NCERT Solutions for Exercise 10.2 Class 10 Maths Chapter 10

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

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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.

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NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2
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NCERT Solutions Subject Wise

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.

Circle Exercise 10.1

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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

(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) $70 \degree$

(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.

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:

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.

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

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

NCERT Solutions For Class 10 Maths Chapter 10 Exercise 10.2

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