
1.Number System
14
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11

Lecture1.12

Lecture1.13

Lecture1.14


2.Polynomials
10
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10


3.Coordinate Geometry
8
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8


4.Linear Equations
8
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7

Lecture4.8


5.Euclid's Geometry
7
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7


6.Lines and Angles
10
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8

Lecture6.9

Lecture6.10


7.Triangles
11
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9

Lecture7.10

Lecture7.11


8.Quadrilaterals
13
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11

Lecture8.12

Lecture8.13


9.Area of Parallelogram
11
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11


10.Constructions
7
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7


11.Circles
11
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7

Lecture11.8

Lecture11.9

Lecture11.10

Lecture11.11


12.Heron's Formula
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Surface Area and Volume
16
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9

Lecture13.10

Lecture13.11

Lecture13.12

Lecture13.13

Lecture13.14

Lecture13.15

Lecture13.16


14.Statistics
15
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11

Lecture14.12

Lecture14.13

Lecture14.14

Lecture14.15


15.Probability
8
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

NCERT Solutions – Circles Exercise
Exercise 10.1
Q.1 Fill in the blanks :
(i)The center of a circle lies in ______ of the circle.(exterior/interior)
(ii) A point , whose distance from the center of a circle is greater than its radius lies in _______of the
circle (exterior/interior)
(iii) The longest chord of a circle is a _________ of the circle.
(iv) An arc is a ______ when its ends are the ends of a diameter.
(v) Segment of a circle is the region between an arc and _____of the circle.
(vi) A circle divides the plane, on which it lies, in _______ parts.
Sol.
(i) interior
(ii) exterior
(iii) diameter.
(iv) semi circle
(v) chord
(vi) three parts.
Q.2 Write true or false : Give reasons for your answers :
(i) Line segment joining the centre to any point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal chords.
(iii) If a circle is divided into three equal arcs each is a major arc.
(iv) A chord of a circle, which is twice as long as its radius is a diameter of the circle.
(v) Sector is the region between the chord and its corresponding arc.
(vi) A circle is a plane figure.
Sol.
(i) True : As ‘radius’ is used in two sense – in the sense of a line segment and also in the sense of a length.
(ii) False : As an infinite number of points lie on the circle so we get infinite number of chord on joining any two points.
(iii) False : Since these arcs are equal in length so they are called equal arc.
(iv) True : As diameter is the longest chord and its length is twice the radius of the circle.
(v) False : As the region between an arc and the two radii, joining the center to end points of the are is known as a sector.
(vi) True : Since a circle is the locus of a point which moves in a plane in such a way that its distance from a given fixed point in the plane is always constant. So, a circle lies in a plane.
Exercise 10.2
Q.1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centers.
Sol.
