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NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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Triangles, circles and constructions combined have 15 marks weightage in class 10 final board exam.

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Here you will get the detailed NCERT solutions for class 10 maths by clicking on the link.

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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.

How many chapters are there in the class 10 maths ?

There are 15 chapters in the class 10 maths NCERT.

NCERT Solutions for Class 10 Maths Chapter 10 Circles 2021

NCERT Solutions for Class 10 Maths Chapter 10 Circles

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.

1594292623722

Q2 Fill in the blanks :
(1) A tangent to a circle intersects it in_________ point (s).
(2) A line intersecting a circle in two points is called a _____.
(3) A circle can have ________ parallel tangents at the most.
(4) The common point of a tangent to a circle and the circle ______

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.

(d) point of contact
The common point of a tangent and a circle.

Q3 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the center O at a point Q so that OQ = 12 cm. Length PQ is :

(A) 12 cm

(B) 13 cm

(D) $\sqrt { 119}$ cm.

Answer:

The correct option is (d) = $\sqrt { 119}$ cm

1594292996735
It is given that the radius of the circle is 5 cm. OQ = 12 cm
According to question,
We know that $\angle QPO=90^0$
So, triangle OPQ is a right-angle triangle. By using Pythagoras theorem,
$PQ =\sqrt{OQ^2-OP^2}=\sqrt{144-25}$
$=\sqrt{119}$ cm

Q4 Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Answer:

1594293094431
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

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

(D) 24.5 cm

Answer:

The correct option is (A) = 7 cm
capture
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

asdfasde

(A) $60 \degree$

(B) $70 \degree$

(D) $9 0 \degree$

Answer:

The correct option is (b)

1594293296951
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°

(D) 80°

Answer:

The correct option is (A)
1594293381602
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:

1594293543765
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:

1594293656560
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:

1594293766356
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:

1594293901543
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

1594294016205

Answer:

1594294152195
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:

1594294251028
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:

1594294380467
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:

1594294663376
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.

adsfad

Answer:

Exercise 10.2, Q. 12 - A triangle ABC is drawn to circumscribe a circle

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:

1594294831132
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.

Chapter No.	Chapter Name
Chapter 1	NCERT solutions for class 10 maths chapter 1 Real Numbers
Chapter 2	NCERT solutions for class 10 maths chapter 2 Polynomials
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	NCERT solutions for class 10 maths chapter 6 Triangles
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	NCERT solutions for class 10 maths chapter 11 Constructions
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	NCERT solutions for class 10 maths chapter 14 Statistics
Chapter 15	NCERT solutions for class 10 maths chapter 15 Probability

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NCERT Solutions for Class 10 Maths Chapter 10 Circles

NCERT solutions for class 10 maths chapter 10 Circles Excercise: 10.1

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NCERT solutions for class 10 maths chapter 10 Circles Excercise: 10.2

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