NCERT Solutions for Exercise 10.4 Class 9 Maths Chapter 10

NCERT Solutions for Exercise 10.4 Class 9 Maths Chapter 10 - Circles

Edited By Ramraj Saini | Updated on Apr 23, 2024 12:10 PM IST

NCERT Solutions for Class 9 Maths Exercise 10.4 Chapter 10 Circles- Download Free PDF

NCERT Solutions for Class 9 Maths Exercise 10.4 - We'll start with a basic definition of a circle and then go over the theorems discussed in chapter 10 of Class 9 Math Exercise 10.4. Any closed shape with all points connected at the same distance from the centre is called a circle. Any point equidistant from any of the circle's boundaries is the centre of the circle. Radius is a Latin word that means 'ray,' but it refers to the line segment that connects the circle's centre and edge. Any line that begins or ends at the circle's centre and connects to any point on the circle's border is defined as the radius of the circle.

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NCERT Solutions for Class 9 Maths Exercise 10.4 Chapter 10 Circles- Download Free PDF
NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.4
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Benefits of NCERT Solutions for Class 9 Maths Exercise 10.4
Key Features of Exercise 10.4 Class 9 Maths
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

NCERT Solutions for Exercise 10.4 Class 9 Maths Chapter 10 - Circles

The 9th class maths exercise 10.4 answers consists of six questions expertly developed by Careers360 subject matter specialists. These class 9 maths chapter 10 exercise 10.4 provide students with complete help by providing extensive, step-by-step explanations. PDF versions are also freely accessible for download, increasing accessibility and convenience.

Along with NCERT book Class 9 Maths chapter 9 exercise 10.4 the following exercises are also present.

**Please be aware that this chapter has been renumbered as Chapter 9 in the CBSE Syllabus for the academic year 2024-25.

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NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.4

Q1 Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ . Find the length of the common chord.
Answer:

Given: Two circles of radii $\small 5\hspace{1mm}cm$ and $\small 3\hspace{1mm}cm$ intersect at two points and the distance between their centres is $\small 4\hspace{1mm}cm$ .

To find the length of the common chord.

Construction: Join OP and draw $OM\perp AB\, \, and \, \, \, ON\perp CD.$

1640237228523

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

$\therefore PM\perp AB$

$\angle PMA=90 \degree$

Let, PM = x , so QM=4-x

In $\triangle$ APM, using Pythagoras theorem

$AM^2=AP^2-PM^2$ ...........................1

Also,

In $\triangle$ AQM, using Pythagoras theorem

$AM^2=AQ^2-MQ^2$ ...........................2

From 1 and 2, we get

$AP^2-PM^2=AQ^2-MQ^2$

$\Rightarrow 3^2-x^2=5^2-(4-x)^2$

$\Rightarrow 9-x^2=25-16-x^2+8x$

$\Rightarrow 9=9+8x$

$\Rightarrow 8x=0$

$\Rightarrow x=0$

Put,x=0 in equation 1

$AM^2=3^2-0^2=9$

$\Rightarrow AM=3$

$\Rightarrow AB=2AM=6$

Q2 If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.

Construction : Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

Proof :

1640237267424

In $\triangle$ OMP and $\triangle$ ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

$\angle$ OMP = $\angle$ ONP (Both are right angled)

Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AM = CN$ ......................3

Adding 1 and 3, we have

AM + PM = CN + PN

$\Rightarrow AP = CP$

Subtract 4 from 2, we get

AB-AP = CD - CP

$\Rightarrow PB = PD$

Q3 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer:

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. $\angle$ OPM= $\angle$ OPN

Proof :

Construction: Join OP and draw $OM\perp AB\, \, \, \, and\, \, \, ON\perp CD.$

In $\triangle$ OMP and $\triangle$ ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

$\angle$ OMP = $\angle$ ONP (Both are right-angled)

Thus, $\triangle$ OMP $\cong$ $\triangle$ ONP (By RHS rule)

$\angle$ OPM= $\angle$ OPN (CPCT)

Q4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that $\small AB=CD$ (see Fig. $\small 10.25$ ).

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.

To prove : AB = CD

Construction: Draw $OM\perp AD$

Proof :

1640237314606

BC is a chord of the inner circle and $OM\perp BC$

So, BM = CM .................1

(Perpendicular OM bisect BC)

Similarly,

AD is a chord of the outer circle and $OM\perp AD$

So, AM = DM .................2

(Perpendicular OM bisect AD )

Subtracting 1 from 2, we get

AM-BM = DM - CM

$\Rightarrow AB = CD$

Q5 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius $\small 5m$ drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is $\small 6m$ each, what is the distance between Reshma and Mandip?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw $OL\perp RS$ .

