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NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10 Circles are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 10.2 which is an exercise followed by exercise 10.1 includes the concept of circles, there are many numerical problems with the number of tangents from a point. If we talk about tests and exams, this is an important part to cover. These concepts are easy to understand and can be worked accordingly. Students can find NCERT solutions for class 10 Maths here.
NCERT solutions for exercise 10.2 Class 10 Maths chapter 10 Circles covers problems on the topics like the concept of finding radius and distance from one point on the circle and other points outside the circle. 10th class Maths exercise 10.2 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.
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Circles Class 10 Chapter 10 Circles Exercise: 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.
Some other topics of exercise 10.2 Class 10 Maths are included in this section. Firstly NCERT solutions for Class 10 Maths chapter 10 exercise 10.2 Contains basic questions which is to represent the problems of finding radii using distance formula and Pythagoras Theorem. End Questions of Class 10 Maths chapter 10 exercise 10.2 belongs to finding the distance between two tangents and the relation of line and circle. In NCERT syllabus Class 10 Maths chapter 10 exercise 10.2 also covers problems of co-centric circles and numerical related to chords and centres circumscribing a circle.
Also Read| Circles Class 10th Notes
Also see-
Subject Wise NCERT Exemplar Solutions
A curved line whose ends meet and all points on the line are at the same distance from the centre or a path that revolves around a central point or a group of items arranged is called a circle.
The shortest distance between two parallel tangents of the circle is Diameter i.e. twice the radius of given circle
a circle can be formed by three points but the condition is the points must be non-collinear non-parallel.
The sum of interior angles of a quadrilateral is 360 degrees.
The degree 3 polynomial is referred to as Cubic polynomials is a type of polynomial in which there are
If in a circle if two tangents are parallel and another tangent from parallel tangent one cuts parallel tangent 2 forming two triangles. If these triangles have three sides in common then these triangles are equal by sss congruence.
The line intersecting circle at two points is called a Secant
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.
Yes, you can definitely apply for diploma courses after passing 10th CBSE. In fact, there are many diploma programs designed specifically for students who have completed their 10th grade.
Generally, passing 10th CBSE with a minimum percentage (often 50%) is the basic eligibility for diploma courses. Some institutes might have specific subject requirements depending on the diploma specialization.
There is a wide range of diploma courses available in various fields like engineering (e.g., mechanical, civil, computer science), computer applications, animation, fashion design, hospitality management, and many more.
You can pursue diplomas at various institutions like:
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