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

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]

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

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.

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.

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.

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.

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

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° – A, 90° – B and 90° – C.

Solution.

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.

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.

(i) Let bisector of ∠A meet the circumcircle of Δ ABC at M.

Join BM and CM.

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