Given : AB and CD are two equal chords of a circle with center at O.
To prove : ∠AOB=∠COD
Proof : In Δs AOB and COD, we have
AO = CO [Radii of the same circle]
BO = DO [Radii of the same circle]
and AB = CD [Given]
Therefore by SSS criterion of congruence, we have
ΔAOB≅ΔCOD
⇒ ∠AOB=∠COD [C.P.C.T.]
Q.2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Sol.
Given : AB and CD are two chords such that angles subtended by these chords at the centre of the circle are equal.
i.e. ∠AOB=∠COD
To prove : AB = CD
Proof : In Δs AOB and COD we have
AO = CO [Radii of the same circle]
BO = DO [Radii of the same circle]
and ∠AOB=∠COD
Therefore By SAS criterion of congruence, we have
ΔAOB≅ΔCOD
⇒ AB = CD [C.P.C.T.]
Exercise 10.3
Q.1 Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Sol.
On drawing a different pairs of circles, we find that they have two points in common. Now we shall prove that a pair of circle cannot intersect each other at more than two points.
Let the two circles cut each other at three points. But from the three points only one circle passes. Hence, two circles if intersect each other then they will intersect at two points. These two points are the ends of their common chords. This chord can be a common chord of a number of circles which will pass through the end points of this common chord.
Q.2 Suppose you are given a circle. Give a construction to find its centre.
Sol. Steps of Construction :
1. Take 3 points A, B and C on the circumference of the circle.
2. Join AB and BC.
3. Draw PQ and RS, the perpendicular bisectors of AB and BC, which intersect each other at O. Then,O is the centre of the circle.
Q.3 If two circles intersec at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Sol.
Given : Two circles, with centres O and O’ intersect, at two points A and B so that AB is the common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect AB at M.
To prove : OO’ is the perpendicular bisector of AB.
Construction : Draw line segment OA, OB, O’A and O’B.
Proof : Δs OAO’ and OBO’ we have
OA = OB [Radii of the same circle]
O’A = O’B [Radii of the same circle]
and OO’ = OO’
By SSS criterion of congruence, we have
ΔOAO′≅ΔOBO′
⇒ ∠AOO′=∠BOO′
⇒ ∠AOM=∠BOM … (1)
[Since ∠AOO′=∠AOMand∠BOM=∠BOO′]
In Δs AOM and BOM, we have
OA = OB [Radii of the same circle]
∠AOM=∠BOM [From (1)]
and OM = OM [Common]
Therefore by SAS criterion of congruence, we have
ΔAOM≅ΔBOM
⇒ AM = BM and ∠AMO=∠BMO
But ∠AMO+∠BMO=180∘
Therefore 2∠AMO=180∘⇒∠AMO=90∘
Thus , AM = BM and ∠AMO=∠BMO=90∘
Hence , OO’ is the perpendicular bisector of AB.
Exercise 10.4
Q.1 Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Sol.
Let O and O’ be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be their common chord.
We have OP = 5 cm, O’P = 3 cm and OO’ = 4 cm
Since OP2=PO′2+O′O2 [52=32+42]
⇒ OO’P is a right ∠dΔ, right angled as O’.
Area of ΔOO′P=12×O′P×OO′
=12×3×4
=6sq.units...(1)
Also ,
Area of ΔOO′P=12×OO′×PL
=12×4×PL=2PL … (2)
From (1) and (2), we have
2 × PL = 6 ⇒ PL = 3
We know that when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord i.e., OO’ is the perpendicular bisector of PQ.
Therefore PQ = 2 × PL = (2 × 3) cm = 6 cm
Q.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.
Sol.
Given : AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove : (i) AP = PD (ii) PB = CP.
Construction : Draw OM ⊥ AB , ON ⊥ CD.
Join OP,
AM = MB = 12 AB [Perpendicular from centre bisects the chord]
CN = ND = 12 CD [Perpendicular from centre bisects the chord]
