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NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.1
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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.2
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
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.3
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.4

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.5
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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.6

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.7

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.8

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.9
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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.10

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.11

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
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.12

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.13

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.14

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.
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.15
NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6.16

We hope the NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6, drop a comment below and we will get back to you at the earliest.

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