ML Aggarwal Solutions Class 9 Math 15th Chapter Circle Exercise 15.1
ML Aggarwal Understanding ICSE Mathematics Class 9 Solutions Fifteenth Chapter Circle Exercise 15.1. APC Solution Class 9 Exercise 15.1.
Exercise 15.1
(1) Calculate the length of a chord which is at a distance 12cm from the centre of a circle of radius 13cm.
Solution:
Given, radius of circle DA = 13cm
Distance of circle from chord AB = 12cm
∴ In △ADC
∠ACD = 90° [OC ⊥ AB; property of circle chord]
By Pythagoras theorem,
∴ AO2 = AC2 + OC2
Or, 132 = AC2 + 122
Or, AC = √169-144
Or, AC = √25 = 5cm
∴ AB = AC × 2 = 5 × 2 = 10cm
[∵ OC⊥AB bisect the chord]
(2) A chord of length 48cm is drawn in a circle of radius 25cm. Calculate its distance from the centre of the circle.
Solution:
Given, length of chord AB = 48cm
Radius OA = 25cm
Draw : OC ⊥ AB
In △AOC, by Pythagoras theorem,
AD2 = AC2 + OC2
Or, 252 = (AB/2)2 + OC2
Or, OC2 = 252 – (48/2)2
Or, OC = √625 – 576
Or, OC = √49 = 7cm
[∵ Perpendicular drawn from centre to chord bisects the chord]
∴ Distance of chord AB from circle is 7cm.
(3) A chord of length 8cm is at a distance 3cm from the centre of the circle. Calculate the radius of the circle.
Solution:
Given, length of chord AB = 8cm
Distance from circle OC = 3cm
Here, OC ⊥ AB
In △AOC, by Pythagoras theorem,
AO2 = AC2 + DC2
Or, AO2 = (8/2)2 + 32 [∵ OC bisect AB, OC ⊥ AB, chord is bisect]
Or, AO = √16+9 = √25 = 5a
∴ Radius of circle = AO = 5cm
(4) Calculate the length of a chord which is at a distance 6cm from the centre of a circle of diameter 20cm.
Solution:
Given, distance from centre OC = 6cm
Diameter of circle = 20cm
∴ Radius OA of circle = 20/2 = 10cm
In △AOC, ∠ACO = 90 °, AC = CB
[1 from center to chord bisects the chord]
∴ By Pythagoras theorem,
OA2 = OC2 + AC2
Or, 102 = 62 + AC2
Or, AC = √100 – 36 = √64 = 8
∴ Length of chord
AB = 2×AC = 8×2
= 16cm
(5) A chord of length 16 cm is at a distance 6cm from the centre of the circle. Find the length of the chord of the same circle which is at a distance 8cm from the centre.
Solution:
Given, length of chord AB = 16cm
Distance of chord AB from centre OC
= 6cm
∴ In △ACO, ∠ACO = 90°
AC = 1/2 × AB [∵ OC from centre bisects the chord AB]
Or, AC = 16/2 = 8cm
∴ By Pythagoras theorem,
AO2 = OC2 + AC2
Or, AO = √62 + 82 = √36+64 = √100 = 10cm
∴ Radius of circle AO = PO = 10cm
Now, distance of chord PQ from centre x = 8cm
∴ In △POR, ∠PRO = 90°, [1 or tan circle]
∴ By Pythagoras theorem,
PR2 + OR2 = PO2
Or, PR2 = 102 – 82
Or, PR = √100 – 64 = √36 = 6cm
∴ Length of chord PQ
= 2×PR [⊥ or bisect the chord]
= 2×6 = 12 cm
(6) In a circle of radius 5cm, AB and CD are two parallel chords of length 8cm and 6cm respectively. Calculate the distance between the chords, if they are on
(i) The same side of the centre
(ii) The opposite sides of the centre
Solution:
(i) Given, length of chord AB = 8cm
Length of chord CD = 6cm
AB||CD
∴ O – P – Q is a triangle line
∵ OP ⊥ AB & OQ ⊥ CD
OA = OC = 5cm [radius]
Now, In △APO, ∠APO = 90°, AB/2 = 8/2 = 4 [∵⊥ OP from centre bisect the chord]
