Selina Concise Mathematics Class 10 ICSE Solutions Circles

Selina Concise Mathematics Class 10 ICSE Solutions Circles

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles

Circles Exercise 17A – Selina Concise Mathematics Class 10 ICSE Solutions

Circles Class 10 Question 1.
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 1
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 2

Question 2.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 3
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 4

Question 3.
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠ OCA,
(ii) ∠OAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 6

Question 4.
In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 7
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 8

Question 5.
In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 9
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 10
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 11

Question 6.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and O2 are the centres of two circles.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 12
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 13

Question 7.
In the figure given beow, find :
(i) ∠ BCD,
(ii) ∠ ADC,
(iii) ∠ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 14
Show steps of your workng.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 15

Question 8.
In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :
(i) ∠ ACB,
(ii) ∠ OBC,
(iii) ∠ OAB,
(iv) ∠CBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 16
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 17

Question 9.
Calculate :
(i) ∠ CDB,
(ii) ∠ ABC,
(iii) ∠ ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 18
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 19

Question 10.
In the figure given below, ABCD is a eyclic quadrilateral in which ∠ BAD = 75°; ∠ ABD = 58° and ∠ADC = 77°. Find:
(i) ∠ BDC,
(ii) ∠ BCD,
(iii) ∠ BCA.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 20
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 21

Question 11.
In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find :
(i) ∠ ADB
(ii) ∠ AEB
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 22
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 23

Question 12.
Given—∠ CAB = 75° and ∠ CBA = 50°. Find the value of ∠ DAB + ∠ ABD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 24
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 25

Question 13.
ABCD is a cyclic quadrilateral in a circle with centre O.
If ∠ ADC = 130°; find ∠ BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 26
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 27

Question 14.
In the figure given below, AOB is a diameter of the circle and ∠ AOC = 110°. Find ∠ BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 28
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 29

Question 15.
In the following figure, O is centre of the circle,
∠ AOB = 60° and ∠ BDC = 100°.
Find ∠ OBC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 30

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 31

Question 16.
ABCD is a cyclic quadrilateral in which ∠ DAC = 27°; ∠ DBA = 50° and ∠ ADB = 33°.
Calculate :
(i) ∠ DBC,
(ii) ∠ DCB,
(iii) ∠ CAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 32
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 33

Question 17.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°. Find the number of degrees in:
(i) ∠DCE;
(ii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 34
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 35

Question 17 (old).
In the figure given below, AB is diameter of the circle whose centre is O. Given that:
∠ ECD = ∠ EDC = 32°.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 36
Show that ∠ COF = ∠ CEF.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 37

Question 18.
In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 38
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 39

Question 19.
In the following figure,
(i) if ∠BAD = 96°, find BCD and
(ii) Prove that AD is parallel to FE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 40
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 41

Question 20.
Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 42
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 43

Question 21.
In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 44
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 45

Question 22.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 46

Question 23.
The figure given below, shows a circle with centre O. Given: ∠ AOC = a and ∠ ABC = b.
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 47
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 48

Question 24.
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 49

Question 24 (old).
ABCD is a quadrilateral inscribed in a circle having ∠A = 60°; O is the centre of the circle. Show that: ∠OBD + ∠ODB = ∠CBD + ∠CDB
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 50

Question 25.
In the figure given RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°
Calculate:
(i) ∠RNM;
(ii) ∠NRM.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 51
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 52

Question 26.
In the figure given alongside, AB || CD and O is the centre of the circle. If ∠ ADC = 25°; find the angle AEB. Give reasons in support of your answer.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 54

Question 27.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at Cand D. Prove that AC is parallel to BD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 55
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 56

Question 28.
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 57

Question 29.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 58
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 59

Question 30.
In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 60
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 61

Question 31.
In the figure, O is the centre of the circle, ∠AOE = 150°, DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 62
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 63

Question 32.
In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 64
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 65

Question 33.
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that ∠APB = a°. Calculate, in terms of a°, the value of:
(i) obtuse ∠AOB
(ii) ∠ACB
(iii) ∠ADB.
Give reasons for your answers clearly.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 66
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 67

