NCERT Solutions for Exercise 10.6 Class 9 Maths Chapter 10

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

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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Circles Class 9 Chapter 10 Exercise: 10.6
More About NCERT Solutions for Class 9 Maths Exercise 10.6
Benefits of NCERT Solutions for Class 9 Maths Exercise 10.6
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

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.

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.

Answer:

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

To prove : $\angle$ PAQ = $\angle$ PBQ

Proof : In $\triangle$ APQ and $\triangle$ BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, $\triangle$ APQ $\cong$ $\triangle$ BPQ (By SSS)

$\angle$ PAQ = $\angle$ PBQ (CPCT)

Q2 Two chords AB and CD of lengths $\small 5\hspace {1mm}cm$ and $\small 11\hspace {1mm}cm$ 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 $\small 6\hspace {1mm}cm$ , find the radius of the circle.

Answer:

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

To find Radius (OA).

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

1640237834315

Proof: CD is a chord of circle and $OM \perp CD$

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 $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$5.5 ^2+x^2=2.5^2+(6-x)^2$

$\Rightarrow 30.25+x^2=6.25+36+x^2-12x$

$\Rightarrow 30.25-42.25=-12x$

$\Rightarrow -12=-12x$

$\Rightarrow x=1$

From 2, we get

$OC^2=5.5^2+1^2=30.25+1=31.25$

$\Rightarrow OC=\frac{5}{2}\sqrt{5} cm$

OA = OC

Thus, the radius of the circle is $\frac{5}{2}\sqrt{5} cm$

Q3 The lengths of two parallel chords of a circle are $\small 6\hspace {1mm}cm$ and $\small 8\hspace {1mm}cm$ . If the smaller chord is at distance $\small 4\hspace {1mm}cm$ from the centre, what is the distance of the other chord from the centre?

Answer:

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

To find: Length of ON

Construction: Draw $OM \perp CD \, \, and \, \, \, ON\perp AB$

Proof :

1640237861219

Proof: CD is a chord of circle and $OM \perp CD$

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 $\triangle$ OCM , using Pythagoras,

$OC ^2=CM^2+OM^2$ .............................1

and

In $\triangle$ OAN , using Pythagoras,

$OA ^2=AN^2+ON^2$ .............................2

From 1 and 2,

$CM ^2+OM^2=AN^2+ON^2$ (OC=OA =radii)

$\Rightarrow 3 ^2+4^2=4^2+(4-x)^2$

$\Rightarrow 9+16=16+16+x^2-8x$

$\Rightarrow 9=16+x^2-8x$

$\Rightarrow x^2-8x+7=0$

$\Rightarrow x^2-7x-x+7=0$

$\Rightarrow x(x-7)-1(x-7)=0$

$\Rightarrow (x-1)(x-7)=0$

$\Rightarrow x=1,7$

So, x=1 (since $x\neq 7> OM$ )

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 $\small \angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : $\angle ABC = \frac{1}{2}(\angle AOC-\angle DOE)$

Construction: Join AC and DE.

Proof :

1640237883312

Let $\angle$ ADC = x , $\angle$ DOE = y and $\angle$ AOD = z

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

$\angle$ AOC + $\angle$ DOE + $\angle$ AOD + $\angle$ EOC = $360 \degree$

$\Rightarrow x+y+z+z=360 \degree$

$\Rightarrow x+y+2z=360 \degree$ .........................................1

In $\triangle$ OAD ,

OA = OD (Radii of the circle)

$\angle$ OAD = $\angle$ ODA (angles opposite to equal sides )

$\angle$ OAD + $\angle$ ODA + $\angle$ AOD = $180 \degree$

$\Rightarrow 2\angle OAD+z=180 \degree$

$\Rightarrow 2\angle OAD=180 \degree-z$

$\Rightarrow \angle OAD=\frac{180 \degree-z}{2}$

$\Rightarrow \angle OAD=90 \degree-\frac{z}{2}$ .............................................................2

Similarly,

$\Rightarrow \angle OCE=90 \degree-\frac{x}{2}$ .............................................................3

$\Rightarrow \angle OED=90 \degree-\frac{y}{2}$ ..............................................................4

$\angle$ ODB is exterior of triangle OAD . So,

$\angle$ ODB = $\angle$ OAD + $\angle$ ODA

$\Rightarrow \angle ODB=90 \degree-\frac{z}{2}+z$ (from 2)

$\Rightarrow \angle ODB=90 \degree+\frac{z}{2}$ .................................................................5

similarly,

$\angle$ OBE is exterior of triangle OCE . So,

$\angle$ OBE = $\angle$ OCE + $\angle$ OEC

$\Rightarrow \angle OEB=90 \degree-\frac{z}{2}+z$ (from 3)

$\Rightarrow \angle OEB=90 \degree+\frac{z}{2}$ .................................................................6

From 4,5,6 ;we get

$\angle$ BDE = $\angle$ BED = $\angle$ OEB - $\angle$ OED

$\Rightarrow \angle BDE=\angle BED=90 \degree+\frac{z}{2}-(90-\frac{y}{2})=\frac{y+z}{2}$

