NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.4.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Circles |

Exercise |
Ex 11.4 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.4

**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:**

Let O and O’ be the centres of the circles of radii 5 cm and 3 cm, respectively.

Let AB be their common chord.

Given, OA = 5 cm, O’A = 3cm and OO’ = 4 cm

∴ AO’^{2} + OO’^{2} = 3^{2} + 4^{2} = 9 + 16- 25 = OA^{2}

∴ OO’A is a right angled triangle and right angled at O’

Area of ∆OO’A = \(\frac { 1 }{ 2 }\) x O’A x OO’

= \(\frac { 1 }{ 2 }\) x 3x 4= 6sq units …(i)

Also, area of ∆OO’A = \(\frac { 1 }{ 2 }\) x OO’ x AM

= \(\frac { 1 }{ 2 }\) x 4 x AM =2 AM …(ii)

From Eqs. (i) and (ii), we get

2AM = 6 ⇒ AM = 3

Since, when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord.

∴ AB = 2 x AM= 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 segments of the other chord.**

**Solution:**

**Given:** MN and AS are two chords of a circle with centre O, AS and MN intersect at P and MN = AB

**To prove:** MP = PB and PN = AP

**Construction:** Draw OD ⊥ MN and OC ⊥ AB.

Join OP

**Proof :** ∵ DM = DN = \(\frac { 1 }{ 2 }\) MN (Perpendicular from centre bisects the chord)

and AC = CB = \(\frac { 1 }{ 2 }\) AB (Perpendicular from centre bisects the chord)

MD = BC and DN = AC (∵ MN = AS)…(i)

in ∆ODP and ∆OPC

OD = OC (Equal chords of a circle are equidistant from the centre)

∠ ODP = ∠OCP

OP = OP (Common)

∴ RHS criterion of congruence,

∆ ODP ≅ ∆ OCP

∴ DP = PC (By CPCT)…(ii)

On adding Eqs. (i) and (ii), we get

MD + DP = BC + PC

MP = PB

On subtracting Eq. (ii) from Eq. (i), we get

DN – DP = AC – PC

PN = AP

Hence, MP = PB and PN = AP are proved.

**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:** RQ and MN are chords of a with centre O. MN and RQ intersect at P and MN = RQ

**To prove:** ∠ OPC = ∠ OPB

**Construction:** Draw OC ⊥ RQ and OB ⊥ MN.

Join OP.

**Proof:** In ∆ OCP and ∆ OBP, we get

∠ OCP = ∠ OBP (Each = 90°)

OP = OP (Common)

OC = OB (Equal chords of a circle are equidistant from the centre)

∴ By RHS criterion of congruence, we get

∆ OCP ≅ ∆ OBP

∴ ∠ OPC = ∠ OPB (By CPCT)

**Question 4.**

**If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).**

**Solution:**

Let OP be the perpendicular from O on line l. Since, the perpendicular from the centre of a circle to a chord

Now, BC is the chord of the smaller circle and OP ⊥ BC.

∴ BP = PC ……(i)

Since, AD is a chord of the larger circle and OP ⊥ AD.

∴ AP = PD …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

AP – BP = PD – PC

⇒ AB = CD

Hence proved.

**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 6 m each, what is the distance between Reshma and Mandip?**

**Solution:**

Let O be the centre of the circle and Reshma, Salma and Mandip are represented by the points Ft, S and M, respectively.

Let RP = xm.

From Eqs. (i) and (ii), we get

Hence, the 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:**

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = x

Therefore, ∆ PQR is an equilateral triangle. Drawn altitudes PC, QD and RN from vertices to the sides of a triangle and intersect these altitudes at the centre of a circle M.

As PQR is an equilateral, therefore these altitudes bisects their sides.

In ∆ PQC,

PQ^{2} = PC^{2} + QC^{2} (By Pythagoras theorem)

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