AM = ND and MB = CN [Since AB = CD (given)]
In Δs OMP and ONP , we have
OM = ON [Equal chords of a circle are equidistant from the centre]
∠OMP=∠ONP [Since Each = 90º]
OP = OP [Common]
By RHS’ criterion of congruence,
ΔOMP≅ΔONP
⇒ MP = PN … (2) [C.P.C.T.]
Adding (1) and (2) , we have
AM + MP = ND + PN ⇒ AP = PD
Subtracting (2) from (1), we have
MB – MP = CN – PN ⇒ PB = CP
Hence (i) AP = PD and (ii) PB = CP
Q.3 If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with th chlords.
Sol.
Given : AB and CD are chords of a circle with centre O, AB and CD intersect at P and AB = CD.
To prove : ∠OPE=∠OPF
Construction : Draw OE ⊥ AB and OF ⊥ CD. Join OP.
In Δs OEP and OFP , we have
∠OEP=∠OFP [Since Each = 90º]
OP = OP [Common]
OE = OF [ Equal chords of a circle are equidistant from the centre]
Therefore by RHS criterion of congruence
ΔOEP≅ΔOFP
⇒ ∠OPE=∠OPF [C.P.C.T.]
Q.4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see figure)
Let OM be perpendicular from O on line l. We know that the perpendicular from the centre of a circle to a chord, bisects the chord.
Since BC is a chord of the smaller circle and OM ⊥ BC
Therefore BM = CM … (1)
Again AD is a chord of the larger circle and OM ⊥ AD.
Therefore AM = DM … (2)
Subtracting (1) from (2) we get
AM – BM = DM – CM ⇒ AB = CD.
Q.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?
Sol.
Let the three girls Reshma, Salma and Mandip are standing on the circle of radius 5 cm at points B, A and C respectively.
We know that if AB and AC are two equal chords of a circle, then the centre of the circles lies on the bisector of ∠BAC.
Here AB = AC = 6 cm. So the bisector of ∠BAC passes through the centre O i.e. OA is the bisector of ∠BAC.
Since the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio 6 : 6 = 1 : 1 i.e., M is the middle point of BC.
Now, M is the mid point of BC ⇒ OM ⊥ BC.
In right ∠dΔABM , we have
AB2=AM2+BM2
⇒ 36=AM2+BM2
⇒ BM2=36−AM2 … (1)
In the right Δ OBM, we have
OB2=OM2+BM2
⇒ 25=(OA−AM)2+BM2
⇒ BM2=25−(OA−AM)2
⇒ BM2=25−(5−AM)2 … (2)
From (1) and (2) , we get
36−AM2=25−(5−AM)2
⇒ 11−AM2+(5−AM)2=0
⇒ 11−AM2+25−10AM+AM2=0
⇒ 10AM=36
⇒ AM=3.6
Putting AM = 3.6 in (1) we get
BM2=3.6−(3.6)2=36−12.96
⇒ BM=36−12.96−−−−−−−−−√=23.04−−−−√=4.8cm
⇒ BC=2BM=2×4.8=9.6cm
Hence, the distance between Reshma and Mandip = 9.6 cm
Q.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 th e length of the string of each phone.
Sol.
Let ABC is an equilateral triangle of side 2x metres.
Clearly, BM=BC2=2x2=xmetres
In right ∠dΔABM
AM2=AB2−BM2
=(2x)2−x2=4x2−x2=3x2
⇒ AM=3–√x
Now , OM = AM – OA (3x−−√−20)m
In right ∠dΔOBM, we have
OB2=BM2+OM2
⇒ 202=x2+(3x−−√−20)2
⇒ 400=x2+3x2−403x−−√+400
⇒ 4x2−403x−−√=0
⇒ 4x(x−103–√)=0
Since x≠0 Therefore x−103–√=0
⇒ x=103–√
Now, BC=2BM=2x=203–√
Hence, the length of each string =203–√m
Exercise 10.5
Q. 1 In figure A, B and C are three points on a circle with centre O such that ∠BOC = 30º and ∠AOB = 60º. If D is a point on the circle other than the arc ABC, find ∠ADC.
Since are ABC makes ∠AOC=∠AOB+∠BOC = 60º + 30º = 90º at the centre of the circle and ∠ADC at a point on the remaining part of the circle.
Therefore ∠ADC=12(∠AOC)=12×90o=45o
Q. 2 A chord of a circle is equal to the radius of the circle. Find th eangle subtended by the chord at a point on the minor arc and also at a point on the major are.
Sol.
Let PQ be chord. Join OP and OQ.
It is given that PQ = OP = OQ (Since Chord = radius)
