Class 9 NCERT Solutions Maths Chapter 9: Circles

Ex. 9.1

Ex. 9.2

Ex. 9.3

Circles Exercise Ex. 9.1

Solution 1

A circle is a collection of points which are equidistant from a fix point. This fix point is called as the centre of circle and this equal distance is called as radius of circle. And thus shape of a circle depends on the radius of the circle.

So, if we try to superimpose two circles of equal radius, one each other both circles will cover each other.
So, two circles are congruent if they have equal radius.

Now consider two congruent circles having centre O and O' and two chords AB and CD of equal lengths

Now in

AOB and

CO'D
AB = CD           (chords of same length)
OA = O'C           (radii of congruent circles)
OB = O'D           (radii of congruent circles)

AOB

CO'D (SSS congruence rule)

AOB =

CO'D (by CPCT)
Hence equal chords of congruent circles subtend equal angles at their centres.

Solution 2

Let us consider two congruent circles (circles of same radius) with centres as O and O'.

AOB and

CO'D

AOB =

CO'D       (given)
OA = O'C           (radii of congruent circles)
OB = O'D           (radii of congruent circles)

AOB

CO'D (SSS congruence rule)

AB = CD (by CPCT)
Hence, if chords of congruent circles subtend equal angles at their centres then chords are equal.

Circles Exercise Ex. 9.2

Solution 1

Let radius of circle centered at O and O' be 5 cm and 3 cm respectively.
   OA = OB = 5 cm
   O'A = O'B = 3 cm
   OO' will be the perpendicular bisector of chord AB.

AC = CB

    Given that OO' = 4 cm
   Let OC be x. so, O'C will be 4 - x
   In

OAC
OA² = AC² + OC²

5² = AC² + x²

25 - x² = AC² ... (1)
In

O'AC
O'A² = AC² + O'C²

3² = AC² + (4 - x)²

9 = AC² + 16 + x² - 8x

AC² = - x² - 7 + 8x       ... (2)
From equations (1) and (2), we have
   25 - x² = - x² - 7 + 8x
           8x = 32
   x = 4
So, the common chord will pass through the centre of smaller circle i.e. O'. and hence it will be diameter of smaller circle.

Now, AC² = 25 - x² = 25 - 4² = 25 - 16 = 9

AC = 3 m

The length of the common chord AB = 2 AC = (2

3) m = 6 m

Solution 2

Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.

Draw perpendiculars OV and OU on these chords.
In

OVT and

OUT
OV = OU (Equal chords of a circle are equidistant from the centre)

OVT =

OUT (Each 90^o)
OT = OT (common)

OVT

OUT (RHS congruence rule)

VT = UT (by CPCT) ... (1)

It is given that
PQ = RS ... ... ... ... (2)

PV = RU                ... ... ... ... (3)
    On adding equations (1) and (3), we have
    PV + VT = RU + UT

PT = RT                          ... ... ...   ... (4)
    On subtracting equation (4) from equation (2), we have
    PQ - PT = RS - RT

QT = ST                          ... ... ... ... (5)
    Equations (4) and (5) shows that the corresponding segments of
    chords PQ and RS are congruent to each other.

Solution 3

     Let PQ and RS are two equal chords of a given circle and there are intersecting each other at point T.
     Draw perpendiculars OV and OU on these chords.
     In

OVT and

OUT
OV = OU (Equal chords of a circle are equidistant from the centre)

OVT =

OUT (Each 90^o)
OT = OT (common)

OVT

OUT (RHS congruence rule)

OTV =

OTU (by CPCT)

Hence, the line joining the point of intersection to the centre makes equal angles with the chords.

Solution 4

Let us draw a perpendicular OM on line AD.

Here, BC is chord of smaller circle and AD is chord of bigger circle.
We know that the perpendicular drawn from centre of circle bisects the chord.

BM = MC ... (1)

   And AM = MD      ... (2)
   Subtracting equations (2) from (1), we have
   AM - BM = MD - MC

AB = CD

Solution 5

Draw perpendiculars OA and OB on RS and SM respectively.
Let R, S and M be the position of Reshma, Salma and Mandip respectively.

AR = AS =

= 3cm

   OR = OS = OM = 5 m    (radii of circle)
    In OAR
   OA² + AR² = OR²
   OA² + (3 m)² = (5 m)²
   OA² = (25 - 9) m² = 16 m²
    OA = 4 m
   We know that in an isosceles triangle altitude divides the base, so in

RSM

RCS will be of 90^o and RC = CM
Area of

ORS =

RC = 4.8

RM = 2RC = 2(4.8)= 9.6

So, distance between Reshma and Mandip is 9.6 m.

Solution 6

    Given that AS = SD = DA
            So, ASD is a equilateral triangle
             OA (radius) = 20 m.
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.
We also know that median intersect each other at the 2: 1. As AB is the median of equilateral triangle ABC, we can write

AB = OA + OB = (20 + 10) m = 30 m.

ABD

AD² = AB² + BD²
AD² = (30)² +

So, length of string of each phone will be

Circles Exercise Ex. 9.3

Solution 1

We may observe that

AOC =

AOB +

BOC
= 60^o + 30^o
= 90^o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

Solution 2

OAB
AB = OA = OB = radius

OAB is an equilateral triangle.

So, each interior angle of this triangle will be of 60^o

AOB = 60^o

Now,

In cyclic quadrilateral ACBD

ACB +

ADB = 180^o (Opposite angle in cyclic quadrilateral)

ADB = 180^o - 30^o = 150^o
So, angle subtended by this chord at a point on major arc and minor arc are 30^o and 150^o respectively.