Proof:

1640237339300 In $\triangle$ ORS,

OS = OR and $OL\perp RS$ (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In $\triangle$ OLS, by pytagoras theorem,

$OL^2=OS^2-SL^2$

$\Rightarrow OL^2=5^2-3^2=25-9=16$

$\Rightarrow OL=4$

In $\triangle$ ORK and $\triangle$ OMK,

OR = OM (Radii)

$\angle$ ROK = $\angle$ MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

$\triangle$ ORK $\cong$ $\triangle$ OMK (By SAS)

RK = MK (CPCT)

Thus, $OK\perp RM$

area of $\triangle$ ORS = $\frac{1}{2}\times RS\times OL$ ...............................1

area of $\triangle$ ORS = $\frac{1}{2}\times OS\times KR$ .............................2

From 1 and 2, we get

$\frac{1}{2}\times RS\times OL$ $=\frac{1}{2}\times OS\times KR$

$\Rightarrow RS\times OL=OS\times KR$

$\Rightarrow 6\times 4=5\times KR$

$\Rightarrow KR=4.8 cm$

Thus, $RM =2 KR=2\times 4.8 cm=9.6 cm$

Q6 A circular park of radius $\small 20m$ is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP $\perp$ SD

Proof :

1640237370579

Let AS = SD = AD = 2x cm

In $\triangle$ ASD,

AS = AD and AP $\perp$ SD

So, SP = PD = x cm

In $\triangle$ OPD, by Pythagoras,

$OP^2=OD^2-PD^2$

$\Rightarrow OP^2=20^2-x^2=400-x^2$

$\Rightarrow OP=\sqrt{400-x^2}$

In $\triangle$ APD, by Pythagoras,

$AP^2=AD^2-PD^2$

$\Rightarrow (AO+OP)^2+x^2=(2x)^2$

$\Rightarrow (20+\sqrt{400-x^2})^2+x^2=4x^2$

$\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2$

$\Rightarrow 800+40\sqrt{400-x^2}=4x^2$

$\Rightarrow 200+10\sqrt{400-x^2}=x^2$

$\Rightarrow 10\sqrt{400-x^2}=x^2-200$

Squaring both sides,

$\Rightarrow 100(400-x^2)=(x^2-200)^2$

$\Rightarrow 40000-100x^2=x^4-40000-400x^2$

$\Rightarrow x^4-300x^2=0$

$\Rightarrow x^2(x^2-300)=0$

$\Rightarrow x^2=300$

$\Rightarrow x=10\sqrt{3}$

Hence, length of string of each phone $= 2x=20\sqrt{3}$ m

Benefits of NCERT Solutions for Class 9 Maths Exercise 10.4

Equal Chords and Their Distances from the Center is the subject of exercise 10.4 in Class 9 Math.
NCERT syllabus Class 9 Maths chapter 10 exercise 10.4 introduces us to theorems related to the equal chords in a circle.
Understanding the principles from chapter 10 exercise 10.4 in Class 9 Arithmetic will help us grasp the theorems of equal chords and their distance.

Key Features of Exercise 10.4 Class 9 Maths

9th class maths exercise 10.4 answers is a comprehensive exercise that covers various concepts related to circles, including chords, secants, and tangents.
Class 9 maths ex 10.4 offers a variety of problems with different levels of complexity. This allows students to practice and strengthen their understanding of circle-related concepts.
Solutions to the problems in class 9 ex 10.4 are presented in a step-by-step format. This helps students follow the solution process and understand the concepts involved.
Solutions for class 9 maths ex 10.4 are often available in PDF format. This enables students to download and access them offline for convenient studying.
Exercise 10.4 aligns with the prescribed syllabus, covering all relevant topics and concepts related to circles in Class 9 Maths.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

1. What is the main concept of NCERT solutions for Class 9 Maths exercise 10.4?

In this exercise, we learn about the chords and their related theorems such as Chords that are equidistant from the circle's centre have the same length.

2. What is the angle subtended by the largest chord of the circle on circumference?

The largest chord subtends the right angle on the circumference.

3. The number of equal chords in the circle is finite?

NO, because equal chords can be infinite.

4. If the chord is equal to the radius of the circle, what angle will it subtend at a point on the major arc?

The subtended angle of the chord at a position on the major arc is 30. An equilateral triangle has 60 degrees on each side. An arc of a circle at its centre has double the angle of any other point of the circle.

5. Every chord is also the diameter of a circle. Is it also true that the converse of this statement is true?

Because its endpoints are on the circle's circumference, every diameter is a chord. It's the longest chord that runs through the circle's centre.

However, as every chord does not pass through the centre, every chord cannot be a diameter.

6. What is the definition of con cyclic points?

Concyclic points are a group of points that all lie in the same circle.

7. When a circle is divided in three equal arcs, each arc is major arc. Is it true or false?

False, because a major arc's measure is more than 180° and equal to 360° minus the minor arc's measure with the same endpoints.

8. The length of perpendicular OE on CD = 5 cm when two equal chords AB and CD of a circle. When OF is perpendicular on AB. What is the length of OF?

The length of OF is also 5cm because Chords that are equidistant from the circle's centre have the same length.

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