∴ By Pythagoras theorem,
AO2 = AP2 + OP2
Or, 52 = 42 + OP2
Or OP = √25 – 16, √9 = 3cm
In △COQ, ∠CQO = 90°, CQ = CD/2 [∵⊥ OQ bisect chord CD]
= 6/2 = 3
∴ By Pythagoras theorem,
CQ2 + OQ2 = CD2
Or, 32 + OQ2 = 52
Or, OQ = √25-9 = √16 = 4 cm
∴ Distance between chord AB & CD = OQ – OP
= 4 – 3 = 1cm
(ii) Given, length of chord AB = 8cm
Length of chord CD = 6cm
AB||CD
∴ P – O – Q is A straight line
⊥ to CD & AB
Now, In △AOQ, ∠AQD = 90°, AQ = AB/2 = 8/2 = 4
∴ By Pythagoras theorem,
AO2 = OQ2 + AQ2
Or, 52 = OQ2 + 42
Or, OQ = √25-16 = √9 = 3
∴ Distance from AB to CD
= OQ + OP
= 3 + 4 = 7cm
[∵ ⊥ OQ bisect chord AB]
In △COP, ∠CPO = 90
CP = 1/2 CD = 1/2 × 6 = 3
[∵ ⊥ OP bisect CD]
∴ By Pythagoras theorem
OC2 = OP2 + CP2
Or, 52 = OP2 + 32
Or, QP = √25-9 = √16 = 4 cm
(7) (a) In the figure (i) given below, O is the centre of the circle, AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD.
AB = 24 cm, OM = 5cm, ON = 12cm. Find the:
(i) Radius of the circle
(ii) Length of chord CD.
(b) In the figure (ii) given below, CD is a diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3cm, find the radius of the circle.
Solution:
(a) Given, OM⊥AB, ON⊥CD
AB = 24cm, OM = 5cm, ON = 12cm
Construction: Join OA&OC
In △AON, ∠AMO = 90° OM⊥AB
AM = AB/2 [OM⊥AB bisect chord AB]
= 24/2 = 12 cm
∴ By Pythagoras theorem, (i) ∴ Radius of circle OA = 13cm
OA2 = AM2 + OM2
Or, OA2 = 122 + 52
Or, OA = √144+25 = √169 = 13
Now, In △CON, ∠CNO = 90° [ON⊥CD]
CN = CD/2, ON = 12cm, CO = Radius = 13 cm
∴ By Pythagoras theorem,
CO2 = ON2 + CN2
Or, CN2 = 132 – 122
Or, CN = √169-144 = √25 = 5 cm
∴ Length of chord CD = CN×2 = 5×2 = 10cm
[∵ ON⊥CD bisect CD]
(b)
Given, AE = BE = 4 cm
CD is the diameter that cut chord AB into two equal halves.
∴ ∠OEB = ∠DEA = 90°
[∵⊥ from centre bisects the chord]
CE = 3cm, construction : join OB
In △OBE, ∠OEB = 90°
∴ By Pythagoras theorem
OB2 = OE2 + BE2
Or, OB2 = (OC – CE)2 + 42
Or, OB2 = OC2 – 20C CB + CE2 + 42
Or, OB2 = OC2 – 20C × 3 + 32 + 16
Or, OB2 – OC2 = 60C + 9 + 16
Here, OB = OC = radius of circle.
Let, OB = OC = r
∴ r2 – r2 = – 6 × r + 25
Or, 6r = 25
Or, r = 25/6 = 4 1/6 cm
∴ radius of circle = r = 4 1/6
(8) In the adjoining figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15cm, find the distance MN between the two chords of length 24 cm and 18cm respectively.
Solution:
Give, radius of the circle r = 15cm
AB = 24cm, CD = 18cm
Now, construction : join OA & OC
In △AOM, ∠ANO = 90°
AM = AB/2 = 24/2 = 12cm [⊥MN bisects chord AB]
∴ By Pythagoras theorem,
OA2 = OM2 + AM2
Or, 152 = OM2 + 122
Or, OM = √225-144 = √81 = 9cm
In △CON, ∠CNO = 90°, CN = CD/2 = 18/2 = 9cm [∵ ⊥MN bisect chord CD]
∴ By Pythagoras theorem,
OC2 = ON2 + CN2
Or, 152 = ON2 + 92
Or, ON = √225-81 = √144 = 12cm
∴ MN = ON + OM = 12+9 = 21cm
(9) AB and CD are two parallel chords of a circle of lengths 10cm and 4cm respectively. If the chords lie on the same side of the centre and the distance between them is 3cm, find the diameter of the circle.
