## Circles

**Chapter 10**

**Exercise 10.4**

**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:**

PA = 5cm

PQ = 4cm

AQ = 3cm

Also, AB = 2AM

Let, MQ = x.

Consider the ΔAPM,

^{PA2 }= PM^{2}+AM^{2}

⇒ 5^{2 }= (4-x)^{2}+AM^{2}

⇒ 25 = 16+x^{2}-8x+AM^{2}

∴ AM^{2} = 9-x^{2}+8x — (i)

Now consider ΔAMQ,

AQ^{2 }= AM^{2}+MQ^{2}

⇒ 3^{2 }= AM^{2}+x^{2}

∴ AM^{2} = 9-x^{2} — (ii)

By equating equation (i) and equation (ii),

9 -x^{2}+8x = 9-x^{2}

⇒ 8x = 0

⇒ x = 0

Now, put the value of x in equation (i)

AM^{2} = 9-0^{2}

⇒ AM= 3cm

∴ The length of the cord i.e. AB = 2AM

So, AB = 2×3 = 6cm

**EX 10.4 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,

∠OME = ∠ONE [Each equal to 90°]

OM = ON [Proved]

OE = OE [Common hypotenuse]

∴ ∆OME ≅ ∆ONE [RHS congruence criteria]

⇒ ME = NE [C.P.C.T.]

Adding AM on both sides,

⇒ AM + ME = AM + NE

⇒ AE = DN + NE = DE

∵ AB = CD

⇒ AM = DN

⇒ AE = DE …(i)

Now, AB – AE = CD – DE

⇒ BE = CE …….(ii)

From (i) and (ii),

AE = DE and CE = BE

**EX 10.4 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 90°]

∴ ∆OME ≅ ∆ONE [ RHS congruence criteria]

⇒ ∠OEM = ∠OEN [C.P.C.T.]

⇒ ∠OEA = ∠OED

**EX 10.4 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).**

**Solution:**

Given: A line AB is intersecting two circles with the common centre O.

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 ⊥ AD.

Proof: For the inner circle,

OM ⊥ AD [By construction]

∴ BM = CM …(i) [Perpendicular from the centre to the chord bisects the chord]

For the outer circle,

OM ⊥ AD [By construction]

∴ AM = DM …(ii) [Perpendicular from the centre to the chord bisects the chord]

Subtracting (ii) from (i), we have

AM – BM = DM – CM

⇒ AB = CD

**EX 10.4 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?**

**Solution:**

**EX 10.4 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: Points A, S and D are the positions of Ankur, Syed and David.

∴ AS = SD = AD

Radius of circular park = 20m , hence AO = SO = DO = 20

Consturction: Draw AP ⊥ SD

Proof: Let AS = SD = AD = 2*x *cm

In ∆ASD,

AS = AD and AP ⊥ SD {By construction}

SP = PD = *x* cm