Solution 3

Consider PR as a chord of circle.
Take any point S on major arc of circle.
Now PQRS is a cyclic quadrilateral.

PQR +

PSR = 180^o (Opposite angles of cyclic quadrilateral)

PSR = 180^o - 100^o = 80^o
We know that angle subtended by an arc at centre is double the angle subtended by it any point on the remaining part of the circle.

POR = 2

PSR = 2 (80^o) = 160^o

POR
OP = OR (radii of same circle)

OPR =

ORP (Angles opposite equal sides of a triangle)

OPR +

ORP +

POR = 180^o (Angle sum property of a triangle)

OPR + 160^o= 180^o
2

OPR = 180^o - 160^o = 20^o

OPR = 10^o

Solution 4

ABC

BAC +

ABC +

ACB = 180^o (Angle sum property of a triangle)

BAC + 69^o + 31^o = 180^o

BAC = 180^o - 100º

BAC = 80^o

BDC =

BAC = 80^o (Angles in same segment of circle are equal)

Solution 5

CDE

CDE +

DCE =

CEB (Exterior angle)

CDE + 20^o = 130^o

CDE = 110^o
But

BAC =

CDE (Angles in same segment of circle)

BAC = 110^o

Solution 6

For chord CD

CBD =

CAD (Angles in same segment)

CAD = 70^o

BAD =

BAC +

CAD = 30^o + 70^o = 100^o

BCD +

BAD = 180^o (Opposite angles of a cyclic quadrilateral)

BCD + 100^o = 180^o

BCD = 80^o
In

ABC

AB = BC (given)

BCA =

CAB (Angles opposite to equal sides of a triangle)

BCA = 30^o
We have

BCD = 80^o

BCA +

ACD = 80^o
30^o +

ACD = 80^o

ACD = 50^o

ECD = 50^o

Solution 7

Let ABCD a cyclic quadrilateral having diagonals as BD and AC intersecting each other at point O.

(Consider BD as a chord)

BCD +

BAD = 180^o (Cyclic quadrilateral)

BCD = 180^o - 90^o = 90^o

(Considering AC as a chord)

ADC +

ABC = 180^o (Cyclic quadrilateral)
90^o +

ABC = 180^o

ABC = 90^o
Here, each interior angle of cyclic quadrilateral is of 90^o. Hence it is a rectangle.

Solution 8

Consider a trapezium ABCD with AB | |CD and BC = AD Draw AM

CD and BN

CD
In

AMD and

BNC
AD = BC (Given)

AMD =

BNC (By construction each is 90^o)
AM = BM (Perpendicular distance between two parallel lines is same)

AMD

BNC (RHS congruence rule)

ADC =

BCD (CPCT) ... (1)

BAD and

ADC are on same side of transversal AD

BAD +

ADC = 180^o ... (2)

BAD +

BCD = 180^o [Using equation (1)]
This equation shows that the opposite angles are supplementary.
So, ABCD is a cyclic quadrilateral.

Solution 9

    Join chords AP and DQ
   For chord AP

PBA =

ACP       (Angles in same segment)       ... (1)
   For chord DQ

DBQ =

QCD       (Angles in same segment)       ... (2)
   ABD and PBQ are line segments intersecting at B.

PBA =

DBQ       (Vertically opposite angles)       ... (3)
   From equations (1), (2) and (3), we have

ACP =

QCD

Solution 10

Consider a

ABC
Two circles are drawn while taking AB and AC as diameter.
Let they intersect each other at D and let D does not lie on BC.
Join AD

ADB = 90^o (Angle subtend by semicircle)

ADC = 90^o (Angle subtend by semicircle)

BDC =

ADB +

ADC = 90^o + 90^o = 180^o
Hence BDC is straight line and our assumption was wrong.
Thus, Point D lies on third side BC of

ABC

Solution 11

ABC

ABC +

BCA +

CAB = 180^o (Angle sum property of a triangle)

90^o +

BCA +

CAB = 180^o

BCA +

CAB = 90^o ... (1)
In

ADC

CDA +

ACD +

DAC = 180^o (Angle sum property of a triangle)

90^o +

ACD +

DAC = 180^o

ACD +

DAC = 90^o ... (2)
Adding equations (1) and (2), we have

BCA +

CAB +

ACD +

DAC = 180^o

(

BCA +

ACD) + (

CAB +

DAC) = 180^o

BCD +

DAB = 180^o ... (3)
But it is given that

B +

D = 90^o + 90^o = 180^o ... (4)
From equations (3) and (4), we can see that quadrilateral ABCD is having sum of measures of opposite angles as 180^o.
So, it is a cyclic quadrilateral.
Consider chord CD.
Now,

CAD =

CBD (Angles in same segment)

Solution 12

Let ABCD be a cyclic parallelogram.

A +

C = 180^o   (Opposite angle of cyclic quadrilateral)   ... (1)
   We know that opposite angles of a parallelogram are equal

A =

C and

B =

D
From equation (1)

A +

C = 180^o

A +

A = 180^o

A = 90^o
Parallelogram ABCD is having its one of interior angles as 90^o, so, it is a rectangle.

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Class 9 NCERT Solutions Maths Chapter 9: Circles

Circles Exercise Ex. 9.1

Solution 1

Solution 2

Circles Exercise Ex. 9.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Circles Exercise Ex. 9.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

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