Solution:
Given, AB = 10cm, CD = 4cm
EF = 3cm
Construction join : OA & OC
OE⊥AB & CD
Now, In △AOF, ∠AFO = 90 [∵ OE⊥AB]
AF = AB/2 = 10/2 = 5cm [⊥ OE bisects chord AB
∴ Let, OA = OC = r [Radius of circle]
∴ By Pythagoras theorem,
OA2 = AF2 + OF2
Or, r2 = 52 + OF2
or, OF2 = r2 – 52
Or, OF2 = r2 – 25 — (i)
In △COE, ∠CEO = PO
[∵ OE ⊥ CD]
CE = CD/2 = 4/2 = 2 cm
[∵ ⊥CE bisect chords]
∴ By Pythagoras theorem,
OC2 = CE2 + OE2
Or, r2 = 22 + OE2
Or, OE2 = r2 – 4 —- (ii)
Now, subtracting equation (ii) from equation (i) we get,
OE2 – OF2 = 22 – 4 – 22 + 25
or, (OE + OF) (OE – OF) = 21
Or, (OE + OF) × EF = 21
Or, (OE + OF) × 3 = 21
Or, OE + OF = 7 —– (iii)
We know,
OE – OF = EF
Or, OE – OF = 3 —- (iv)
Now putting the value of OE in equation (ii) we get
52 = r2 – 4
Or, r2 = 25 + 4
Or, r = √29
∴ Radius of circle = √29 cm
∴ Diameter = 2 × r
= 2 × √29
= 2√29 cm
(10) ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12√5 cm and BC = 24 cm, find the radius of the circle
Solution:
Given, △ABC is an isosceles △.
AB = AC = 12√2 cm
BC = 24 cm
Construction : join AD⊥BC through O.
Joint OB
∴ Let, OB = OA = r [Radius]
∴ In △ABD, ∠ADB = 90° [∵ AD⊥BC]
BD = BC/2 = 24/2 = 12cm [∵ ⊥ AD bisect chord BC]
∴ By Pythagoras theorem,
AB2 = BD2 + AD2
Or, (12√2)2 = 122 + AD2
Or, AD = √288 – 14
Or, AD = √144 = 12cm
∴ Radius of circle = r = 12cm
Now, In △BOD, ∠BDO = 90°
∴ OD = AD = OA = 212 – r
BD = 12cm
∴ By Pythagoras theorem,
OB2 = BD2 + OD2
Or, r2 = 122 + (12 – r)2
Or, r2 = 144 + 144 – 24r + 22
Or, 24r = 288
Or, r = 12cm
(11) An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.
Solution:
Given, △ABC is an equilateral △.
AB = AC = BC = 6cm
Let, radius of circles be r
∴ OB = r = OA
Construction : join AD⊥BC
∴ 0 lies on AD.
In △ABD, ∠ADB = 90°
BD = BC/2 = 6/2 = 3 cm [∵ ⊥AD bisect chord BC]
∴ By Pythagoras theorem,
AB2 = BD2 + AD2
Or, 62 = 32 + AD2
Or, AD = √36 – 9
Or, AD = √27
Or, AD = 3√3cm
In △BOD, ∠BDO = 90°
BD = 3cm, OD = AO – OA
∴ By Pythagoras theorem
OB2 = BD2 + OD2
Or, r2 = 32 + (3√3 – r)2
Or, r2 = 32 + 27 – 6√3r + r2
Or, 6√3r = 36
Or, r = 36/6√3 = 2√3 cm
∴ Radius of circle = 2√3 cm
(12) AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.
