For previous exercise – Circles – Exercise 10.3 – Class IX

**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:

In ∆AOO’,

AO^{2} = 5^{2} = 25

AO’^{2} = 3^{2} = 9

OO’^{2} = 4^{2} = 16

AO’^{2} + OO’^{2} = 9 + 16 = 25 = AO^{2}

⇒∠AO’O = 90˚ [By converse of Pythagoras theorem]

Similarly, ∠BO’O = 90˚

∠AO’B = 90˚ + 90˚ = 180˚

Therefore, ∠AO’B is a straight line whose midpoint is O’.

⇒ AB = (3 + 3)cm = 6 cm

**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: Let AB and CD be the two equal chords intersecting at point X.

⇒AB = CD

To prove:

Corresponding segments are equal i.e., AX = DX

and BX = CX

Proof: We draw OM ⊥AB and ON ⊥CD

So, AM = BM = ^{1}/_{2} AB

and DN = CN = ^{1}/_{2} CD

We know, AB = CD

⇒^{1}/_{2} AB = ^{1}/_{2} CD

⸫AM = DN ………………………..….(1)

and MB = CN …………………………(2)

In ∆OMX and ∆ONX,

∠OMX = ∠ONX = 90˚

OX = OX (common)

OM = ON [AB and Cd are equal chords and equal chords are equidistant from the centre)

⸫∆OMX ≅ ∆ONX (RHS congruence rule)

⸫MX = NX (CPCT) …………….(3)

On adding (1) and (3) we get,

AM + MX = DN + NX

AX = DX

On subtracting (2) and (3), we get,

BM – MX = CN – MX

BX = CX

Therefore, AX = DX and BX = CX

**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 AB and CD are two equal chords of a circle which meets at E within the centre 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 = 90˚

⇒∠LEO = ∠MEO (CPCT)

**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).**

Solution:

Given: Two concentric circles with centre O and a line intersects the circles at A, B, C and D

To prove:

AB = CD

Proof:

Let two circles be C_{1} and C_{2} and line be *l.*

We draw Op perpendicular to line *l*.

In circle C_{1},

OP ⊥BC (As OP is perpendicular to line l)

So, OP bisects BC

i.e., BP = CP …………(1) (perpendicular drawn from centre of a circle to a chord bisects the chord)

In circle C_{2},

OP ⊥ AD (As Op is perpendicular to line l)

So, OP bisects AD

i.e., AP = DP …………(2) (perpendicular drawn from centre of a circle to a chord bisects the chord)

On subtracting (2) and (1)

AP – BP = DP – CP

⇒ AB = CD

**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?**

Solution:

Let Reshma, Salma and Mandip be represented by R, S and M respectively.

Draw OL ⊥ RS,

OL^{2} = OR^{2} – RL^{2}

OL^{2} = 5^{2} – 3^{2} [RL = 3 m, because OL ⊥ RS]

= 25 – 9 = 16

OL = √(16) = 4

Now, area of triangle ORS = ^{1}/_{2} × KR × 05

= ^{1}/_{2} × KR × 05

Also, area of ∆ORS = ^{1}/_{2} × RS × OL = ^{1}/_{2} × 6 × 4 = 12 m^{2}

⇒^{1}/_{2} × KR × 5 = 12

⇒ KR = ^{12×2}/_{5} = ^{24}/_{5} = 4.8 m

⇒ RM = 2KR

⇒ RM = 2 × 4.8 = 9.6 m

Hence, distance between Reshma and Mandip is 9.6 m

**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.**

Solution:

Let Ankur, Syed and David be represented by A, S and D respectively (as shown on the figure)

Let PD = SP = SQ = QA = AR = RD = x

In ∆OPD,

OP^{2} = 400 – x^{2}

⇒ OP = √(400 – x^{2})

⇒ AP = 2√(400 – x^{2}) + √(400 – x^{2}) [∵ AP divides the median in the ratio 2 : 1]

= 3√(400 – x^{2})

In ∆APD,

PD^{2} = AD^{2} – DP^{2}

⇒ x^{2} = (2x)^{2} – (3√(400 – x^{2})^{2}

⇒ x^{2} = 4x^{2} – 9(400 – x^{2})

⇒ x^{2} = 4x^{2} – 3600 + 9x^{2}

⇒ 12x^{2} = 3600

⇒ x^{2} = ^{3600}/_{12} = 300

⇒ x = 10√3

Thus, SD = 2x = 2 × 10√3 = 20√3

∴ ASD is an equilateral triangle.

⇒ SD = AS = AD = 20√3

Hence, length of the string of each phone is 20√3 m

For next exercise – Circles – Exercise 10.5 – Class IX

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