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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.
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.
Given: Two circles of radii and intersect at two points and the distance between their centres is .
To find the length of the common chord.
Construction: Join OP and draw
Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.
Let, PM = x , so QM=4-x
In APM, using Pythagoras theorem
...........................1
Also,
In AQM, using Pythagoras theorem
...........................2
From 1 and 2, we get
Put,x=0 in equation 1
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
Proof :
In OMP and ONP,
AP = AP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
OMP = ONP (Both are right angled)
Thus, OMP ONP (By SAS rule)
PM = PN..........................1 (CPCT)
AB = CD ............................2(Given )
......................3
Adding 1 and 3, we have
AM + PM = CN + PN
Subtract 4 from 2, we get
AB-AP = CD - CP
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. OPM= OPN
Proof :
Construction: Join OP and draw
In OMP and ONP,
AP = AP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
OMP = ONP (Both are right-angled)
Thus, OMP ONP (By RHS rule)
OPM= OPN (CPCT)
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
Proof :
BC is a chord of the inner circle and
So, BM = CM .................1
(Perpendicular OM bisect BC)
Similarly,
AD is a chord of the outer circle and
So, AM = DM .................2
(Perpendicular OM bisect AD )
Subtracting 1 from 2, we get
AM-BM = DM - CM
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 .
Proof:
In ORS,
OS = OR and (by construction )
So, RL = LS = 3cm (RS = 6 cm )
In OLS, by pytagoras theorem,
In ORK and OMK,
OR = OM (Radii)
ROK = MOK (Equal chords subtend equal angle at centre)
OK = OK (Common)
ORK OMK (By SAS)
RK = MK (CPCT)
Thus,
area of ORS = ...............................1
area of ORS = .............................2
From 1 and 2, we get
Thus,
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 SD
Proof :
Let AS = SD = AD = 2x cm
In ASD,
AS = AD and AP SD
So, SP = PD = x cm
In OPD, by Pythagoras,
In APD, by Pythagoras,
Squaring both sides,
Hence, length of string of each phone m
NCERT solutions Class 9 Maths exercise 10.4 – This exercise includes some important theorems about two equal chords for the examination point of view. We'll start with a definition of a chord.
The theorems are:
A circle's equal chords (or congruent circles' equal chords) are equidistant from the centre (or centres).
Chords that are equidistant from the circle's centre have the same length.
Also Read| Circles Class 9 Notes
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.
Also, See
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.
The largest chord subtends the right angle on the circumference.
NO, because equal chords can be infinite.
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.
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.
Concyclic points are a group of points that all lie in the same circle.
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.
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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