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NCERT Solutions for Class 10 Maths Chapter 10 Circles provided here. The ncert solutions have been carefully created by Maths experts at Careers360 keeping in mind of latest syllabus and pattern of CBSE 2023-24. Circles class 10 solutions are a vital resource for students preparing for the Class 10 board exams. These solutions provide step-by-step, comprehensive and detailed answers for all questions in the NCERT Class 10 Maths textbook and are arranged in an exercise-wise format to aid students' preparation and understanding. It is advisable to complete the NCERT Class 10 Maths syllabus at the earliest. To achieve high marks on the exam, students are encouraged to use these NCERT Solutions chapter 10 maths class 10 as part of their study materials. They are free to download using the link given below in this page.
Also See
When a circle with equation x2 + y2 = a2 is intersected by a line y = mx + c, the equation of Tangent Line to Circle: y = mx ± a √(1 + m2)
Relationship between Tangent and Radius - The tangent to a circle is perpendicular to the radius through the point of contact.
Length of Tangent from External Point: 2√(r2 - d2)
Where r is the circle's radius and d is the distance from the external point to the circle's center.
Free download NCERT Solutions for Class 10 Maths Chapter 10 Circles PDF for CBSE Exam.
Circles class 10 NCERT solutions Excercise: 10.1
Q1 How many tangents can a circle have?
Answer:
The lines that intersect the circle exactly at one single point are called tangents. In a circle, there can be infinitely many tangents.
(a) one
A tangent of a circle intersects the circle exactly in one single point.
(b) secant
It is a line that intersects the circle at two points.
(c) Two,
There can be only two parallel tangents to a circle.
(d) point of contact
The common point of a tangent and a circle.
(A) 12 cm
(B) 13 cm
(C) 8.5 cm
(D) cm.
The correct option is (d) = cm
It is given that the radius of the circle is 5 cm. OQ = 12 cm
According to question,
We know that
So, triangle OPQ is a right-angle triangle. By using Pythagoras theorem,
cm
AB is the given line and the line CD is the tangent to a circle at point M and parallels to the line AB. The line EF is a secant parallel to the AB
NCERT solutions circles class 10 Excercise: 10.2
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
The correct option is (A) = 7 cm
Given that,
The length of the tangent (QT) is 24 cm and the length of OQ is 25 cm.
Suppose the length of the radius OT be cm.
We know that is a right angle triangle. So, by using Pythagoras theorem-
OT = 7 cm
(A)
(B)
(C)
(D)
The correct option is (b)
In figure,
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180
(A) 50°
(B) 60°
(C) 70°
(D) 80°
The correct option is (A)
It is given that, tangent PA and PB from point P inclined at
In triangle OAP and
OBP
OA =OB (radii of the circle)
PA = PB (tangents of the circle)
Therefore, by SAS congruence
By CPCT,
Now, OPA = 80/4 = 40
In PAO,
= 50
Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Let line and line
are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents and
respectively.
therefore,
||
{
1 &
2 are alternate angles}
In the above figure, the line AXB is the tangent to a circle with center O. Here, OX is the perpendicular to the tangent AXB ( ) at point of contact X.
Therefore, we have,
BXO +
YXB =
OXY is a collinear
OX is passing through the center of the circle.
Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since APO = 90 0
Therefore, APO is a right-angle triangle. By using Pythagoras theorem;
In the above figure, Pq is the chord to the larger circle, which is also tangent to a smaller circle at the point of contact R.
We have,
radius of the larger circle OP = OQ = 5 cm
radius of the small circle (OR) = 3 cm
OR PQ [since PQ is tangent to a smaller circle]
According to question,
In OPR and
OQR
PRO =
QRO {both
}
OR = OR {common}
OP = OQ {both radii}
By RHS congruence OPR
OQR
So, by CPCT
PR = RQ
Now, In OPR,
by using pythagoras theorem,
PR = 4 cm
Hence, PQ = 2.PR = 8 cm
To prove- AB + CD = AD + BC
Proof-
We have,
Since the length of the tangents drawn from an external point to a circle are equal
AP =AS .......(i)
BP = BQ.........(ii)
AS = AP...........(iii)
CR = CQ ...........(iv)
By adding all the equations, we get;
AP + BP +RD+ CR = AS +DS +BQ +CQ
(AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
AB + CD = AD + BC
Hence proved.
To prove- AOB =
Proof-
In AOP and
AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, AOP
AOC
and by CPCT, PAO =
OAC
..................(i)
Similarly, from OBC and
OBQ, we get;
QBC = 2.
OBC.............(ii)
Adding eq (1) and eq (2)
PAC +
QBC = 180
2( OBC +
OAC) = 180
( OBC +
OAC) = 90
Now, in OAB,
Sum of interior angle is 180.
So, OBC +
OAC +
AOB = 180
AOB = 90
hence proved.
To prove -
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And ,
(since tangents and radius are perpendiculars)
According to question,
In quadrilateral PAOB,
OAP +
APB +
PBO +
BOA =
90 + APB + 90 +
BOA = 360
Hence proved .
Q11 Prove that the parallelogram circumscribing a circle is a rhombus.