Question 34.
In the given figure, O is the centre of the circle and ∠ ABC = 55°. Calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 68
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 69

Question 35.
In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that ∠BCD = 2∠ABE
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 70
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 71

Question 36.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate:
(i) ∠ DAB,
(ii) ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 72
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 73

Question 37.
∠ In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠ EAB = 63°; calculate:
(i) ∠EBA,
(ii) BCD.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 75

Question 38.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°; calculate:
(i) ∠ DAB,
(ii) ∠ DBA,
(iii) ∠ DBC,
(iv) ∠ ADC.
Also, show that the ∆AOD is an equilateral triangle.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 77

Question 39.
In the given figure, I is the incentre of the ∆ ABC. Bl when produced meets the circumcirle of ∆ ABC at D. Given ∠BAC = 55° and ∠ ACB = 65°, calculate:
(i) ∠DCA,
(ii) ∠ DAC,
(iii) ∠DCI,
(iv) ∠AIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 78
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 79

Question 40.
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ABC = 2 ∠APQ
(ii) ∠ACB = 2 ∠APR
(iii) ∠QPR = 90° – \(\frac{1}{2}\)BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 80
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 81

Question 40 (old).
The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB are produced to meet at F. If ∠BEC = 42° and ∠BAD = 98°; calculate:
(i) ∠AFB,
(ii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 82
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 83

Question 41.
Calculate the angles x, y and z if: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 84
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 85

Question 42.
In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate:
(i) Angle ABC
(ii) Angle BEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 86
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 87

Question 43.
In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, and r in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 88
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 89

Question 44.
In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80° and angle ACE = 10°. Calculate:
(i) Angle BEC;
(ii) Angle BCD;
(iii) Angle CED.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 90
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 91

Question 45.
In the given figure, AE is the diameter of circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 92
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 93

Question 46.
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 94
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 95

Question 47.
Use the given figure to find
(i) ∠BAD
(ii) ∠DQB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 96
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 97

Question 48.
In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:
(i) ∠COB
(ii) ∠DOC
(iii) ∠DAC
(iv) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 98
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 99

Question 49.
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 100
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 101

Question 50.
In the given figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°; calculate ∠RTS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 102
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 103

Question 51.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°; calculate
(i) ∠RPQ
(ii) ∠STP.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 104
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 105

Question 52.
AOD = 60°; calculate the numerical values of:
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°; calculate the numerical values of:
(i) ∠ABD,
(ii) ∠DBC,
(iii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 106
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 107

Question 53.
In the given figure, the centre of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40″; find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 108
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 109

Question 54.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°; find:
(i) ∠BCD,
(ii) ∠ACB.
Hence, show that AC is a diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 110
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 111

Question 55.
In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 112

Question 56.
The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 196

Question 57.
In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠ MAD =x and ∠BAC = y.
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that : x = y
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 198

Question 61 (old).
In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 199

Circles Exercise 17B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 200

Question 2.
In the following figure, AD is the diameter of the circle with centre 0. chords AB, BC and CD are equal. If ∠DEF = 110°, calculate:
(i) ∠ AFE,
(ii) ∠FAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 201

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 202

Question 3.
If two sides of a cycli-quadrilateral are parallel; prove thet:
(i) its other two side are equal.
(ii) its diagonals are equal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 203

Question 4.
The given figure show a circle with centre O. also, PQ = QR = RS and ∠PTS = 75°. Calculate:
(i) ∠POS,
(ii) ∠ QOR,
(iii) ∠PQR.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 205

Question 5.
In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠ AOB,
(ii) ∠ ACB,
(iii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 206
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 207

Question 6.
In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 208

Question 7.
In the given figure. AB = BC = CD and ∠ABC = 132°, calculate:
(i) ∠AEB,
(ii) ∠ AED,
(iii) ∠COD.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 210

Question 8.
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find:
(i) ∠ CAB,
(ii) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 211
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 212

Question 9.
The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 213
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 214

Question 10.
In the given figire, BD is a side of a regularhexagon, DC is a side of a regular pentagon and AD is adiameter. Calculate:
(i) ∠ ADC
(ii) ∠BAD,
(iii) ∠ABC
(iv) ∠ AEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 215
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 216
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 217

Circles Exercise 17C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the given circle with diametre AB, find the value of x.

