NCERT Solutions for Class 10 Maths Chapter 10 Circles
NCERT Solutions for Class 10 Maths Chapter 10 Circles: Students must refer to the NCERT Class 10 Maths solutions chapter 10 to understand the chapters in a better way. NCERT Solutions for Class 10 Maths Chapter 10 Circles introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. In this chapter, students will study the different conditions that arise when a line and a circle are given in a plane. Here you will get NCERT solutions for Class 6 to 12 for science and Maths. NCERT Solutions for Class 10 Maths Chapter 10 Circles will help the students to prepare for the exams in a strategic way. It is advisable to complete the NCERT Class 10 Maths syllabus at the earliest. Students must also refer to the NCERT Class 10 Maths books for the exam preparation.
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NCERT solutions for class 10 maths chapter 10 Circles 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.
Answer:
(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.
Answer:
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
Answer:
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 for class 10 maths chapter 10 Circles Excercise: 10.2
(A) 7 cm
(B) 12 cm
(C) 15 cm
(D) 24.5 cm
Answer:
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)
Answer:
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°
Answer:
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.
Answer:
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}
Answer:
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.
Answer:
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;
Answer:
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
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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.
Answer:
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
Answer:
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 class 10 maths chapter 10 Circles are solved by subject experts to help students in their preparation keeping step by step marking in the mind. This chapter introduces some complex and important terms like tangents, tangents to a circle, number of tangents from a point on a circle. In this chapter, we will study the different conditions that arise when a line and a circle are given in a plane. This chapter has fundamental concepts that are important for students in their future studies. To solve these types of situations, we will learn the approach to apply the concept of the tangent to a circle in NCERT solutions for class 10 maths chapter 10 Circles.
NCERT solutions for class 10 Maths Chapter Wise
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 | NCERT solutions class 10 maths chapter 10 Circles |
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 |
How to use NCERT solutions for Class 10 Maths Chapter 10 Circles?
It 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 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 chapter 10 circles.
Lastly, solve the previous years question papers related to the particular chapter.
NCERT solutions of class 10 subject wise
All the best!
Frequently Asked Question (FAQs) - NCERT Solutions for Class 10 Maths Chapter 10 Circles
Question: Which is the official website of NCERT ?
Answer:
NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.
Question: What is the weightage of chapter circles in CBSE 10 board final exam ?
Answer:
Triangles, circles and constructions combined have 15 marks weightage in class 10 final board exam.
Question: Where can I find the complete solutions of NCERT class 10 maths ?
Answer:
Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.
Question: How is the NCERT solution helpful in the CBSE board exams?
Answer:
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.
Question: How many chapters are there in the class 10 maths ?
Answer:
There are 15 chapters in the class 10 maths NCERT.
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Questions related to CBSE Class 10th
Age criteria to appear in class 10 CBSE exam
Hello Bishal
According to guidelines of CBSE, minimum age to appear for class x must be 14 years. There is no upper limit to appear for class x cbse board.
A candidate can appear for maximum three attempts.
Some candidates give private exams or sometimes students fail in standard ix then they privately appear for class x then their age must be more than 14 years. Sometimes students appear for x class after one year gap of passing class ix then also their age would be 15 or 16 as there is no upper limit age.
My child is supposed to make another attempt at compartment exam of 10th cbse Missed last date for applying. since the boards are postponed to May 4th can I pay her fees and apply now.
Hello sir I'm sorry to inform you but now you're not eligible to fill the registration form as last date to fill the registration Form was 9th December 2020. Yes examination gets postponed to May but portal to fill the registration form is not open , in case if the registration form portal will open again you can fill the registration form and make the payment.
Feel free to comment if you've any doubt
Good luck
i want to know the sample paper and marking scheme of 10 th science
Hello there,
To get the sample papers for CBSE Class 10th, you should visit the official CBSE website. There you can get all the papers. Solving the previous year papers, will be very beneficial for you. So, solve as much papers as you can.
As per the latest CBSE Class 10 exam pattern 2020-21, the theory paper will be of 80 marks, while 20 marks will be for internal assessment. To know more about exam pattern, please visit the link given below.
https://www.google.com/amp/s/school.careers360.com/articles/cbse-10th-exam-pattern/amp
Check out the papers here:
https://www.google.com/amp/s/school.careers360.com/articles/cbse-class-10-sample-papers/amp
will the 2021 board exams for grade 10 conducted online or offline
Hello,
The 10th CBSE board exam will be conducted in offline mode only, These board exams will not be conducted in online mode. Central Board of Secondary Education conducts class 10th board examination for regular and private students. The exams of CBSE 10th 2021 will be conducted in February and March 2021. Candidates can fill CBSE 10th application form 2021 online on cbse.nic.in from October 24 to November 11, 2020.
is unit exam marks taken into consideration for cbse internal marks grade 10.
Hey,
Your concern completely depends upon school. School management plans the division of marks. For few schools they must consider even class test on the other hand some just focus on mid and end . So the division of marks vary from school to school. What the school do is continuous access the student and then give marks accorrdingly. In my opinion focus on each and every test. Apart from internals it work for you in finals as well.
All the best