To prove - the parallelogram circumscribing a circle is a rhombus
Proof-
ABCD is a parallelogram that circumscribes a circle with center O.
P, Q, R, S are the points of contacts on sides AB, BC, CD, and DA respectively
AB = CD .and AD = BC...........(i)
It is known that tangents drawn from an external point are equal in length.
RD = DS ...........(ii)
RC = QC...........(iii)
BP = BQ...........(iv)
AP = AS .............(v)
By adding eq (ii) to eq (v) we get;
(RD + RC) + (BP + AP) = (DS + AS) + (BQ + QC)
CD + AB = AD + BC
2AB = 2AD [from equation (i)]
AB = AD
Now, AB = AD and AB = CD
AB = AD = CD = BC
Hence ABCD is a rhombus.
Consider the above figure. Assume center O touches the sides AB and AC of the triangle at point E and F respectively.
Let the length of AE is x.
Now in ,
(tangents on the circle from point C)
(tangents on the circle from point B)
(tangents on the circle from point A)
Now AB = AE + EB
Now
Area of triangle
Now the area of
Area of
Area of
Now Area of the = Area of
+ Area of
+ Area of
On squaring both the side, we get
Hence
AB = x + 8
=> AB = 7+8
=> AB = 15
AC = 6 + x
=> AC = 6 + 7
=> AC = 13
Answer- AB = 15 and AC = 13
Given- ABCD is a quadrilateral circumscribing a circle. P, Q, R, S are the point of contact on sides AB, BC, CD, and DA respectively.
To prove-
Proof -
Join OP, OQ, OR and OS
In triangle DOS and
DOR,
OD =OD [common]
OS = OR [radii of same circle]
DR = DS [length of tangents drawn from an external point are equal ]
By SSS congruency, DOS
DOR,
and by CPCT, DOS =
DOR
.............(i)
Similarily,
...............(2, 3, 4)
SImilarily,
Hence proved.
NCERT solutions for circle class 10 are solved by subject experts to help students in their preparation keeping step by step marking in the mind. Circles Class 10 introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. Class 10 Maths ch 10, we will study the different conditions that arise when a line and a circle are given in a plane. Circle class 10 has fundamental concepts that are important for students in their future studies.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
Circle is the chapter, the logic and the memory both are tested equally. First, you should learn the theorems, terminologies, and concepts of circles.
Once you have done the theorems, go through some examples of the NCERT textbook.
When you have done the above-said points, then solve the practice exercises of Circles Class 10 where your understanding of concepts will be tested.
While solving the practice exercises, you can take the help of NCERT solutions for Class 10 Maths hapter 10 circles.
Lastly, solve the previous years question papers related to the particular chapter.
The NCERT Solutions for class 10 maths chapter 10 Circles include several key features that help students to improve their understanding and practice of the material.
These circle class 10 solutions include answers for all exercise questions in the NCERT Class 10 Maths textbook.
These solutions help students to practice challenging questions, align with the updated CBSE syllabus and cover possible exam questions, provide guidance for important drawings.
These solutions aid students in memorizing important formulas and calculation methods.
Happy study!
Triangles, circles and constructions combined have 15 marks weightage in Class 10 final board exam. To obtain a good score in the CBSE 10th board exam follow NCERT book and the previous board papers. More questions can be practiced from NCERT exemplar.
The key concepts covered in NCERT Solutions for circle chapter class 10 include the introduction to circles, the properties of tangents to a circle, the calculation of the number of tangents from a point on a circle, and a summary of the entire chapter. Students can also use chapter 10 class 10 maths solutions in PDF format.
As most of the questions in CBSE board exam are directly asked from NCERT textbook, so must solve all the problems given in the NCERT textbook. NCERT solutions are not only important when you are stuck while solving the problems but you will get to know how to answer in the board exam in order to get good marks in the board exam.
The NCERT Solutions for chapter 10 class 10 maths includes answers to all exercise questions in the NCERT Class 10 Maths textbook. Additionally, they provide guidance for practicing challenging questions, which will enhance students' comprehension of the topics covered in the chapter.
It is essential to study all the questions provided in NCERT Solutions for Class 10 Maths Chapter 10 as they may be present in CBSE board exams and class tests. By familiarizing yourself with these questions, you will be well-prepared for your upcoming exams. In NCERT Solutions, class 10 Maths chapter 10 question answer are comprehensively discussed. Therefore students can use these recourses to score well in the exam.
Exam Date:01 January,2025 - 14 February,2025
Exam Date:01 January,2025 - 14 February,2025
Hello
Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.
1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.
2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.
3. Counseling and Seat Allocation:
After the KCET exam, you will need to participate in online counseling.
You need to select your preferred colleges and courses.
Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.
4. Required Documents :
Domicile Certificate (proof that you are a resident of Karnataka).
Income Certificate (for minority category benefits).
Marksheets (11th and 12th from the Karnataka State Board).
KCET Admit Card and Scorecard.
This process will allow you to secure a seat based on your KCET performance and your category .
check link for more details
https://medicine.careers360.com/neet-college-predictor
Hope this helps you .
Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.
Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.
hello Zaid,
Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.
best of luck!
According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.
You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.
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