## MP Board Class 9th Maths Solutions Chapter 10 Circles Ex 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.

Solution:

Given

Let O and O_{1} be the centre of bigger and smaller circle respectively

OA = OB = 5 cm

O_{1}A – O_{1}B = 3 cm

OO_{1} = 4 cm

To find: AB.

Construction:

Join OA, OB, O_{1}A and O_{1}B join AB also.

In ∆OAO_{1} and ∆OBO_{1}

OA = OB (Radii of a circle)

O_{1}A = O_{1}B (Radii of a circle)

OO_{1} = OO_{1} (Common)

so ∆OAO_{1} = ∆OBO_{1} (By SSS)

and so ∠1 = ∠2 (By CPCT)

In ∆OCA and ∆OCB,

OA = OB (Radii of a circle)

∠1 = ∠2 (Proved)

OC = OC (Common)

∆OCA = ∆OCB (By SAS)

so AC = BC (By CPCT)

and ∠ACO = ∠BCO (By CPCT)

∠ACO + ∠BCO = 180° (LPA’s)

⇒ ∠ACO + ∠BCO = 180°

2∠ACO = 180°

∠ACO = 90°

ar (OAO_{1}) = \(\frac{1}{2}\) x OO_{1} x AC

= \(\frac{1}{2}\) x 4 x AC = 2ACcm^{2} …..(i)

In ∆QAO_{1}, a = 5 cm, bc = 4 cm, c = 3 cm

12

s = \(\frac{5+4+3}{2}\) = \(\frac{12}{2}\) = 6 cm

s – a = 6 – 5 = 1 cm

s – b = 6 – 4 = 2 cm

s – c = b – 3 = 3 cm

ar(OAO_{1}) = \(\sqrt{s(s-a)(s -b)(s- c)}\).

= \(\sqrt{6x1x2x3}\)

= 6 cm^{2} …..(ii)

From (i) and (ii), we get

2AC =6

AC = 3 cm

Now, AB = 2AC (∴ AC = BC)

= 2 x 3 = 6 cm.

Method II. By Construction:

Geometrically, AB is the diameter of the circle of radius 3 cm as it passes through centre O_{1}

AB = 2 x 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 segment of the other chord.

Solution:

Given

C (O, r) is a circle in which AB and CD are two equal chords which intersect at P.

To prove:

CP = BP and AP = DP.

Construction:

Draw OE and OF perpendiculars on AB and CD respectively. Join OP.

Proof:

In ∆OPF and ∆OPE,

OP = OP (Common)

OE = OF (∴ AB = CD)

∠F = ∠E (Each 90°)

∆OPF = ∆OPE (By RHS)

and so PE = PF …..(1) (By CPCT)

AB = CD (Given)

\(\frac{1}{2}\) AB = \(\frac{1}{2}\) CD

BE = CF

and AE = DF …..(2)

Adding (1) and (2), we get,

PE + AE = PF + DF

∴ AP = DP

Subtracting (1) and (2) we get,

BE – PE = CF – PF

∴ BP = CP

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

AB and CD are two equal chords of a circle which intersect at E.

To prove:

∠1 = ∠2

Construction:

Draw OL ⊥ AB and OM ⊥ CD. Join OE.

Proof:

In ∆OLE and ∆OME,

OE = OE

OL = OM (∴ AB = CD)

∠L = ∠M (Each 90°)

∆OLE = ∆OME (By RHS)

and so∠1 = ∠2 (By 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. adjacent)

Solution:

Given:

C (O, r) and C (O, r) are two concentric circles.

To prove: AB = CD

Construction: Draw OP ⊥ AD.

Proof:

In circle I, AD is the chord and OP ⊥ AD.

AP = DP …(1)

In circle II, BC is the Chord and OP L BC.

∴ BP = CP …(2)

Subtracting (1) and (2), we get

AP – BP = DP – CP

AB = CD

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:

Given:

OR = OM = 5 m and SR = SM = 6 m.

To find: MR.

Constrution:

Join OR, OM and OS. Draw ON ⊥ SR. In AORS and AOMS,

OS = OS (Common)

RS = MS (Given)

OR = OM (Given)

∆ORS = ∆OMS (By SSS)

and ∠1 = ∠2 (By CPCT)

SP = SP (Common)

SR = SM (Given)

∠1 = ∠2 (Proved)

∆SPR = ∆SPM (By SAS)

and so PR = PM (By CPCT)

and ∠3 = ∠4 (By CPCT)

∠3 + ∠4 = 180° (LPA’s)

2∠3 = 180°

∠3 = \(\frac{180^{\circ}}{2}\) = 90°

ar (∆OSR) = \(\frac{1}{2}\) x OS x PR …(i)

= \(\frac{1}{2}\) x 5 x PR

ON ⊥ SR

RN = \(\frac{1}{2}\) SR

(Perpendicular drawn from the centre of a circle to a chord bisects the chord)

= \(\frac{6}{2}\) = 3m

In ∆ONR ON^{2} = \(\sqrt{O R^{2}-N R^{2}}\) (Using Pythagoras Theorem)

= \(\sqrt{5^{2}-3^{2}}\) = \(\sqrt{4^{2}}\) = 4m

ar (∆OSR) = \(\frac{1}{2}\) x SR x ON

= \(\frac{1}{2}\) x 6 x \(\frac{1}{2}\) x 4 = 12m^{2} …..(ii)

From (i) and (ii), we get

PR = \(\frac{2×12}{2}\) = 4.8m

MR = 2 PR

= 2 x 4.8

= 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:

Given: OS = OA = 20 m and AS = SD = AD

To find: AS, SD and AD.

Construction:

Draw AE ⊥ SD. Join OS.

Let AS = SD = AD = 2x (say)

In equilateral ∠ASD, AE ⊥ SD

⇒ E is the mid-point of SD

SE = \(\frac{2x}{2}\) = x

In ∆AES, AS^{2} = AE^{2} + SE^{2}

(2x)^{2} = AE^{2} + x^{2}

4x^{2} – x^{2} = AE^{2}

AE = \(\sqrt{3x^{2}}\) = \(\sqrt{3}\)

OE = AE – AO

= (\(\sqrt{3}\) – 20)m

In ∆OES, OS^{2} = OE^{2} + SE^{2}

(20)^{2} = [(\(\sqrt{3}\)x) 20]^{2} + x^{2}

400 = (\(\sqrt{3}\)x)^{2} – 2 x \(\sqrt{3}\)x × 20 + (20)^{2} + x^{2}

= 3x^{2} + 400 – 40\(\sqrt{3}\)x + x^{2}

400 – 400 = 4x^{2} – 40\(\sqrt{3}\)x

0 = 4x^{2} – 40\(\sqrt{3}\)x

40\(\sqrt{3}\)x = 4x^{2}

40\(\frac { 40\sqrt { 3 } }{ 4 } \) = x

x = 10\(\sqrt{3}\)m .

2x = 2 x 10\(\sqrt{3}\) = 20\(\sqrt{3}\)m

AS = SD = AD = 20\(\sqrt{3}\)m.