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|Exercise||Class 9 Chapter 11 Circles Exercise 11.6|
|Number of Questions Solved||10|
NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.6
NCERT TEXTBOOK EXERCISES
Question 1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Given: Two circles with centres 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)
∴ By SSS criterion, we get
∆ OCO’ ≅ ∆ ODO’
Hence, ∠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 centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Let O be the centre of the given circle and let its radius be cm.
Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.
Let ON = a cm
∴ OM = (6 – a) cm
Join OA and OC.
Then, OA = OC = b c m
Since, the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm
In ∆OAN and ∆OCM, we get
OA2 = ON2 + AN2
OC2 = OM2 + CM2
⇒ b2 = a2 + (2.5)2
and, b2 = (6-a)2 + (5.5)2 …(i)
So, a2 + (2.5)2 = (6 – a)2 + (5.5)2
⇒ a2 + 6.25= 36-12a + a2 + 30.25
⇒ 12a = 60
⇒ a = 5
On putting a = 5 in Eq. (i), we get
b2 = (5)2 + (2.5)2
= 25 + 6.25 = 31.25
So, r = √31.25= 5.6cm (Approx.)
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 centre, what is the distance of the other chord from the centre ?
Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of circle.
Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.
∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.
PM = MQ = 1/2 PQ = 6/2 = 3 cm
and RN = NS = 1/2 RS = 8/2 = 4 cm
In ∆OPM, we have
OP2 = OM2 + PM2
⇒ a2 =42 + 32 = 16 + 9 = 25
⇒ a = 5
In ∆ORN, we have
⇒ OR2 = ON2 + RN2
⇒ a2 = ON2 + (4)2
⇒ 25 = ON2 + 16
⇒ ON2 = 9
⇒ ON = 3cm
Hence, the distance of the chord PS from the centre is 3 cm.
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 the difference of the angles subtended by the chords AC and DE at the centre.
Since an exterior angle of a triangle is equal to the sum of the interior opposite angles.
∴ In ∆BDC, we get
∠ADC = ∠DBC + ∠DCB …(i)
Since, angle at the centre is twice at a point on the remaining part of circle.
∴ ∠DCE = 1/2 ∠DOE
⇒ ∠DCB = 1/2 ∠DOE (∵ ∠DCE = ∠DCB)
∠ADC = 1/2 ∠AOC
∴ 1/2 ∠AOC = ∠ABC + 1/2 ∠DOE (∵ ∠DBC = ∠ABC)
∴ ∠ABC = 1/2 (∠AOC – ∠DOE)
Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.
Question 5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Given: PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.
To prove: A circle drawn on PQ as diameter will pass through O.
Construction: Through O, draw MN || PS and EF || PQ.
Proof : ∵ PQ = SR ⇒ 1/2 PQ = 1/2 SR
So, PN = SM
Similarly, PE = ON
So, PN = ON = NQ
Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.
Question 6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Since, ABCE is a cyclic quadrilateral, therefore
∠AED+ ∠ABC= 180°
(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)
∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)
So, ∠ADE + ∠ABC = 180°
(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)
From Eqs. (i) and (ii), we get
∠AED + ∠ABC = ∠ADE + ∠ABC
⇒ ∠AED = ∠ADE
∴ In ∆AED We have
∠AED = ∠ADE
So, AD = AE
(∵ Sides opposite to equal angles of a triangle are equal)
Question 7. AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
(i) Let BD and AC be two chords of a circle bisect at P.
In ∆APB and ∆CPD, we get
PA = PC ( ∵ P is the mid-point of AC)
∠APB = ∠CPD (Vertically opposite angles)
and PB = PD (∵ P is the mid-point of BD)
∴ By SAS criterion
∆CPD ≅ ∆APB
∴ CD= AB (By CPCT) …(I)
∴ BD divides the circle into two equal parts. So, BD is a diameter.
Similarly, AC is a diameter.
(ii) Now, BD and AC bisect each other.
So, ABCD is a parallelogram.
Also, AC = BD
∴ ABCD is a rectangle.
Question 8. Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – 1/2 A, 90° – 1/2 B and 90° – 1/2 C.
∵ ∠EDF = ∠EDA + ∠ADF
∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.
∴ ∠EDA = ∠EBA
and similarly ∠ADF and ∠FCA are the angles in the same segment and hence
Question 9. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Let O’ and O be the centres of two congruent circles.
Since, AB is a common chord of these circles.
∴ ∠BPA = ∠BQA
(∵ Angle subtended by equal chords are equal)
⇒ BP = BQ
Question 10. In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.
(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.
Join BM and CM.
∴ ∠MBC = ∠MAC (Angles in same segment)
and ∠BCM = ∠BAM (Angles in same segment)
But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)
∴ ∠MBC = ∠BCM
So, MB = MC (Sides opposite to equal angles are equal)
So, M must lie on the perpendicular bisector of BC
(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.
Since, M lies on perpendicular bisector of BC.
∴ BM = CM
∠MBC = ∠MCB
But ∠MBC = ∠MAC (Angles in same segment)
and ∠MCB = ∠BAM (Angles in same segment)
So, from Eq. (i),
∠BAM = ∠CAM
AM is the bisector of A.
NCERT Solutions for Class 9 Maths Chapter 11 Circles Exercise 11.6 PDF
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