Solution:
Given, AB diameter of circle
AM = 18 cm, BM = 8 cm
∴ AB = AM + BM = 18 + 8 = 26 cm
∴ Radius r = AB/2 = 26/2 = 13 cm
Construction join : CD through M
CD ⊥ AB
[∵ CD is the shortest chord through M]
∴ In △COM, ∠CMO = 90°
CO = r = 13 cm
OM = OB – BM = r – 8 = 13 – 8 = 5 cm
∴ by Pythagoras theorem,
OC2 = CM2 + OM2
Or, r2 = CM2 + 52
Or, 132 = CM2 + 52
Or, CM = √169-25
Or, CM = √144 = 12 cm
∴ CD = 2 × CM [∵ ⊥ AB bisect the chord CD]
Or, CD = 2 × 12
= 24 cm
∴ Length of shortest chord CD = 24 cm
(13) A rectangle with one side of length 4 cm in inscribed in a circle of diameter 5 cm. Find the area of the rectangle.
Solution:
Given, ABCD is a rectangle.
Construction : join AC diagonal
AC = 4cm
∴ In △ABC, ∠ABC = 40° [Angle of rectangle]
∴ By Pythagoras theorem,
AB2 + BC2 = AC2
Or, 42 + BC2 = 52
Or, BC = √25-16 = √9 = 3cm
∴ Area of rectangle AB × BC = 3 × 4 = 12cm2
(14) The length of the common chord of two intersecting circles is 30cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.
Solution:
Given, length of common chord AB = 30 cm
Construction : join PA & QA radius of two circles.
PQ ⊥ AB
∴ In △PMA ∠PMA = 90° [PQ⊥AB]
AM = AB/2 = 30/2 = 15cm [⊥ PQ bisect chord AB]
PA = radius = 25cm (Given)
∴ In △ by Pythagoras theorem,
PA2 = PM2 + AM2
Or, 252 = PM2 + 152
Or, PM = √625 – 225
Or, PM = √400
Or, PM = 20 cm
In △QMA, ∠QMA = 90o
OA = radius = 17cm (Given)
AM = 15 cm
∴ By Pythagoras theorem,
QA2 = QM2 + AM2
Or, 172 = QM2 + 152
Or, QM = √289 – 225
Or, QM = √64 = 8cm
∴ PQ = PM + QM = 20 + 8 = 28cm
∴ Distance between centre of two circle = PQ = 28c
(15) The line joining mid-points of two chords of a circle passes through its centre. Prove that the chords are parallel.
Solution:
Given, E, F are mid-point of AB & CD.
EF passes through O.
∴ OE ⊥ AB [line through mid-point of chord is ⊥ the centre]
OF ⊥ CD [line through mid-point of chord is ⊥ the centre]
EOF is a straight line given,
∴ ∠OEA = 90° [∵ OE⊥AB]
∠OFC = 90° [∵ OF⊥CD]
∴ ∠OEA + ∠OFC = 90° + 90° = 180
∴ Sum of Conjacent angles are 180°
∴ AB||CD
(16) If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.
Solution:
Given, AB||CD
EF is the diameter.
EF⊥AB
∴ ∠ANM = 90° [∵ EF⊥AD]
∴ ∠CMN = ∠ANM = 90° [alternate angle, AB||CD]
∴ EF⊥CD
∴ M is the mid-point of CD [∵ EF diameter ⊥ CD bisects chord CD]
∴ EF bisects CD. (Proved)
(17) In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.
Solution:
Given, △ABC is an equilateral △.
AD, BE&CP are medians of △ABC
O is the point of intersection of medians.
∴ O is the centroid of △ABC —- (i)
Now, In △ABD,
∠B = 60° [Angle of equilateral △]
∠BAD = ∠A/2 = 60/2 = 30 [∵ median AD bisect vertex angle ∠A of equilateral △ABC]
∴ ∠B + ∠BAD + ∠ADB = 180° [Angle sum property of △]
Or, 60° + 30° + ∠ADB = 180°
Or, ∠ADB = 180° – 90 = 90°
∴ AD is the perpendicular bisector of BC of △ABC
Similarly, In △BEC
BE is the perpendicular bisector of side AC △AOC.
In △AFC
CF is the perpendicular bisector of side AB of △ABC
∴ Perpendicular bisectors AD, BE & CF metals at point O.
∴ Point O is the circumcenter of △ABC.
But, point O is also the centroid of △ABC [From (i)]
∴ The centroid & the circumcenter of △ABC coinside.
(18) (a) In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that CA = 20D.
