Class IX Math
NCERT Solution for Circles
NCERT Solution for Circles
EXERCISE 10.4
Q1. 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 chrod.
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 teh common chord.
∴ ∠OLP = ∠OLP = 90°
PL = PQ
Now in right ∠OLP,
PL2 + OL2 = OP2
⇒ PL2 + (4 – x)2 = 52
⇒ PL2 = 52 – (4 – x)2
⇒ PL2 = 25 – 16 – x2 + 8x
⇒ PL2 = 9 – x2 + 8x …(1)
Again, in ΔO’LP,
PL2 = 32 – x2 = 9 – x2 …(2)
From (1) and (2), 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 = 3cm
PQ = PL + LQ = 3 cm + 3 cm = 6 cm
Thus, the required length of the common chord = 6 cm.
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.
Solution: We have a circle with centre O. Equal chords AB and CD intersect at E.
To prove that AE = DE and CE = BE, draw OM ⊥ AB and ON ⊥ CD.
Since, AB = CD [Given]
∴OM=ON [Equal chords are equidisitant from teh centre.]
Now, in right ΔOME and right ΔONE,
OM = ON [Proved]
OE = OE [Common]
∴ ΔOME ≌ ΔONE [RHS criterion]
ME = NE [c.p.c.t]
⇒ AE = DE …(1)
Since AB = CD [Given]
∴ AB – AE = CD – DE
⇒ CE = BE …(2)
From (1) and (2), we have
AE = DE and CE = BE
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.
Solution: We have a circle with centre O and equal chords AB and CD are intersecting at E. OE is joined.
A To prove that ∠1 = ∠2,let us draw OM ⊥ AB and ON ⊥ CD.
In right ΔOME and right ΔONE,
OM = ON [Equal chords are equidistant from the centre.]
OE = OE [Common]
∴ ΔOME ≌ ΔONE [RHS criterion]
⇒ Their corresponding parts are equal.
∴ ∠OEM = ∠OEN
or ∠OEA = ∠OED
Class IX EXERCISE 10.4
Q4. 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: We have two circles with the common centre O.
A line ‘l’ intersects the outer circle at A and D and the inner circle at B and C. To prove that AB = CD,
let us draw OM ⊥ l.
For the outer circle,
⇒ OM ⊥ l [Construction]
∴ AM = MD [Perpendicular from the centre to the chord bisects the chord]
…(1)
For the inner circle,
OM ⊥ l [Construction]
∴BM = MC [Perpendicular from the centre to the chord bisects the chord] …(2)
Subtracting (2) from (1), we have
AM – BM = MD – MC
AB = CD
Q5. 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 hall 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 are positioned at R, S and M, respectively on the circle of radius 5 m.
RS = SM = 6 m [Given]
Equal chords of a circle subtend equal angles at the centre.
∠1 = ∠2
In ΔPOR and ΔPOM,
OP = OP [Common]
OR = OM [Radii of the same circle]
∠1 = ∠2 [Proved]
∴ ΔPOR ≌ΔPOM [SAS criterion]
⇒ Their corresponding parts are equal.
∴ PR = PM and ∠OPR = ∠OPM
∵ ∠OPR + ∠OPM = 180° [Linear pairs]
∴ ∠OPR = ∠OPM = 90°
⇒ OP⊥RM
Now, in A RSP and A MSP, [6 m each]
RS=MS [Common]
SP = SP
∠RSP = ∠MSP
∴ΔRSP ≌ΔMSP [SAS criterion]
But ∠RSP + ∠MSP = 180°
⇒∠RPS = ∠MSP = 90° [Each]
∴ SP passes through O.
Let OP = x m
∴ SP = (5 – x) m
Now, in right ΔOPR,
x2 + RP2 = 52
In right ΔSPR,
(5 – x)2 + RP2 = 62
From (1), RP2 = 52 – x2
From (2), RP2 = 62 – (5 – x)2
∴ 52 – x2 = 62 – (5 – x)2
⇒ 25 – x2 = 36 –[25 – 10x + x2]
⇒ 25 – x2 – 36 –[25 – 10x + x2 = 0
⇒ –10x + 14 = 0 ⇒ 10x = 14
Now, RP2 = 52 – x2 ⇒RP2 = 25 – (1.4)2
⇒ RP2 = 25 – 1.96 = 23.04 m
∴ RM = 2 RP = 2 × 4.8 m = 9.6 m
Thus, distance between Reshma and Mandip is 9.6 m.
Q6. A circular park of radius 20 m is situated in a colony. David are sitting at equal distance on its boundary each having a toy each other. Find the length of the string of each phone.
Sol: In the figure, let Ankur, Syed and David are sitting at A, S
AS = SD = AD
i.e. DASD is an equilateral triangle.
Let the length of each side of the equilateral triangle is 2x metres.
Let us draw AM ⊥ SD.
Since DASD is an equilateral,
∴ AM passes through O.