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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 219
∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a A)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180° ⇒ x + 120° = 180°
∴ x= 180° – 120° = 60° Ans.

Question 1.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 220
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 221

Question 2.
In the given figure, ABC is a triangle in which ∠ BAC = 30° Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.
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Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 223

Question 3.
Prove that the circle drawn on any one a the equalside of an isoscele triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 224

Question 3 (old).
The given figure show two circles with centres A and B; and radii 5 cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 225
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 226

Question 4.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠ CBE = 65°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 227
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 228

Question 5.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 229

Question 6.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 230
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 231

Question 7.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Provet that the points B, C, E and D are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 232

Question 7 (old).
Chords AB and CD of a circle intersect each other at point P such that AP = CP.
Show that: AB = CD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 233
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 234

Question 8.
In the given rigure, ABCD is a cyclic eqadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ ADC = 92°, ∠ FAE = 20°; determine ∠ BCD. Given reason in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 235
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 236

Question 9.
If I is the incentre of triangle ABC and Al when produced meets the cicrumcircle of triangle ABC in points D. if ∠ BAC = 66° and ∠ = 80o.calculate:
(i) ∠ DBC
(ii) ∠ IBC
(iii) ∠ BIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 237
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 238
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 239

Question 10.
In the given figure, AB = AD = DC = PB and ∠ DBC = x°. Determine, in terms of x:
(i) ∠ ABD,
(ii) ∠ APB.
Hence or otherwise, prove thet AP is parallel to DB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 240
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 241

Question 11.
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠ CQE are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 242
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 243

Question 12.
In the given, AB is the diameter of the circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 244
If ∠ ADC = 32°, find angle BOC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 245

Question 13.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ prouduced meet at point A: whereas sides PQ and SR produced meet at point B.
If ∠A: ∠B = 2 : 1;find angles A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 246
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 247

Question 17 (old).
If the following figure, AB is the diameter of a circle with centre O and CD is the chord with lengh equal radius OA.
If AC produced and BD produed meet at point p; show that ∠APB = 60°
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 248
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 249
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 250

Question 14.
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 251
If the bisector of angle A meet BC at point E and the given circle at point F, prove that:
(i) EF = FC
(ii) BF =DF
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 252

Question 15.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point e; whereas sides BC and AD produced meet at point F. I f ∠ DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 253
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 254

Question 16.
The following figure shows a cicrcle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimetre of the cyclic quadrilateral PORS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 255
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 256

Question 17.
In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD.prove that:
(i) arc BC = arc DB
(ii) AB is bisector of ∠ CAD.
Further if the lenghof arc AC is twice the lengthof arc BC find :
(a) ∠ BAC
(b) ∠ ABC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 257
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 258
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 259

Question 18.
In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;
(i) ∠ BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ ADC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 260
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 261

Question 19.
In the given figure, ∠ACE = 43° and ∠ = 62°; find the values of a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 262
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 263

Question 20.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠ BAC = 25°
Find
(i) ∠ CAD
(ii) ∠ CBD
(iii) ∠ ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 264
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 265

Question 21.
ABCD is a cyclic quadrilateral of a circle with centre o such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle..if AD and BC produced meet at P, show that APB =60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 266

Question 22.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 267
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 268

Question 23.
In the figure, given below, AD = BC, ∠ BAC = 30° and ∠ = 70° find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 269
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 270

Question 24.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 271

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 272

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 273

Question 25.
In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of
i. ∠BCD
ii. ∠BOD
iii. ∠OBD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 274
Solution:
∠DAE and ∠DAB are linear pair
So,
∠DAE + ∠DAB = 180°
∴∠DAB = 110°

Also,
∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC
∴∠BCD = 70°
∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD…angles subtended by an arc on the centre and on the circle
∴∠BOD = 140°

In ΔBOD,
OB = OD……radii of same circle
So,
∠OBD =∠ODB……isosceles triangle theorem
∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle
2∠OBD = 40°
∠OBD = 20°

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