Class 9 - Mathematics
Circles - Exercise 10.4
Top Block 1
Exercise 10.4
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.
Answer :
In ∆AOO′,
AO2 = 52 = 25
AO′2 = 32 = 9
OO′2 = 42 = 16
AO′2 + OO′2 = 9 + 16 = 25 = AO2
⇒ ∠AO′O = 900 [By converse of Pythagoras theorem]
Similarly, ∠BO′O = 900.
⇒ ∠AO′B = 900 + 900 = 1800
⇒ AO′B is a straight line whose mid-point is O.
⇒ AB = 3 + 3 = 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.
Answer :
Given: AB and CD are two equal chords of a circle which meet at E.
To prove: AE = CE and BE = DE
Construction: Draw OM ⊥ AB and ON ⊥ CD and join OE.
Proof: In ∆OME and ∆ONE,
OM = ON [Equal chords are equidistant]
OE = OE [Common]
∠OME = ∠ONE [Each equal to 90°]
So, ∆OME ≅∆ONE [by RHS axiom]
⇒ EM = EN ……..(i) [by CPCT]
Now AB = CD [Given]
⇒ AB/2 = CD/2
⇒ AM = CN ………(ii) [Perpendicular from centre bisects the chord]
Adding equation (i) and (ii), we get
EM + AM = EN + CN
⇒ AE = CE …………….(iii)
Now, AB = CD ……….(iv)
⇒ AB – AE = CD – AE [From equation (iii)]
⇒ BE = CD – CE
Mddle block 1
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.
Answer :
Given: AB and CD are two equal chords of a circle which meet at E within the circle and a line
PQ joining the point of intersection to the centre.
To Prove: ∠AEQ = ∠DEQ
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Proof: In ∆OLE and ∆OME, we have
OL = OM [Equal chords are equidistant]
OE = OE [Common]
∠OLE = ∠OME [Since each = 900]
So, ∆OLE ≅∆OME [by RHS congruence]
⇒ ∠LEO = ∠MEO [CPCT]
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 Fig. 10.25).
Answer :
Given: A line AD intersects two concentric circles at A, B, C and D, where O is the centre of these circles.
To prove: AB = CD
Construction: Draw OM ⊥ AD.
Proof: AD is the chord of larger circle.
So, AM = DM ………(i) [OM bisects the chord]
BC is the chord of smaller circle
So, BM = CM ……..(ii) [OM bisects the chord]
Subtracting equation (ii) from equation (i), we get
AM – BM = DM – CM
⇒ 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 6m each, what is the distance between Reshma and Mandip?
Answer :
Let Reshma, Salma and Mandip be represented by R, S and M respectively.
Now, Draw OL ⊥ RS.
OL2 = OR2 – RL2
⇒ OL2 = 52 – 32 [RL = 3 m, because OL ⊥ RS]
⇒ OL2 = 25 – 9
⇒ OL2 = 16
⇒ OL = √16
⇒ OL = 4
Now, area of triangle ORS = (1/2) * KR * OR
= (1/2) * KR * 5
Also, area of ∆ORS = (1/2) * RS * OL
= (1/2) * 6 * 4
= 12 m2
Now, (1/2) * KR * 5 = 12
⇒ KR = (12 * 2)/5
⇒ KR = 24/5
⇒ KR = 4.8 m
Now, RM = 2KR
⇒ RM = 2 * 4.8
⇒ RM = 9.6 m
Hence, distance between Reshma and Mandip is 9.6 m.
Question : 6: A circular park of radius 20m 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.
Answer :
Let Ankur, Syed and David be represented by A, S and D respectively.
Let PD = SP = SQ = QA = AR = RD = x
In ∆OPD,
OP2 = 400 – x2
⇒ OP = √(400 – x2)
⇒ AP = 2√(400 – x2) + √(400 – x2) [Since centroid divides the median in the ratio 2 : 1]
⇒ AP = 3√(400 – x2)
Now, in ∆APD,
PD2 = AD2 – DP2
⇒ x2 = (2x)2 – {3√(400 – x2)}2
⇒ x2 = 4x2 – 9(400 – x2)
⇒ x2 = 4x2 – 3600 + 9x2
⇒ x2 = 13x2 – 3600
⇒ 13x2 – x2 = 3600
⇒ 12x2 = 3600
⇒ x2 = 3600/12
⇒ x2 = 300
⇒ x = √300
⇒ x = 10√2
Now, SD = 2x = 2 * 10√3 = 20√3
Hence, ASD is an equilateral triangle.
So, SD = AS = AD = 20√3
Hence, the length of the string of each phone is 20√3 m.