NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.

- 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 10.6 |

Number of Questions Solved |
10 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Ex 10.6 Class 9 Maths Question 1.

Prove that the line of centers of two intersecting circles subtends equal angles at the two points of intersection.

Solution.

Given Two circles with centers O and O’ which intersect each other at C and D.

To prove: ∠OCO’ = ∠ODO’

Construction: Join OC, OD, O’C and O’ D.

Proof: In ∠OCO’ and ∠ODO’, we have

OC = OD [radii of the same circle]

O’C = O’D [radii of the same circle]

OO’ = OO’ [common sides]

∴ Δ OCO’ = Δ ODO’ [by SSS congruence rule]

Then, ∠OCO’ = ∠ODO’ [by CPCT]

Ex 10.6 Class 9 Maths Question 2.

Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its center. If the distance between AB and CD is 6 cm, then find the radius of the circle.

Solution.

Let 0 be the center of the given circle and its radius be b cm.

AB = 5 cm and CD = 11 cm are two chords on opposite sides of its center.

Draw ON ⊥ AB and OM ⊥ CD.

Since, ON ⊥ AB, OM ⊥ CD, and AB||CD, therefore the points N, O and M are collinear.

Given, distance between AB and CD = MN = 6 cm

Let ON=a cm

OM=(6-a) cm

Ex 10.6 Class 9 Maths Question 3.

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the center, what is the distance of the other chord from the center?

Solution.

Let PQ and RS be two parallel chords of a circle with center O such that PQ = 6 cm and RS = 8 cm.

Let a be the radius of the circle. Draw ON ⊥ RS and OM ⊥ PQ.

Since, PQ || RS, and ON ⊥ RS and OM ⊥ PQ.

Therefore, points O, N, and M are collinear. v OM = 4 cm and M and N are the mid-points of PQ and RS, respectively.

Ex 10.6 Class 9 Maths Question 4.

Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half of the difference of the angles subtended by the chords AC and DE at the center.

Solution.

Ex 10.6 Class 9 Maths Question 5.

Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Solution.

Given: Let PQRS be a rhombus and PR and SQ are its two diagonals which bisect each other at right angles at 0.

To prove: A circle drawn on PQ as diameter will pass through O.

Construction: Through O, draw MN || PS and EF || PQ.

Ex 10.6 Class 9 Maths Question 6.

ABCD is a parallelogram. The circle through A, B, and C intersect CD (produced, if necessary) at E. Prove that AE = AD.

Solution.

Ex 10.6 Class 9 Maths Question 7.

AC and BD are chords of a circle which bisects each other. Prove that

(i) AC and BD are diameters,

(ii) ABCD is a rectangle.

Solution.

(i) Let BD and AC be two chords of a circle bisect at P.

In Δ ASP and Δ CPD, we have PA = PC

Ex 10.6 Class 9 Maths Question 8.

Bisectors of angles A, B and C of the triangle ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the Δ DEF are 90° – \(\cfrac { 1 }{ 2 } \) A, 90° – \(\cfrac { 1 }{ 2 } \)B and 90° – \(\cfrac { 1 }{ 2 } \)C.

Solution.

Ex 10.6 Class 9 Maths Question 9.

Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn, so that P and Q lie on the two circles. Prove that BP = BQ.

Solution.

Ex 10.6 Class 9 Maths Question 10.

In any Δ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the Δ ABC.

Solution.

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