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Edited By safeer | Updated on Jul 29, 2022 06:39 PM IST

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.

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

** Answer: **

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)

** Answer: **

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

To find Radius (OA).

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,

(OC=OA =radii)

From 2, we get

OA = OC

Thus, the radius of the circle is

** Answer: **

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,

(OC=OA =radii)

So, x=1 (since )

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

Hence, second chord is 3 cm away from centre.

** Answer: **

Given : AD = CE

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

In OAD ,

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,

ODB = OAD + ODA

(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

** Answer: **

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.

** Answer: **

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

** Answer: **

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

Also AD || BC ...............2

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.

** Answer: **

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.

** Answer: **

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

Adding 1 and 2,we have

1+ 2= 3+ 4

,

and

Similarly, and

** Answer: **

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 )

** Answer: **

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.

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.

**Also Read| **Circles Class 9 Notes

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**

1. What are the main things you learned from NCERT solutions for Class 9 Maths exercise 10.6?

This exercise deals with all the perspectives that associate the difficult questions regarding circles and their properties.

2. How can you describe a circle?

A circle can be characterized as a closed (shut figure), two-dimensional bent (curved) shape, with zero corners to such an extent that each point of the circumference of the circle is equidistance from the proper centre of the circle.

3. How many circles can be drawn from three non-colinear points?

If three non-colinear points are given then there is one and only one circle that passes through all of the points.

4. What is the measurement of the opposite angle in a cyclic quadrilateral if one of them is 135?

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

The other angle =180-135=45

5. What is a tangent?

Tangent is a straight line that passes through only one point of the circle’s circumference.

6. Can we get any chord that has length greater than the diameter of the circle?

No, we can’t get any chord that is greater or equal to the diameter of the circle as the diameter is the largest chord.

7. In a cyclic quadrilateral the opposite angles are at the ratio of 2:3 then find the angles?

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

Sum of ratio = 5

Angles are:

2/5 times 180 and 3/5 times 180

The angles are 72 and 108

8. How many tangents can be drawn from a given point of the circle’s circumference?

One and only one tangent is possible through one given point of the circumference.

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