$\Rightarrow \angle BDE+\angle BED=y+z$ ..................................................7

In $\triangle$ BDE ,

$\angle$ DBE + $\angle$ BDE + $\angle$ BED = $180 \degree$

$\Rightarrow \angle DBE +y+z=180 \degree$

$\Rightarrow \angle DBE =180 \degree-(y+z)$

$\Rightarrow \angle ABC =180 \degree-(y+z)$ ...................................................8

Here, from equation 1,

$\frac{x-y}{2}=\frac{360 \degree-y-2x-y}{2}$

$\Rightarrow \frac{x-y}{2}=\frac{360 \degree-2y-2x}{2}$

$\Rightarrow \frac{x-y}{2}=180 \degree-y-x$ ...................................9

From 8 and 9,we have

$\angle ABC=\frac{x-y}{2}=\frac{1}{2}(\angle AOC-\angle DOE)$

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

Answer:

Given : ABCD is a rhombus.

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

Proof :

1640237931704

ABCD is rhombus.

Thus, $\angle AOC = 90 \degree$ (diagonals of a rhombus bisect each other at $90 \degree$ )

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 $\small AE=AD$ .

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 :

1640237957576

$\angle$ ADC = $\angle$ 3 , $\angle$ ABC = $\angle$ 4, $\angle$ ADE = $\angle$ 1 and $\angle$ AED = $\angle$ 2

$\angle 3+\angle 1=180 \degree$ .................1(linear pair)

$\angle 2+\angle 4=180 \degree$ ....................2(sum of opposite angles of cyclic quadrilateral)

$\angle$ 3 = $\angle$ 4 (oppsoite angles of parallelogram )

From 1 and 2,

$\angle$ 3+ $\angle$ 1 = $\angle$ 2 + $\angle$ 4

From 3, $\angle$ 1 = $\angle$ 2

From 4, $\triangle$ AQB, $\angle$ 1 = $\angle$ 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 :

1640237982446

In $\triangle$ ABD and $\triangle$ CDO,

AO = OC (Given )

$\angle$ AOB = $\angle$ COD (Vertically opposite angles )

BO = DO (Given )

So, $\triangle$ ABD $\cong$ $\triangle$ CDO (By SAS)

$\angle$ BAO = $\angle$ DCO (CPCT)

$\angle$ BAO and $\angle$ DCO are alternate angle and are equal .

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

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

From 1 and 2,

$\angle A+\angle C=180 \degree$ ......................3(sum of opposite angles)

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

From 3 and 4,

$\angle A+\angle A=180 \degree$

$\Rightarrow 2\angle A=180 \degree$

$\Rightarrow \angle A=90 \degree$

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 :

1656935732366

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

$\angle A=90 \degree$ (proved in (i))

A parallelogram with one angle $90 \degree$ , 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 $\small 90^{\circ}-\frac{1}{2}C$ , $\small 90^{\circ}-\frac{1}{2}B$ and $\small 90^{\circ}-\frac{1}{2}A$

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 $\small 90^{\circ}-\frac{1}{2}C$ , $\small 90^{\circ}-\frac{1}{2}B$ and $\small 90^{\circ}-\frac{1}{2}A$

Proof :

1640238007328

$\angle$ 1 and $\angle$ 3 are angles in same segment.therefore,

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

and $\angle$ 2 = $\angle$ 4 ..................2

Adding 1 and 2,we have

$\angle$ 1+ $\angle$ 2= $\angle$ 3+ $\angle$ 4

$\Rightarrow \angle D=\frac{1}{2}\angle B+\frac{1}{2}\angle C$ ,

$\Rightarrow \angle D=\frac{1}{2}(\angle B+\angle C)$

$\Rightarrow \angle D=\frac{1}{2}(180 \degree+\angle C)$

and $\Rightarrow \angle D=\frac{1}{2}(180 \degree-\angle A)$

$\Rightarrow \angle D=90 \degree-\frac{1}{2}\angle A$

Similarly, $\Rightarrow \angle E=90 \degree-\frac{1}{2}\angle B$ and $\angle F=90 \degree-\frac{1}{2}\angle C$

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 $\small BP=BQ$ .

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 :

1640238028954

AB is a common chord in both congruent circles.

$\therefore \angle APB = \angle AQB$

In $\triangle BPQ,$

$\angle APB = \angle AQB$

$\therefore BQ = BP$ (Sides opposite to equal of the triangle are equal )

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

Answer:

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

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

1640238078439

Let $\angle$ ABD = $\angle$ 1 , $\angle$ ADC = $\angle$ 2 , $\angle$ DCB = $\angle$ 3 , $\angle$ CBD = $\angle$ 4

$\angle$ 1 and $\angle$ 3 lies in same segment.So,

$\angle$ 1 = $\angle$ 3 ..........................1(angles in same segment)

similarly, $\angle$ 2 = $\angle$ 4 ......................2

also, $\angle$ 1= $\angle$ 2 ..............3(given)

From 1,2,3 , we get

$\angle$ 3 = $\angle$ 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.

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 of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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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