The document Ex 10.4 NCERT Solutions- Circles Class 9 Notes | EduRev is a part of the Class 9 Course Class 9 Mathematics by Full Circle.

All you need of Class 9 at this link: Class 9

**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:** 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 the common chord.

∴ ∠OLP = ∠OLP = 90º

PL = PQ

Now in right ΔOLP,

PL^{2} + OL^{2 }= OP^{2}

⇒ PL^{2} + (4 - x)^{2} = 5^{2}

⇒ PL^{2 }= 5^{2} - (4 - x)^{2}

⇒ PL^{2 }= 25 - 16 - x^{2} + 8x

⇒ PL^{2} = 9 - x^{2 }+ 8x ...(1)

Again, in ΔO'LP,

PL^{2 }= 32 - x^{2} = 9 - x^{2} ...(2)

From (1) and (2),

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

⇒ 8x = 0

⇒ x= 0

⇒ L and O' coincide.

∴ PQ is a diameter of the smaller circle.

⇒ PL = 3 cm

But PL = LQ

∴ LQ = 3 cm

∴ PQ = PL + LQ

= 3 cm + 3 cm = 6 cm

Thus, the required length of the common chord = 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. 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 equidistant from the centre]

Now, in right ΔOME and right ΔONE,

OM = ON [Proved]

OE = OE [Common]

∴ ΔOME ≌ ΔONE [RHS criterion]

⇒ ME = NE [c.p.c.t.]

⇒ (1/2) AE = (1/2) DE [Perpendiculars from the centre bisect the chord.]

⇒ 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

**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:** We have a circle with centre O and equal chords AB and CD are intersecting at E. OE is joined.

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

**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 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 = M D [∵ Perpendicular from the centre 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

**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: **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 = 180º [Linear pairs]

∴ ∠OPR = ∠OPM = 90º

⇒ OP ⊥ RM

Now, in ΔRSP and ΔMSP, RS = MS (6 m each)

SP = SP (Common)

∠RSP = ∠MSP

∴ ΔRSP ≌ ΔMSP [SAS criterion]

⇒ RP = MP and ∠RPS = ∠MPS [c.p.c.t]

But ∠RPS + ∠MPS = 180°

⇒ ∠RPS = ∠MPS = 90º (Each)

∴ SP passes through O.

Let OP = x m

∴ SP = (5 - x) m

Now, in right ΔOPR, x^{2} + RP^{2} = 5^{2} …(1)

In right ΔSPR, (5 - x)^{2} + RP^{2} = 6^{2} …(2)

From (1),

RP^{2} = 5^{2} - x^{2}

From (2),

RP^{2} = 6^{2} - (5 - x)^{2}

∴ 5^{2} - x^{2} = 6^{2 }- (5 - x)^{2}

⇒ 25 - x^{2} = 36 - [25 - 10x + x^{2}]

⇒ 25 - x^{2} - 36 + 25 - 10x + x^{2} = 0

⇒ -10x + 14 = 0

⇒ 10x = 14

⇒ x= (14/10) = 1.4

Now, RP^{2} = 5^{2} - x^{2 }

⇒ RP^{2} = 25 - (1.4)^{2}

⇒ RP^{2} = 25 - 1.96 = 23.04 m

∴ RP =√(23.04) = 4.8 m

∴ RM = 2 RP = 2 x 4.8 m = 9.6 m

Thus, distance between Reshma and Mandip is 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:** In the figure, let Ankur, Syed and David are sitting at A, S and D respectively such that AS = SD = AD

i.e.ΔASD is an equilateral triangle.

Let the length of each side of the equilateral triangle is 2x metres.

Let us draw AM ⊥ SD.

Since ΔASD is an equilateral,

∴ AM passes through O.

⇒ SM = (1/2) SD = (1/2) (2x)

⇒ SM = x

Now, in right ΔASM, AM^{2} + SM^{2} = AS^{2}

⇒ AM^{2 }= AS^{2} - SM^{2}

= (2x)^{2 }-x^{2} = 4x^{2} - x2 = 3x^{2}

⇒ AM = 3x

Now, OM = AM - OA = ( 3x - 20) m

Again, in right ΔOBM, we have

OS^{2 }= SM^{2} + OM^{2}

⇒ 20^{2} = x^{2} + (3x - 20)^{2}

⇒ 400 = x^{2} + 3x^{2} - 40^{3} x + 400

⇒ 4x^{2} - 40^{3} x= 0

⇒ 4x (x - 10^{3}) = 0

⇒ x = 0 or x = 10^{3}

But x = 0 is not required.

∴ x= 10^{3} m

Now, SD = 2x m = 2 x 10^{3} m

= 20^{3} m.

Thus, the length of the string of each phone = 20^{3} m.

**ANGLE SUBTENDED BY AN ARC OF A CIRCLE**

**REMEMBER**

- Equal chords make congruent arcs and congruent arcs make equal chords of a circle.
- If two chords of a circle are equal, then they subtend equal angles at the centre.
- The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
- Angles in the same segment of a circle are equal. Angle in a semicircle is a right angle.
- If a line segment joining two points subtend equal angles at two other points lying on the same side of the line containing the line segment the four points lie on a circle (i.e. they are concyclic).
- A quadrilateral is called cyclic, if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
- If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

132 docs

### Long Answer Type Questions- Circles

- Doc | 1 pages
### Ex 10.5 NCERT Solutions- Circles

- Doc | 7 pages
### Ex 10.6 NCERT Solutions- Circles

- Doc | 8 pages
### Value Based Questions- Circles

- Doc | 1 pages

- Short Answer Type Questions- Circles
- Doc | 4 pages
- Ex 10.3 NCERT Solutions- Circles
- Doc | 2 pages