(b) In the figure (ii) given below, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ⊥AC prove that PB = QC
Solution:
(a) Given, OD⊥AB
O is the centre circle
∴ D is the mid-point of AB —– (i)
[∵ ⊥OD from centre bisect chord AB]
BC is the diameter of circle
O is the centre of circle
∴ OB = OC [Two radius of circle]
∴ O is the mid-point of BC —- (ii)
∴ In △ABC,
1/2 by mid-point theorem,
OD = 1/2 CA [from (i) & (ii)]
Or, CA = 20D (Proved)
(b)
Given, AB = AC
OP ⊥ AB, QC ⊥ AC
N is the mid-point of chord AC
N is the mid-point of chord AB
[∵ OP⊥AB, QC⊥AC, ⊥ is from centre bisect the chords]
∴ CN = 1/2 AE
BN = 1/2 AB
Or, BM = 1/2 AC [∵ AB = AC
Or, BM = CN —- (i)
Now, OM = ON [∵ ⊥ distance from centre to chords of equal length and equal]
Now, OQ = OP [radius of circle]
Or, ON + QN = OM + PN
Or, QN = PN [∵ ON = OM] —– (ii)
In △QNC & △PNB,
(i) BM = CN [From (i)]
(ii) QN = PM [From (ii)]
(iii) ∠ONC = ∠PMB = 90° [OP⊥AB, OQ⊥AC
∴ △QNC ≅ △PMB by SAS
∴ QC = PB [Corresponding congruent sides]
(19) (a) In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D Prove that AB = CD.
(b) In the figure (ii) given below, chord AB and CD of a circle with centre O intersect at E, If OE bisects ∠AED, prove that AB = CD.
Solution:
(a) Given, line & intersect two concentric circles at point A, B, C, D.
Construction : join OM ⊥ line l
∴ BM = CM [∵ ⊥OM bisect chord BC] —– (i)
AM = DM [∵ ⊥OM bisect chord AD] —– (ii)
Now, AB = AM – BM —- (iii)
CD = DM – CM — (iv)
Subtracting equation (iv) from equation (iii) we get,
AB – CD = AM – BM – DM + CM
Or, AB – CD = AM – BM – AM + BM [From (i) & (ii)]
Or, AB – CD = 0
Or, AB = C (Proved)
(b) Given, OE bisect ∠AED
In △OME & △ONE
Construction : OM⊥AB, ON⊥CD.
(i) ∠ONE = ∠ONE = 90°
(ii) ∠OEM = OEN [∵ OE bisect ∠AED]
(iii) OE is the common side
∴ △OME ≅ △ONE by AAS
∴ OM = ON [Corresponding congruent sides]
∴ Distance from chord AB from centre OD = OM is equal to distance from chord CD from centre O = ON
∴ Chord AB = chord CD
(20) (a) In the figure (i) given below, AD is a diameter of a circle with centre O. If AB||CD, prove that AB = CD.
(b) In the figure (ii) given below, AB and CD are equal chords of a circle with centre O. If AB and CD meet at E (outside the circle) prove that:
(i) AE = CE
(ii) BE = DE
Solution:
(a) Given, AD is the diameter of circle AB||CD
Construction : join OM⊥CD, ON⊥AB
∴ In △AON & △DOM
(i) OA = OD [Radius]
(ii) ∠ODN = ∠OAN [alternate angle, AB||CD]
(iii) ∠OMD = ∠ONA = 90°
∴ △AON ≅ △DOM by AA.S
∴ OM = ON [Corresponding congruent sides]
∴ Distance from chord AB to centre O = ON is equal to distance from chord CD to centre O = OM is equal
∴ AB = CD
(B) Given, AB = CD
Construction : join : OM⊥AB, ON⊥CD
OE
∴ In △OME & △ONE
(i) ON = OM [Chords of some length me equidistant from centre of circle]
(ii) ∠ONE = ∠ OME = 90°
(iii) OE is the common side
∴ △OME ≅ △ONE by R.H.S
∴ NE = ME [Corresponding congruent sides]
(i) Now, N is the mid-point of CD [∵ OM⊥AB, ON⊥CD]
M is the mid-point of AB
∴ CN = 1/2 CD = 1/2 AB = AM [∵ AB = CD]
AE = AM + ME
Or, AE = CN + NE [∵ CN = AM, ME = NE]
Or, AB = CE
(ii) Now, BE = AE – AB
Or, BE = CE – CD [∵ AB = CD & AE = CF]
Or, BE = DE (Proved)
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