# NCERT Solutions for Exercise 10.6 Class 9 Maths Chapter 10 - Circles

The NCERT Solutions for Class 9 Maths exercise 10.6 is a discretionary (not according to the assessment perspective for Class 9) exercise that contains points like equal chords of a circle, cyclic quadrilaterals, angle extended by an arc of a circle and finding lengths of the radius when the distance of chord and chord length is given.

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A few important concepts related to NCERT syllabus Class 9 Maths chapter 10 exercise 10.6 in order to solve this exercise are:

• The sum of any two opposite angles of a cyclic quadrilateral is 180°

• Angles which are present in the same arc of the circle are equal.

• The point subtended by an arc (circular segment) at the circle's centre is twofold the angle subtended by it at the remaining circumference of the circle.

• The chords of equivalent length are consistently equidistant from the centre of the circle

Significant questions in this exercise utilize a mix of at least two of the above ideas.

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Along with NCERT book Class 9 Maths chapter 10 exericse 10.6 the following exercises are also present.

• Circles Exercise 10.1

• Circles Exercise 10.2

• Circles Exercise 10.3

• Circles Exercise 10.4

• Circles Exercise 10.5

## Circles Class 9 Chapter 10 Exercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : PAQ = PBQ

Proof : In APQ and BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, APQ BPQ (By SSS)

PAQ = PBQ (CPCT)

Q2 Two chords AB and CD of lengths and respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is , find the radius of the circle.

Given : AB = 5 cm, CD = 11 cm and AB || CD.

Construction: Draw

Proof :

Proof: CD is a chord of circle and

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In OCM , using Pythagoras,

.............................1

and

In OAN , using Pythagoras,

.............................2

From 1 and 2,

From 2, we get

OA = OC

Thus, the radius of the circle is

Q3 The lengths of two parallel chords of a circle are and . If the smaller chord is at distance from the centre, what is the distance of the other chord from the centre?

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw

Proof :

Proof: CD is a chord of circle and

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In OCM , using Pythagoras,

.............................1

and

In OAN , using Pythagoras,

.............................2

From 1 and 2,

So, x=1 (since )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

To prove :

Construction: Join AC and DE.

Proof :

Let ADC = x , DOE = y and AOD = z

So, EOC = z (each chord subtends equal angle at centre)

AOC + DOE + AOD + EOC =

.........................................1

OA = OD (Radii of the circle)

OAD = ODA (angles opposite to equal sides )

OAD + ODA + AOD =

.............................................................2

Similarly,

.............................................................3

..............................................................4

ODB is exterior of triangle OAD . So,

(from 2)

.................................................................5

similarly,

OBE is exterior of triangle OCE . So,

OBE = OCE + OEC

(from 3)

.................................................................6

From 4,5,6 ;we get

BDE = BED = OEB - OED

..................................................7

In BDE ,

DBE + BDE + BED =

...................................................8

Here, from equation 1,

...................................9

From 8 and 9,we have

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Given : ABCD is a rhombus.

To prove: the circle drawn with AB as diameter passes through the point O.

Proof :

ABCD is rhombus.

Thus, (diagonals of a rhombus bisect each other at )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that .

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove : AE = AD

Proof :

ADC = 3 , ABC = 4, ADE = 1 and AED = 2

.................1(linear pair)

....................2(sum of opposite angles of cyclic quadrilateral)

3 = 4 (oppsoite angles of parallelogram )

From 1 and 2,

3+ 1 = 2 + 4

From 3, 1 = 2

From 4, AQB, 1 = 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

In ABD and CDO,

AO = OC (Given )

AOB = COD (Vertically opposite angles )

BO = DO (Given )

So, ABD CDO (By SAS)

BAO = DCO (CPCT)

BAO and DCO are alternate angle and are equal .

So, AB || DC ..............1

From 1 and 2,

......................3(sum of opposite angles)

A = C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Given: AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram. (proved in (i))

(proved in (i))

A parallelogram with one angle , is a rectangle )

Thus, ABCD is rectangle.

Q8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are , and

Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove : the angles of the triangle DEF are , and

Proof :

1 and 3 are angles in same segment.therefore,

1 = 3 ................1(angles in same segment are equal )

and 2 = 4 ..................2

1+ 2= 3+ 4

,

and

Similarly, and

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that .

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove : BP = BQ

Proof :

AB is a common chord in both congruent circles.

In

(Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Given :In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let ABD = 1 , ADC = 2 , DCB = 3 , CBD = 4

1 and 3 lies in same segment.So,

1 = 3 ..........................1(angles in same segment)

similarly, 2 = 4 ......................2

also, 1= 2 ..............3(given)

From 1,2,3 , we get

3 = 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

## More About NCERT Solutions for Class 9 Maths Exercise 10.6

NCERT solutions Class 9 Maths exercise 10.6 incorporates some significant ideas from the past practices that might end up being essential to tackle a few varieties issues from NCERT solutions Class 9 Maths exercise 10.6

• Equivalent chords of the circle subtend equivalent angle at the focal point (centre) of the circle

• A perpendicular that is drawn from the centre point of the circle to the chord separates the given chord equally (divides the into two)

• Assuming three non-colinear points are given in a plain, there is one and only one circle that goes through these points as a whole.

Aside from all the previously mentioned concepts the other way around of these theorems (hypothesis) are exceptionally basic in addressing some good questions.

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Also Read| Circles Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 10.6

• Exercise 10.6 Class 9 Maths, is based on circles and the majority of the critical properties of circles.

• From Class 9 Maths chapter 10 exercise 10.6 we get to revise and give a final touchup to the whole ideas of this part in a solitary exercise.

• Understanding the concepts from Class 9 Maths chapter 10 exercise 10.6 will make the ideas and questions from better and competitive standards (like Class10) simpler for us.

Also, See

• NCERT Solutions for Class 9 Maths Chapter 10

• NCERT Exemplar Solutions Class 9 Maths Chapter 10

## NCERT Solutions of Class 10 Subject Wise

• NCERT Solutions for Class 9 Maths

• NCERT Solutions for Class 9 Science

## Subject Wise NCERT Exemplar Solutions

• NCERT Exemplar Class 9 Maths
• NCERT Exemplar Class 9 Science