Therefore Δ OPQ is equilateral.
⇒ ∠POQ = 60ºSince are PBQ makes reflex POQ = 360º – 60º = 300º at centre of the circle and ∠PBQ at a point in the minor arc of the circle.
Therefore ∠PBQ=12(reflex∠POQ)
=12×300o=150o
Similarly, ∠PAQ=12(∠POQ)=12 (60º) = 30º
Hence, angle subtended by the chord on the minor are 150º and on the major chord = 30º.
Q.3 In figure ∠PQR = 100º, there P, Q and R are points on a circle with centre O. Find ∠OPR.
Sol.
Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same are at a point on the circumference.
Therefore Reflex ∠POR=2∠PQR
⇒ Reflex ∠POR=2×100∘=200∘
⇒ ∠POR = 360º – 200º = 160º
In Δ OPR, OP = OR [Radii of the same circle]
⇒ ∠OPR=∠ORP
[Angles opp. to equal sides are equal]
and ∠POR= 160º ……… (1) [Proved above]
In ΔOPR,
∠OPR+∠ORP+∠PQR=180∘ [Angle sum property]
⇒ 160∘�+2∠OPR=180∘
⇒ 2∠OPR=180∘�−160∘
⇒ 2∠OPR=20∘
⇒ ∠OPR=10∘
Q.4 In figure ∠ABC= 69º, ∠ACB= 31º, Find ∠BDC.
In Δ ABC,
∠BAC+∠ABC+∠BCA=180o
⇒ ∠BAC+69o+31o=180∘
⇒ ∠BAC=180o−(69o+31o)
=180o−100o=80o
Since angles in the same segment are equal
Therefore ∠BDC=∠BAC=80o
Q.5 In figure A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130º and ∠ECD = 20º. Find ∠BAC.
Sol.
∠CED+∠CEB = 180º [Linear pair]
⇒ ∠CED+ 130º = 180º
⇒ ∠CED= 180º – 130º
= 50º
In Δ ECD, ∠EDC+∠CED+∠ECD = 180º
⇒ ∠EDC+ 50º + 20º = 180º
⇒ ∠EDC= 180º – 50º – 20º
= 110º
⇒ ∠BDC=∠EDC = 110º
Since angles in the same segment are equal
Therefore ∠BAC=∠BDC= 110º.
Q.6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC= 70º, ∠BAC is 30º, find ∠BCD. Further, if AB = BC, find ∠ECD.
Sol.
∠BDC=∠BAC [Angles in the same segment]
⇒ ∠BDC= 30º [Since ∠BAC= 30º (given]
In Δ BCD, we have
∠BDC+∠DBC+∠BCD = 180º [Sum of ∠s of a Δ]
⇒ 30º + 70º + ∠BCD = 180º [Since ∠DBC = 70º, ∠BDC = 30º]
⇒ ∠BCD = 180º – 30º – 70º = 80º
If AB = BC, than ∠BCA=∠BAC = 30º [Angles opp. to equal sides in a Δ are equal]
Now , ∠ECD=∠BCD−∠BCE
= 80º – 30º = 50º
[Since ∠BCD= 80º (found above) and ∠BCE=∠BCA= 30º]
Hence, ∠BCD= 80º and ∠ECD= 50º
Q.7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Sol.
Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quad. ABCD.
To prove : Quadrilateral ABCD is a rectangle.
Solution : – Since all the radii of the same circle are equal
Therefore OA = OB = OC = OD
⇒ OA=OC=12AC
and OB=OD=12BD
⇒ AC = BD
Therefore the diagonals of the quadrilateral ABCD are equal and bisect each other.
⇒ Quadrilateral ABCD is a rectangle.
Q.8 If the non parallel sides of a trapezium are equal prove that it is cyclic.
Sol.
Given : Non – parallel sides AD and BC of a trapezium are equal.
To prove : ABCD is a cyclic trapezium.
Construction : Draw DE ⊥ AB and CF ⊥ AB.
Proof : In order to prove that ABCD is a cyclic trapezium it is sufficient to prove that ∠B+∠D = 180º.
In Δs DEA and CFB, we have
AD = BC [Given]
∠DEA=∠CFB [Each = 90º]
and DE = CF [Distance between two lines is always equal]
Therefore by RHS criterion of congruence, we have
DEA≅CFB
⇒ ∠A=∠Band∠ADE=∠BCF
(Corresponding parts of congruent triangles are equal)
Now, ∠ADE=∠BCF
⇒ 90º + ∠ADE= 90º + ∠BCF
⇒ ∠EDC+∠ADE=∠FCD+∠BCF
[Since ∠EDC= 90º and ∠FCD= 90º]
⇒ ∠ADC=∠BCD
⇒ ∠D=∠C
Thus , ∠A=∠Band∠C=∠D
Therefore , ∠A+∠B+∠C+∠D= 360º
[Since sum of the angles of a quad. is 360º]
⇒ 2∠B+2∠D= 360º
⇒ ∠B+∠D= = 180º
Hence, ABCD is a cyclic trapezium.
Q.9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that ∠ACP=∠QCD
Since angles in the same segment are equal.
Therefore ∠ACP=∠ABP … (1)
and ∠QCD=∠QBD … (2)
