# CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 10 CIRCLES EX – 10.4

## Circles

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.
∵ In two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

∴∠OLP = ∠OLQ = 90° and PL = LQ
Now, in right ∆OLP, we have
PL2 + OL2 = 2
⇒ PL2 + (4 – x)2 = 52
⇒ PL2 = 52 – (4 – x)2
⇒ PL2 = 25 -16 – x2 + 8x
⇒ PL2 = 9 – x2 + 8x …(i)
Again, in right ∆O’LP,
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 = 3cm + 3cm = 6cm
Thus, the required length of the common chord = 6 cm.

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 criteria]
⇒ ME = NE [C.P.C.T.]

Adding AM on both sides, we get
⇒ AM + ME = AM + NE
⇒ AE = DN + NE = DE
∵ AB = CD ⇒ 12AB = 12DC
⇒ AM = DN
⇒ AE = DE …(i)
Now, AB – AE = CD – DE
⇒ BE = CE …….(ii)
From (i) and (ii), we have
AE = DE and CE = BE

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,
OM = ON
[Equal chords are equidistant from the centre]
OE = OE [Common hypotenuse]

∠OME = ∠ONE [Each equal to 90°]
∴ ∆OME ≅ ∆ONE [By RHS congruence criteria]
⇒ ∠OEM = ∠OEN [C.P.C.T.]
⇒ ∠OEA = ∠OED

Question 4.
If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).

Solution:
Given : Two circles with the common centre O.
A line D intersects the outer circle at A and D and the inner circle at B and C.
To Prove : AB = CD.
Construction:
Draw OM ⊥ l.
Proof: For the outer circle,
OM ⊥ l [By construction]
∴ AM = MD …(i)
[Perpendicular from the centre to the chord bisects the chord]

For the inner circle,
OM ⊥ l [By construction]
∴ BM = MC …(ii)
[Perpendicular from the centre to the chord bisects the chord]
Subtracting (ii) from (i), we have
AM – BM = MD – MC
⇒ AB = CD

Question 5.
Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 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 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]

Equal chords of a circle subtend equal angles at the centre.
∴ ∠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 criteria]
∴ 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 criteria]
⇒ ∠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 ∆OPR, we have
x2 + RP2 = 52
RP2 = 52 – x2
In right ∆SPR, we have
(5 – x)2 + RP2 = 62
⇒ RP2 = 62 – (5 – x)2 ……..(ii)
From (i) and (ii), we get
⇒ 52 – x2 = 62 – (5 – x)2
⇒ 25 – x2 = 36 – [25 – 10x + x2]
⇒ – 10x + 14 = 0
⇒ 10x = 14 ⇒ x = 1410 = 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 x 4.8 = 9.6
Thus, distance between Reshma and Mandip is 9.6 m.

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 = 12 SD = 12 (2x)
⇒ SM = x

Now, in right ∆ASM, we have
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 ∆OSM, we have
OS2 = SM2 + OM2 [using Pythagoras theorem]
202 = x2 + (3x−−√ – 20)2
⇒ 400 = x2 + 3x2 – 403x−−√ + 400
⇒ 4x2 = 40 3x−−√
⇒ x = 10√3 m
Now, SD = 2x = 2 x 10√3 m = 20√3 m
Thus, the length of the string of each phone = 20√3 m