NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4.

- Circles Class 9 Ex 10.1
- Circles Class 9 Ex 10.2
- Circles Class 9 Ex 10.3
- Circles Class 9 Ex 10.4
- Circles Class 9 Ex 10.5
- Circles Class 9 Ex 10.6

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Quadrilaterals |

Exercise |
Ex 10.4 |

Number of Questions Solved |
6 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.4

Ex 10.4 Class 9 Maths Question 1.

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Solution.

Let O and O’ be the centers of the circles of radii 5 cm and 3 cm, respectively. Let AB be their common chord

Given, OA = 5 cm, O’ A = 3 cm and 00′ = 4 cm

Then, AO’^{2} + OO’^{2} = 3^{2} + 4^{2} = 9 +16 = 25 = OA^{2
}So, 00′ A is a right angled triangles and right angled at O’.

∴ Area of Δ OO’ A = \(\cfrac { 1 }{ 2 } \) x O’A x OO’ = \(\cfrac { 1 }{ 2 } \) x 3 x 4 = 6 cm^{2} …(i)

Also, we know that when two circles intersect at two points, then their centers lie on the perpendicular bisecto**r** of the common chord.

∴ Area of Δ OO’ A = \(\cfrac { 1 }{ 2 } \) x OO’ x AM = \(\cfrac { 1 }{ 2 } \) x 4 x AM=2AM ………(ii)

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

2AM = 6 ⇒ AM =3 cm

∴ AB = 2 x AM = 2 x 3 = 6 cm

Ex 10.4 Class 9 Maths 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.

Let MN and AB are two chords of a circle with center 0. AB and MN intersect at P and MN = AB

To prove : MP = PB and PN = AP

Construction: Draw OD ⊥ L MN and OC ⊥ AB. Join OP.

Ex 10.4 Class 9 Maths Question 3.

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the center makes equal angles with the chords.

Solution.

Let RQ and MN be two equal chords of a circle with center O.

MN and RQ intersect at P and MN = RQ.

To prove : ∠OPM = ∠OPQ

Ex 10.4 Class 9 Maths Question 4.

If a line intersects two concentric circles (circles with the same center) with center O at A, B, C and D, then prove that AB = CD (See figure).

Solution.

Let a line l intersects two concentric circles with center O at A, B, C and D.

To prove : AB = CD

Construction: Draw OP perpendicular from O on line l.

Proof: We know that the perpendicular drawn from the center of a circle to a chord bisects the chord. Here, BC is the chord of the smaller circle and OP ⊥ BC.

∴ BP = PC

and 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.**

Ex 10.4 Class 9 Maths 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 0 be the center of the circle and Reshma,

Salma and Mandip are represented by the points R, S, and M, respectively.

Draw OP ⊥ RM and ON ⊥ RS.

Ex 10.4 Class 9 Maths 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 be sitting on the points P, Q, and R, respectively on the boundary of circular park.

Clearly, PQ = QR = PR, as they sitting at an equal distance.

Thus, ΔPQR is an equilateral triangle.

Let PQ = QR = PR = x m

Now, draw altitudes PC, QD and RN from vertices to .the sides of a triangle and these altitudes intersect at the center of circle M.

[∵ altitudes of equilateral triangle passes through the circumcenter of the equilateral triangle.]

As Δ PQR is an equilateral triangle, therefore these altitudes bisect their sides.

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