NCERT Solutions for Class 10 Maths Chapter 10 Circles
NCERT Solutions for Class 10 Maths Chapter 10 Circles - In our previous classes, you have learnt that a circle is a closed shape with a collection of points in a plane that are at a specific distance ( called the radius) from a fixed point (called center). NCERT solutions for class 10 maths chapter 10 Circles is covering the in-depth explanations to questions related to a circle. You have also studied important terms related to the circle like segment, arc, sector, chord, etc. In this chapter, there are two exercises with 17 questions in them. 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. 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. This chapter has fundamental concepts that are important for students in their future studies. Circles is a very interesting chapter due to the involvement of geometrical calculations and diagrams. Here you will get NCERT solutions for class 6 to 12 for science and maths. Here you will get NCERT solutions for class 10 also.
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 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 |
NCERT solutions of class 10 subject wise
How to use NCERT solutions for class 10 maths chapter 10 Circles ?
It is the chapter the logic as well as 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 come to the practice exercises where your understanding of concepts will be tested.
While doing the practice exercises, you can take help of NCERT solutions for class 10 maths chapter 10 circles.
After doing all this, the last thing you should solve is past papers related to the particular chapter.
Keep working hard & happy learning!
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 has 15 marks weightage in class 10 final board exam.
Question: Does CBSE provides the solutions of NCERT class 10 ?
Answer:
No, CBSE doesn’t provided NCERT solutions for any class or subject.
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 does the NCERT solutions are helpful in CBSE board exam?
Answer:
As most of the questions in CBSE board exam are directly asked for NCERT textbook, so must solve all the problems given in the NCERT textbook. NCERT solutions are not only important when you 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. Chapter 1- Real Numbers, Chapter 2- Polynomials, Chapter 3- Pair of Linear Equations in Two Variables, Chapter 4- Quadratic Equations, Chapter 5- Arithmetic Progressions, Chapter 6- Triangles, Chapter 7- Coordinate Geometry, Chapter 8- Introduction to Trigonometry, Chapter 9- Some Applications of Trigonometry, Chapter 10- Circles, Chapter 11- Constructions, Chapter 12- Areas Related to Circles, Chapter 13- Surface Areas and Volumes, Chapter 14- Statistics, Chapter 15- Probability are the chapters in the NCERT class 10 maths.
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