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NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10 - Circles

NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10 - Circles

Edited By Ramraj Saini | Updated on Nov 27, 2023 01:49 PM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2

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

Q1 From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the center is 25 cm. The radius of the circle is

(A) 7 cm

(B) 12 cm

(C) 15 cm

(D) 24.5 cm

Answer:

The correct option is (A) = 7 cm
1639544843884 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 l cm.
We know that \Delta OTQ is a right angle triangle. So, by using Pythagoras theorem-

\\OQ^2 = TQ^2+OT^2\\\\ l = \sqrt{25^2-24^2}\\\\OT = l=\sqrt{49}

OT = 7 cm

Q2 In Fig. 10.11, if TP and TQ are the two tangents to a circle with center O so that \angle POQ = 11 0 \degree , then \angle PTQ is equal to

1639544819399

(A) 60 \degree

(B) 70 \degree

(C) 80 \degree

(D) 9 0 \degree

Answer:

The correct option is (b)

15942932969511594293294689
In figure, \angle POQ = 110^0
Since POQT is quadrilateral. Therefore the sum of the opposite angles are 180

\\\Rightarrow \angle PTQ +\angle POQ = 180^0\\\\\Rightarrow \angle PTQ = 180^0-\angle POQ
= 180^0-100^0
= 70^0

Q3 If tangents PA and PB from a point P to a circle with center O are inclined to each other at an angle of 80 \degree , then \angle POA is equal to

(A) 50°

(B) 60°

(C) 70°

(D) 80°

Answer:

The correct option is (A)
15942933816021594293378856
It is given that, tangent PA and PB from point P inclined at \angle APB = 80^0
In triangle \Delta OAP and \Delta OBP
\angle OAP = \angle OBP = 90
OA =OB (radii of the circle)
PA = PB (tangents of the circle)

Therefore, by SAS congruence
\therefore \Delta OAP \cong \Delta OBP

By CPCT, \angle OPA = \angle OPB
Now, \angle OPA = 80/4 = 40

In \Delta PAO,
\angle P + \angle A + \angle O = 180
\angle O = 180 - 130
= 50

Q4 Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Answer:

15942935437651594293541307
Let line p and line q are two tangents of a circle and AB is the diameter of the circle.
OA and OB are perpendicular to the tangents p and q respectively.
therefore,
\angle1 = \angle2 = 90^0

\Rightarrow p || q { \angle \because 1 & \angle 2 are alternate angles}

Q5 Prove that the perpendicular at the point of contact to the tangent to a circle passes through the center.

Answer:

15942936565601594293654097
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 ( OX\perp AXB ) at point of contact X.
Therefore, we have,
\angle BXO + \angle YXB = 90^0+90^0=180^0

\therefore OXY is a collinear
\Rightarrow OX is passing through the center of the circle.

Q6 The length of a tangent from a point A at distance 5 cm from the center of the circle is 4 cm. Find the radius of the circle.

Answer:

15942937663561594293764198
Given that,
the length of the tangent from the point A (AP) is 4 cm and the length of OA is 5 cm.
Since \angle APO = 90 0
Therefore, \Delta APO is a right-angle triangle. By using Pythagoras theorem;

OA^2=AP^2+OP^2
5^2 = 4^2+OP^2
OP=\sqrt{25-16}=\sqrt{9}
OP = 3 cm

Q7 Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

15942939015431594293899175
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 \perp PQ [since PQ is tangent to a smaller circle]

According to question,

In \Delta OPR and \Delta OQR
\angle PRO = \angle QRO {both 90^0 }

OR = OR {common}
OP = OQ {both radii}

By RHS congruence \Delta OPR \cong \Delta OQR
So, by CPCT
PR = RQ
Now, In \Delta OPR,
by using pythagoras theorem,
PR = \sqrt{25-9} =\sqrt{16}
PR = 4 cm
Hence, PQ = 2.PR = 8 cm

Q8 A quadrilateral ABCD is drawn to circumscribe a circle (see Fig. 10.12). Prove that AB + CD = AD + BC

15942940162051594294012805

Answer:

15942941521951594294149641
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
\Rightarrow (AP + BP) + (RD + CR) = (AS+DS)+(BQ + CQ)
\Rightarrow AB + CD = AD + BC

Hence proved.

Q9 In Fig. 10.13, XY and X'Y'are two parallel tangents to a circle with center O and another tangent AB with a point of contact C intersecting XY at A and X'Y' at B. Prove that \angle AOB = 90°.

Answer:

15942942510281594294247906
To prove- \angle AOB = 90^0
Proof-
In \Delta AOP and \Delta AOC,
OA =OA [Common]
OP = OC [Both radii]
AP =AC [tangents from external point A]
Therefore by SSS congruence, \Delta AOP \cong \Delta AOC
and by CPCT, \angle PAO = \angle OAC
\Rightarrow \angle PAC = 2\angle OAC ..................(i)

Similarly, from \Delta OBC and \Delta OBQ, we get;
\angle QBC = 2. \angle OBC.............(ii)

Adding eq (1) and eq (2)

\angle PAC + \angle QBC = 180
2( \angle OBC + \angle OAC) = 180
( \angle OBC + \angle OAC) = 90

Now, in \Delta OAB,
Sum of interior angle is 180.
So, \angle OBC + \angle OAC + \angle AOB = 180
\therefore \angle AOB = 90
hence proved.

Q10 Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the center.