Also ∠ABP=∠QBD … (3) [Vertically opp. angles]
Therefore From (1), (2) and (3) , we have
∠ACP=∠QCD
Q.10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Sol.
Given : Two circles are drawn with sides AB and AC of Δ ABC as diameters. The circle intersect at D.
To prove : D lies on BC.
Construction : Join A and D.
Proof : Since AB and AC are the diameters of the two circles. [Given]
Therefore ∠ADB= 90º [Angles in a semi circle]
and, ∠ADC= 90º [Angles in a semicircle]
Adding we get ∠ADB+∠ADC = 90º + 90º = 180º
⇒ BDC is a straight line.
Hence, D lies on BC.
Q.11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD=∠CBD.
Sol.
Δ ABC and ADC are right ∠d with common hypotenuse AC. Draw a circle with AC as diameter passing through B and D. Join BD.
Clearly, ∠CAD=∠CBD [Since Angles in the same segment are equal]
Q.12 Prove that a cyclic parallelogram is a rectangle.
Sol.
Given : ABCD is a parallelogram inscribed in circle.
To prove : ABCD is a rectangle.
Proof : Since ABCD is a cyclic parallelogram.
Therefore ∠A+∠C = 180º … (1)
But ∠A=∠C … (2)
From (1) and (2), we have
∠A=∠C = 90º
Similarly, ∠B=∠D = 90º
Therefore Each angle of ABCD is of 90º
Hence, ABCD is a rectangle.
Exercise 10.6
Q.1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Sol.
Given : : Two circles with centres A and B, which intersect each other at C and D.
To prove : ∠ACB=∠ADB
Construction : Join AC, AD, BD and BC.
Proof : In Δs ACB and ADB, we have
AC = AD [Radii of the same circle]
BC = BD [Radii of the same circle]
AB = AB [Common]
Therefore by SSS criterion of congruence,
ΔACB≅ΔADB
⇒ ∠ACB=∠ADB [C.P.C.T.]
Q.2 Two chords AB and CD of lengths 5 cm and 11 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.
Sol.
Let O be the centre of the given circle and let its radius be r cm. Draw OP⊥AB and OQ⊥CD. Since OP⊥AB, OQ⊥CD and AB  CD. Therefore, points P, O and Q are collinear. So PQ = 6 cm.
Let OP = x. Then, OQ = (6 – x) cm.
Join OA and OC. Then, OA = OC = r.
Since the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore AP = PB = 2.5 cm and CQ = QD = 5.5 cm.
In right Δs QAP and OCQ, we have
OA2=OP2+AP2 and OC2=OQ2+CQ2
⇒ r2=x2+(2.5)2 …………… (1)
and r2=(6−x)2+(5.5)2 ………….. (2)
⇒ x2+(2.5)2=(6−x)2+(5.5)2
⇒ x2+6.25=36−12x+x2+30.25
⇒ 12x = 60 ⇒ x = 5
Putting x = 5 in (1), we get
r2=52+(2.5)2=25+6.25=31.25
⇒ r=31.25−−−−√ = 5.6 (approx.)
Hence, the radius of the circle is 5.6 cm (approx.)
Q.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.
Sol.
Let AB and CD be two parallel chords of a circle with centre O such that AB = 6 cm and CD = 8 cm.
Let the radius of the circle be r cm.
Draw OP⊥AB and OQ⊥CD. Since AB  CD and OP⊥AB, OQ⊥CD.
Therefore, points O, Q and P are collinear. Clearly OP = 4 cm, and P, Q are midpoints of AB and CD respectively.
Therefore AP=PB=12AB=3 cm
and, CQ=QD=12CD=4 cm
In rt. ∠d ΔOAP, we have
OA2=OP2+AP2
⇒ r2=42+32=16+9=25
⇒ r = 5
In rt. ∠d Δ OCQ, we have
OC2=OQ2+CQ2
⇒ r2=OQ2+42
⇒ 25=OQ2+16
⇒ OQ2=9
⇒ OQ = 3
Hence, the distance of chord CD from the centre is 3 cm.
Q.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.
Sol.
Since an exterior angle of a triangle is equal to the sum of the opposite angles.
In ΔBDC, we have
∠ADC=∠DBC+∠DCB ………… (1)
Since angle at the centre is twice the angle at a point on the remaining part of circle.
Therefore ∠ADC=12∠AOC
and ∠DCB=12∠DOE …………….. (2)
From (1) and (2), we have
12∠AOC=∠ABC+12∠DOE [Since ∠DBC=∠ABC]
⇒ ∠ABC=12(∠AOC−∠DOE)
Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.
Q.5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Sol.
Given : ABCD is a rhombus. AC and BD are its two diagonals which bisect each other at right angles.
To prove : A circle drawn on AB as diameter will pass through O.
Construction : From O draw PQ  AD and EF  AB.
Proof : AB = DC ⇒ 12AB=12DC
AQ = DP
[Since Q and P are midpoints of AB and CD]
Similarly AE = OQ
⇒ AQ = OQ = QB
⇒ A circle drawn with Q as centre and radius AQ passes through A, O and B.
The circle thus obtained is the required circle.
Q.6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Sol.
In order to prove that AE = AD
i.e., ΔAED is an isosceles triangle it is sufficient to prove that ∠AED=∠ADE. Since ABCE is a cyclic quadrilateral.
Therefore ∠AED+∠ABC=180∘ ……. (1)
Now, CDE is a straight line.
⇒ ∠ADE+∠ADC=180∘ ………… (2)
[Since ∠ADC and ∠ABC are opposite angles of a parallellogram i.e. ∠ADC=∠ABC
From (1) and (2), we get
∠AED+∠ABC=∠ADE+∠ABC
⇒ ∠AED=∠ADE
Therefore In ΔAED, we have
∠AED=∠ADE
⇒ AD = AE.
Q.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.
Sol.
(i) Let AB and CD be two chords of a circle with center O.
Let they bisect each other at O.
Join AC, BD, AD and BC.
In Δs AOC and BOD, we have
OA = OB [Since O is the midpoint of AB]
∠AOC=∠BOD [Vert. opp. ∠s]
and, OC = OD [Since O is the midpoint of CD]
By SAS criterion of congruence,
ΔAOC≅ΔBOD
⇒ AC = BD
⇒ arc(AC) = arc(BD) …………. (1)
Similarly, from Δs AOD and BOC, we have
arc(AD) = arc(BC)
From (1) and (2), we have
arc(AC) + arc(AD) = arc(BD) + arc(BC)
⇒ arc(CAD) = arc(CBD)
⇒ CD divides the circle into two parts
⇒ CD is a diameter.
Similarly, AB is a diameter.
(ii) Since ΔAOC≅ΔBOD [Proved above]
⇒ ∠OAC i.e., ∠BAC=∠OBD i.e., ∠ABD
⇒ AC  BD
Again, ΔAOD≅ΔCOB
⇒ AD  CB
⇒ ABCD is a cyclic parallelogram.
⇒ ∠DAC=∠DBA ………….. (3)
[Opp. ∠s of a parallelogram are equal]
Also, ACBD is a cyclic quadrilateral
Therefore ∠DAC+∠DBA=180∘ ……………. (4)
From (3) and (4), we get
∠DAC=∠DBA=90∘
Hence, ABCD is a rectangle.
Q.8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90∘−12A, 90∘−12B, 90∘−12C.
Sol.
We have ∠D = ∠EDF
= ∠EDA+∠ADF
= ∠EBA+∠FCA
[Since ∠EDA and ∠EBA are the angles in the same segment of the circle]
Therefore ∠EDA=∠EBA. Similarly, ∠ADF and ∠FCA are the angles in the same segment and hence, ∠ADF=∠FCA
=12∠B+12∠C
[Since BE is the internal bisector of ∠B and CF is the internal bisector ∠C
∠D=∠B+∠C2
Similarly, ∠E=∠C+∠A2
and ∠F=∠A+∠B2
⇒ ⇒∠D=180∘−∠A2
∠E=180∘−∠B2
and ∠F=180∘−∠C2
[Since ∠A+∠B=180∘]
⇒ ∠D=90∘−∠A2,
∠E=90∘−∠B2
and ∠F=90∘−∠C2
⇒ The angles of the ΔDEF are 90∘−12A, 90∘−12B, 90∘−12C.
Q.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.
Sol.
Let O and O’ be the centre of two congruent circles.
Since AB is a common chord of these circles.
Therefore arc(ACB) = arc(ADB)
∠BPA=∠BQA
⇒ BP = BQ.
Q.10 In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Given : ABC is a triangle inscribed in a circle with centre at O, E is a point on the circle such that AE is the internal bisector of ∠BAC and D is the midpoint of BC.
To Prove : DE is the right bisector of BC i.e. ∠BDE=∠CDE=90∘.
Construction : Join BE and EC.
Sol.
In ΔBDE and ΔCDE, we have
BE = CE
[Since ∠BAE=∠CAE, therefore arc(BE) = arc(CE) ⇒ chord BE = chord CE]
BD = CD [Given]
DE = DE [Common]
Therefore by SSS criterion of congruence,
ΔBDE≅ΔCDE
⇒ ∠BDE=∠CDE [C.P.C.T.]
Also, ∠BDE+∠CDE=180∘ [Linear pari]
Therefore ∠BDE=∠CDE=90∘
Hence, DE is the right bisector of BC.