Three girls Reshma, Salma and Mandip are
playing a game by standing on a circle of radius 5
cm 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 6 cm each, what is
the distance between Reshma and Mandip?
plz plz plz answer this question. This is Vry Vry important fr me to get it's answer right now.
Answers
enu
NCERT Solutions for Class 9 Maths Exercise 10.4
Last Updated: June 18, 2018 by myCBSEguide
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NCERT solutions for Class 9 Maths Circles Download as PDF
NCERT Solutions for Class 9 Maths Exercise 10.4
NCERT Solutions for Class 9 Mathematics Circles
1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.
Ans. Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.
Radius OA = 5 cm, Radius O’A = 3 cm,
Distance between their centers OO’ = 4 cm
In triangle AOO’,
52 = 42 + 32
25 = 16 + 9
25 = 25
Hence AOO’ is a right triangle, right angled at O’.
Since, perpendicular drawn from the center of the circle bisects the chord.
Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.
Therefore length of chord AB = Diameter of circle II
Length of chord AB = 2 x 3 = 6 cm.
NCERT Solutions for Class 9 Maths Exercise 10.4
2. 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.
Ans. Given: Let AB and CD are two equal chords of a circle of centers O intersecting each other at point E within the circle.
To prove: (a) AE = CE
(b) BE = DE
Construction: Draw OM AB, ON CD. Also join OE.
Proof: In right triangles OME and ONE,
OME = ONE =
OM = ON
[Equal chords are equidistance from the centre]
OE = OE [Common]
OMEONE [RHS rule of congruency]
ME = NE [By CPCT] ……….(i)
Now, O is the centre of circle and OM AB
AM = AB
[Perpendicular from the centre bisects the chord] …..(ii)
Similarly, NC = CD ……….(iii)
But AB = CD [Given]
From eq. (ii) and (iii), AM = NC ……….(iv)
Also MB = DN …….…(v)
Adding (i) and (iv), we get,
AM + ME = NC + NE
AE = CE [Proved part (a)]
Now AB = CD [Given]
AE = CE [Proved]
AB – AE = CD – CE
BE = DE [Proved part (b)]
3. 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 chord.
Ans. Given: AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined.
To prove: OEM = OEN
Construction: Draw OM AB and
ON CD.
Proof: In right angled triangles OME and ONE,
OME = ONE [Each ]
OM = ON [Equal chords are equidistant from the centre]
OE = OE [Common]
OMEONE [RHS rule of congruency]
OEM = OEN [By CPCT]
NCERT Solutions for Class 9 Maths Exercise 10.4
4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (See figure)
Ans. Given: Line intersects two concentric circles with centre O at points A, B, C and D.
To prove: AB = CD
Construction: Draw OL
Proof: AD is a chord of outer circle and OL AD.
AL = LD ………(i) [Perpendicular drawn from the centre bisects the chord]
Now, BC is a chord of inner circle and
OL BC
BL = LC …(ii) [Perpendicular drawn from the centre bisects the chord]
Subtracting (ii) from (i), we get,
AL – BL = LD – LC
AB = CD
NCERT Solutions for Class 9 Maths Exercise 10.4
5. Three girls Reshma, Salma and Mandip are standing on a circle of radius 5 m 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 6 m each, what is the distance between Reshma and Mandip?
Ans. Let Reshma, Salma and Mandip takes the position C, A and B on the circle. Since AB = AC
The centre lies on the bisector of BAC.
Let M be the point of intersection of BC and OA.
Again, since AB = AC and AM bisects
CAB.
AM CB and M is the mid-point of CB.
Let OM = then MA =
From right angled triangle OMB,
OB2 = OM2 + MB2
52 = + MB2 ……….(i)
Again, in right angled triangle AMB,
AB2 = AM2 + MB2
62 = + MB2 ……….(ii)
Equating the value of MB2 from eq. (i) and
Hence, from eq. (i),
MB2 = =
= =
MB = = 4.8 cm
BC = 2MB = 2 x 4.8 = 9.6 cm