Answer:

15942943804671594294377915
To prove - \angle APB + \angle AOB = 180^0
Proof-
We have, PA and PB are two tangents, B and A are the point of contacts of the tangent to a circle. And OA\perp PA , OB\perp PB (since tangents and radius are perpendiculars)

According to question,
In quadrilateral PAOB,
\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^0
90 + \angle APB + 90 + \angle BOA = 360
\angle APB + \angle AOB = 180^0
Hence proved
.

Q11 Prove that the parallelogram circumscribing a circle is a rhombus.

Answer:

15942946633761594294660855
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
\Rightarrow 2AB = 2AD [from equation (i)]
\Rightarrow AB = AD
Now, AB = AD and AB = CD
\therefore AB = AD = CD = BC

Hence ABCD is a rhombus.

Q12 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.

Answer:

1639543596459

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 \bigtriangleup ABC ,

CF = CD = 6 (tangents on the circle from point C)

BE = BD = 6 (tangents on the circle from point B)

AE = AF = x (tangents on the circle from point A)

Now AB = AE + EB


\\AB = AE + EB\\\\ => AB = x + 8\\\\ BC = BD + DC\\\\ => BC = 8+6 = 14\\\\ CA = CF + FA\\\\ => CA = 6 + x\\\\

Now
\\s = (AB + BC + CA )/2\\\\ => s = (x + 8 + 14 + 6 +x)/2\\\\ => s = (2x + 28)/2\\\\ => s = x + 14

Area of triangle \bigtriangleup ABC

\\=\sqrt{s*(s-a)*(s-b)*(s-c)}\\\\=\sqrt{(14+x)*[(14+x)-14]*[(14+x)-(6+x)]*[(14+x)-(8+x)]}\\\\=4\sqrt{3(14x+x^2)}
Now the area of \bigtriangleup OBC

\\= (1/2)*OD*BC\\\\ = (1/2)*4*14\\\\ = 56/2 = 28
Area of \bigtriangleup OCA

\\= (1/2)*OF*AC \\\\= (1/2)*4*(6+x) \\\\= 2(6+x) \\\\= 12 + 2x

Area of \bigtriangleup OAB

\\= (1/2)*OE*AB \\\\= (1/2)*4*(8+x) \\\\= 2(8+x) \\\\= 16 + 2x

Now Area of the \bigtriangleup ABC = Area of \bigtriangleup OBC + Area of \bigtriangleup OCA + Area of \bigtriangleup OAB

\\=> 4\sqrt{3x(14+x)}= 28 + 12 + 2x + 16 + 2x \\\\=> 4\sqrt{3x(14+x)} = 56 + 4x \\\\=> 4\sqrt{3x(14+x)} = 4(14 + x) \\\\=> \sqrt{3x(14+x)}] = 14 + x

On squaring both the side, we get

\\3x(14 + x) = (14 + x)^2\\\\ => 3x = 14 + x \:\:\:\:\:\: (14 + x = 0 => x = -14\: is\: not\: possible) \\\\=> 3x - x = 14\\\\ => 2x = 14\\\\ => x = 14/2\\\\ => x = 7

Hence

AB = x + 8

=> AB = 7+8

=> AB = 15

AC = 6 + x

=> AC = 6 + 7

=> AC = 13

Answer- AB = 15 and AC = 13

Q13 Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:

15942948311321594294828094
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-
\\\angle AOB + \angle COD =180^0\\ \angle AOD + \angle BOC =180^0
Proof -
Join OP, OQ, OR and OS
In triangle \Delta DOS and \Delta 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, \Delta DOS \cong \Delta DOR,
and by CPCT, \angle DOS = \angle DOR
\angle c = \angle d .............(i)

Similarily,
\\\angle a = \angle b\\ \angle e = \angle f\\ \angle g =\angle h ...............(2, 3, 4)

\therefore 2(\angle a +\angle e +\angle h+\angle d) = 360^0
\\(\angle a +\angle e) +(\angle h+\angle d) = 180^0\\ \angle AOB + \angle DOC = 180^0
SImilarily, \angle AOD + \angle BOC = 180^0

Hence proved.

More About NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

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

Benefits of NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

  • The NCERT solutions for class 10 Maths chapter 10 exercise 10.2 and the solved example before exercise 10.2 Class 10 Maths are significant since they cover questions from the basic concept of Circle and its relations.
  • If students can answer all of the questions in exercise 10.2 Class 10 Maths, they will have a better understanding of the notion of problem-solving as it is asked in terms of tangents or chords in chapter 10 of Class 10 Maths.
  • Students may receive MCQs, short answer, or long answer questions from the types discussed in Class 10 Maths chapter 10 exercise 10.2 for final exams.

Also see-

NCERT Solutions Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. Define Circle?

 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.

2. What is the shortest distance between two parallel tangents of circle?

The shortest distance between two parallel tangents of the circle is Diameter i.e. twice the radius of given circle

3. From how many points a circle can be formed?

a circle can be formed by three points but the condition is the points must be non-collinear non-parallel.

4. What is the sum of Interior angles of a Quadrilateral ?

The sum of interior angles of a quadrilateral is 360 degrees.

5. What is the name of the polynomial with degree 3?

The degree 3 polynomial is referred to as Cubic polynomials is a type of polynomial in which there are

6. What is SSS Congruence in circle?

 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.

7. What does the line intersecting circle at two points is called?

The line intersecting circle at two points is called a Secant

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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.

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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.

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Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

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Option 1)

2,000 \; J - 5,000\; J

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Option 3)

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Option 4)

20,000 \, \, J - 50,000 \, \, J

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Option 2)

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Option 1)

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Option 2)

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Option 3)

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67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

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Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

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