NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4
NCERT Solutions for Class
9 Maths Chapter 10 Circles Ex 10.4 are the part
of NCERT Solutions for Class 9 Maths. Here you can find the NCERT Solutions for
Chapter 10 Circles Ex 10.4.
Ex 10.4 Class 9 Maths Question 1.
Two circles of radii 5 cm and 3 cm
intersect at two points and the distance between their centres is 4 cm. Find
the length of the common chord.
Solution:
We have two intersecting circles
with centres at O and O’ respectively. Let PQ be the common chord of both the
circles.
Since in two intersecting circles, the line
joining their centres is perpendicular bisector of the common chord.
Now, in right-angled ∆OLP, we have
PL2 + OL2 = OP2
⇒ PL2 + (4 – x)2 = 52
⇒ PL2 = 52 – (4 – x)2
⇒ PL2 = 25 – 16 – x2 + 8x
⇒ PL2 = 9 – x2 + 8x …..…(i)
Again, in right-angled ∆O’LP, we have
PL2 = PO‘2 – LO‘2 = 32 – x2 = 9 – x2 ….…(ii)
From (i) and (ii), we have
9 – x2 + 8x = 9 – x2
⇒ 8x = 0
⇒ x = 0
⇒ L and O’ coincide.
∴ PQ is a diameter of the smaller circle.
⇒ PL = 3 cm
But PL = LQ
∴ LQ = 3 cm
∴ PQ = PL + LQ = 3 cm + 3 cm = 6 cm
Thus, the required length of the common chord = 6 cm.
Ex 10.4 Class 9 Maths Question 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.
Solution:
Given: A circle with centre O and equal chords AB
and CD intersect at E.
To prove: AE = DE and CE = BE
Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OE.
Proof:
Since AB = CD [Given]
∴ OM = ON [Equal chords are
equidistant from the centre]
Now, in ∆OME and ∆ONE, we have
∠OME = ∠ONE [Each
equal to 90°]
OM = ON [Proved above]
OE = OE [Common
hypotenuse]
∴ ∆OME ≅ ∆ONE [By RHS
congruence rule]
⇒ ME = NE [C.P.C.T.]
Adding AM on both sides, we get
⇒ AE = DN + NE = DE [∵ AB = CD ⇒ ½ AB = ½ DC ⇒ AM = DN]
⇒ AE = DE ….…(i)
Now, AB – AE = CD – DE
⇒ BE = CE …….(ii)
From (i) and (ii), we have
AE = DE and CE = BE
Ex 10.4 Class 9 Maths Question 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 chords.
Solution:
Given: A circle with centre O and
equal chords AB and CD are intersecting at E.
To prove: ∠OEA = ∠OED
Construction: Draw OM ⊥ AB and ON ⊥ CD. Join OE.
Proof: In ∆OME and ∆ONE, we have
OM
= ON [Equal
chords are equidistant from the centre]
OE = OE [Common hypotenuse]
∠OME = ∠ONE [Each equal to 90°]
∴ ∆OME ≅ ∆ONE [By RHS congruence rule]
⇒ ∠OEM = ∠OEN [C.P.C.T.]
⇒ ∠OEA = ∠OED
Ex 10.4 Class 9 Maths Question 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).
Solution:
Given: Two circles with the common centre O. A line intersects the outer circle at A and D and the inner circle at B and C.
To prove: AB = CD.
Construction: Draw OM ⊥ AD.
OM ⊥ AD [By construction]
∴ AM = MD ….…(i)
[Perpendicular from the centre to the chord bisects the chord]
For the inner circle,
OM ⊥ BC [By construction]
∴ BM = MC ….…(ii)
[Perpendicular from the centre to the chord
bisects the chord]
Subtracting eq. (ii) from (i), we have
AM – BM = MD – MC
⇒ AB = CD
Ex 10.4 Class 9 Maths Question 5.
Three girls Reshma, Salma and
Mandip are playing a game by 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?
Solution:
Let the three girls Reshma, Salma
and Mandip be positioned at R, S and M, respectively on the circle with centre
O and radius 5 m such that
RS = SM = 6 m [Given]
∴ ∠1 = ∠2
In ∆POR and ∆POM, we have
OP = OP [Common]
OR = OM [Radii of the same circle]
∠1 = ∠2 [Proved above]
∴ ∆POR ≅ ∆POM [By SAS congruence rule]
∴ PR = PM and ∠OPR = ∠OPM [C.P.C.T.]
∵ ∠OPR + ∠OPM = 180° [Linear pair]
∴ ∠OPR = ∠OPM = 90°
⇒ OP ⊥ RM
Now, in ∆RSP and ∆MSP, we have
RS = MS [Each 6 cm]
SP = SP [Common]
PR = PM [Proved above]
∴ ∆RSP ≅ ∆MSP [By SSS congruence rule]
⇒ ∠RPS = ∠MPS [C.P.C.T.]
But ∠RPS + ∠MPS = 180° [Linear pair]
⇒ ∠RPS = ∠MPS = 90°
SP passes through O.
Let OP = x m
∴ SP = (5 – x) m
Now, in right-angled ∆OPR, we have
x2 + RP2 = 52
RP2 = 52 – x2 ……..(i)
In right-angled ∆SPR, we have
(5 – x)2 + RP2 = 62
⇒ RP2 = 62 – (5 – x)2 ……..(ii)
From eq. (i) and (ii), we get
⇒ 52 – x2 = 62 – (5 – x)2
⇒ 25 – x2 = 36 – [25 – 10x + x2]
⇒ 25 – x2 = 36 – 25 + 10x – x2
⇒ 25 + 25 – 36 = 10x
⇒ 10x = 14
⇒ x = 14/10 = 1.4
Now, RP2 = 52 –
x2
⇒ RP2 = 25 – (1.4)2
⇒ RP2 = 25 – 1.96 = 23.04
∴ RP = √23.04 = 4.8
∴ RM = 2RP = 2 × 4.8 = 9.6
Thus, distance between Reshma and Mandip is 9.6
m.
Ex 10.4 Class 9 Maths Question 6.
A circular park of radius 20 m 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.
Solution:
Let Ankur, Syed and David are
sitting at A, S and D respectively in the circular park with centre O such that
AS = SD = DA
i.e., ∆ASD is an equilateral triangle.
Let the length of each side of the equilateral
triangle be 2x.
Draw AM ⊥ SD.
Since ∆ASD is an equilateral triangle.
∴ AM passes through O.
⇒ SM = ½ SD = ½ (2x)
⇒ SM = x
AM2 + SM2 = AS2 [Using Pythagoras theorem]
⇒ AM2 = AS2 – SM2 = (2x)2 –
x2 = 4x2 – x2 = 3x2
⇒ AM = √3x m
Now, OM = AM – OA = (√3x – 20) m
Again, in right-angled ∆OSM, we have
OS2 = SM2 + OM2 [Using Pythagoras theorem]
202 = x2 + (√3x – 20)2
⇒ 400 = x2 + 3x2 –
40√3x + 400
⇒ 4x2 = 40√3x
⇒ x = 10√3 m
Now, SD = 2x = 2 × 10√3 m = 20√3 m
Thus, the length of the string of each phone =
20√3 m
NCERT Solutions for Maths Class 10