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

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

Team Careers360Updated on 29 Jul 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.

This Story also Contains

  1. Circles Class 9 Chapter 10 Exercise: 10.6
  2. More About NCERT Solutions for Class 9 Maths Exercise 10.6
  3. Benefits of NCERT Solutions for Class 9 Maths Exercise 10.6
  4. NCERT Solutions of Class 10 Subject Wise
  5. 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 = $360 ^\circ$

$\Rightarrow x+y+z+z=360 ^\circ$

$\Rightarrow x+y+2z=360 ^\circ$ .........................................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 = $180 ^\circ$

$\Rightarrow 2\angle OAD+z=180 ^\circ$

$\Rightarrow 2\angle OAD=180 ^\circ-z$

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

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

Similarly,

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

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

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

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

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

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

similarly,

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

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

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

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

From 4,5,6 ;we get

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

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

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

In $\triangle$ BDE ,

$\angle$ DBE + $\angle$ BDE + $\angle$ BED = $180 ^\circ$

$\Rightarrow \angle DBE +y+z=180 ^\circ$

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

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

Here, from equation 1,

$\frac{x-y}{2}=\frac{360 ^\circ-y-2x-y}{2}$

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

$\Rightarrow \frac{x-y}{2}=180 ^\circ-y-x$ ...................................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, $\angle AOC = 90 ^\circ$ (diagonals of a rhombus bisect each other at $90 ^\circ$ )

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

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

$\angle 2+\angle 4=180 ^\circ$ ....................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,

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

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

From 3 and 4,

$\angle A+\angle A=180 ^\circ$

$\Rightarrow 2\angle A=180 ^\circ$

$\Rightarrow \angle A=90 ^\circ$

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))

$\angle A=90 ^\circ$ (proved in (i))

A parallelogram with one angle $90 ^\circ$ , 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

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

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

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

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

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

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

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.

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.

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.

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