Question: How many tangents can be drawn from a point outside the circle?
Answer:
From any point outside the circle, we can draw two tangents to the circle.
These two tangents will have equal lengths.
Question: Is the chapter Circles important for Board examinations?
Answer:
Circles is an important chapter for Board examinations as it carries around 68% weightage of the whole paper.
Question: What is the question type distribution for the chapter of Circles in board examinations?
Answer:
Generally, we get MCQ, Fill in the blanks, Short answer, and Long answer questions with a distribution of 1 question for each type. A thorough study and practice from NCERT Exemplar class 10 maths solutions chapter 9 can help you score maximum marks for the same.
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what is the timmg of result declaration of class 10 cbse
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Central Board of Secondary Education may release the CBSE Class 10 result 2021 by July 20, 2021. The officials have not declared any timings exactly when the results will be out. The results will out at any point of the day. Keep an eye on the official website for the latest information. The board publishes CBSE 10th result 2021 online on cbse.nic.in, cbse.gov.in and cbseresults.nic.in. To check the CBSE Class 10th result 2021, students need their roll number, date of birth and other details.
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If you want the soft copies of the books I suggest you to join some telegram groups which provides books, PDFs , question papers etc . It will be beneficial for you .
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Thirdly visit the websites given below
https://ncert.nic.in
https://www.ncertbooks.guru
Hopes this helps.
Wish you luck!
After class 10 result ,how will subjects distribute to students, on the basis of result!?
It depends on the school in which you want admission.
Some schools have the rule that
i. 75%+ can take science.
ii.60%  75% can take commerce.
iii.Below 60% will take humanities.
The choice of choosing stream totally depends on students.
And some schools also check your marks in the subject you want to take admission.
Example If you want to take admission in maths stream they will check your maths and science marks.
If you want to take admission in humanities they will check your social science marks.
Some schools also conduct an entrance exam followed by an interview.
Refer to this link if you are confused about which stream to choose after Class 10th.
https://school.careers360.com/articles/whichstreamselectafterclassx
Feel free to ask if you have any more queries.
All the best!
Board cbse class 10 Is cancelled and promote all student please sir because of corona do not study
Hello aspirant,
No CBSE and ICSE board didn't not cancelled the 10th and 12th exam.
As they had declared that 10th and 12th class exam will not be cancelled and students will have to give it in offline mode in this pandemic also.
Hope this helps you
All the best for your future
CBSE class 10th malayalam can someone please tell me a good one
Adamas World school is the best CBSE school in Kolkata. To get admission please visit https://www.adamasworldschool.org/