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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## Assess NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

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 $l$ 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

(A) $60 \degree$

(B) $70 \degree$

(C) $80 \degree$

(D) $9 0 \degree$

The correct option is (b)

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

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

(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 $\angle APB = 80^0$
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

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}

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.

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;

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

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}

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

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.

To prove- $\angle$ AOB = $90^0$
Proof-
In $\Delta$ AOP and $\Delta$ AOC,
OA =OA [Common]
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.

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,
$\angle$ OAP + $\angle$ APB + $\angle$ PBO + $\angle$ BOA = $360^0$
90 + $\angle$ APB + 90 + $\angle$ BOA = 360
$\angle APB + \angle AOB = 180^0$
Hence proved
.

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.

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

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

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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