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ICSE Class IX Question Bank 2025 : Mathematics circles

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CH 16 : ANGLE PROPERTIES OF CIRCLE CHAPTER 16 ANGLE PROPERTIES OF CIRCLE x = 70 = 35 2 MULTIPLE CHOICE QUESTIONS QUESTION 1. 1. In the following figure O is the centre of circle. what is the value of x ? (a) 42 (c) 26 Thus (a) is correct option. QUESTION 3. 3. (b) 52 (d) 64 In the following figure O is the centre of circle. what is the value of x ? (a) 360 (c) 300 Sol : Sol : In a circle, angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. +AOB = 2+APB = 2 # 30 +AOB = 2+APB = 60 x = 2 # 26 x = Reflex +AOB = 52 Thus () is correct option. = 360 - 60 = 300 Thus (c) is correct option. QUESTION 2. 2. In the following figure O is the centre of circle. what is the value of x ?? QUESTION 4. 4. (a) 35 (c) 45 In the following figure O is the centre of circle. what is the value of x ? (b) 25 (d) 65 Sol : (a) 240 (c) 120 Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Sol : Thus (b) 330 (d) 270 +AOB = 2+APB 70 = 2x (b) 260 (d) 140 +AOB = 2+APB 240 = 2x ICSE CHAPTERWISE PYQ CLASS 10 MATHS PAGE 357 x = 240 = 120 2 Thus (c) is correct option. QUESTION 5. 5. In the following figure O is the centre of circle. what is the value of x ? (a) 30 (c) 15 (b) 40 (d) 50 Sol : Due to radii of same circle, OC = OA Equal side make equal angle. Thus (a) 230 (c) 115 +ACO = +CAO (b) 150 (d) 140 +ACB = x In given figure arc AB subtends +AOB at the centre and +ACB at the remaining part of the circle. Sol : Reflex +AOB = 2+APB x = 2 # 115 Thus = 230 Thus (a) is correct option. 80 = 2x x = 80 2 QUESTION 6. 6. +AOB = 2+ACB = 40 Thus (b) is correct option. In the following figure O is the centre of circle. What is the value of x ? QUESTION 8. 8. (a) 150 (c) 115 (b) 150 (d) 100 Sol : Reflex +AOB = 2+APB = 2 # 130 = 260 x = 360 - 260 = 100 Thus (d) is correct option. QUESTION 7. 7. In the following figure O is the centre of circle. What is the value of x ? In the following figure O is the centre of circle. What is the value of x ? (a) 50 (c) 30 (b) 40 (d) 20 Sol : In DOAC due to radii of same circle OA = OC Equal side make equal angle. Thus +OAC = +ACO = 28 +BAC = +ACO = 28 In the figure, arc BC subtends +BOC at the centre and +BAC at the remaining part of the circle. Thus +BOC = 2+BAC x = 2 # 28 CH 16 : ANGLE PROPERTIES OF CIRCLE = 56 Thus x = 56 . Thus (a) is correct option. QUESTION 11. 11. In the following figure O is the centre of circle. What is the value of x ? QUESTION 9. 9. In the following figure O is the centre of circle. What is the value of x ? (a) 30 (c) 70 (b) 50 (d) 40 Sol : (a) 120 (c) 110 (b) 150 (d) 140 In DABC , +ABC = 40 Due to angle in a semi circle, +BAC = 90 Sum of angles of a triangle is always 180 . Sol : In the figure +AOB = 140 +ACB + +BAC + +ABC = 180 Reflex +AOB = 360 - 140 = 220 Arc ACB subtends +AOB at the centre and +ACB at the remaining part of the circle Thus +AOB = 2+ACB 220 = 2x x = 220 2 = 110 Thus (c) is correct option. QUESTION 10. 10. In the following figure O is the centre of circle. What is the value of x ? (a) 30 (c) 45 (b) 70 (d) 35 Sol : Angle +ACB and +ADB are in the same segment. Thus +ADB = ACB x = 35 Thus (d) is correct option. ) +ACB + 90 + 40 = 180 +ACB + 130 = 180 +ACB = 50 Angle +ADB and +ACB are in same segment of the circle. Thus +ADB = +ACB x = 50 Thus (b) is correct option. QUESTION 12. 12. What is value of angle x in the given figure if +PQR = 28 ? (a) 56 (c) 28 (b) 31 (d) 62 Sol : Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Thus +POR = 2+PQR x = 2 # 28 = 56 ICSE CHAPTERWISE PYQ CLASS 10 MATHS Thus (a) is correct option. QUESTION 13. 13. What is value of angle x in the given figure if +ABC = 35 ? PAGE 359 QUESTION 15. 15. In the following figure O is the centre of circle. What is the value of x ? (a) 30 (c) 55 (b) 40 (d) 50 Sol : (a) 30 (c) 70 (b) 50 (d) 40 In the given circle +DAB = 35 +CBA = 60 Sol : Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Thus +ACB = 30 , +AOC = 2+ABC x = 2 # 35 = 70 Thus (c) is correct option. Due to angles in the same segment we have +ADB = +ACB +ADB = 30 Now in DABD , +DAB + +ADB + +ABD = 180 35 + 30 + 60 + x = 180 QUESTION 14. 125 + x = 180 14. What is value of angle x in the given figure if +ABC = 110 ? x = 55 Thus x = 55 . Thus (c) is correct option. QUESTION 16. 16. In the circle with centre O, if +ADB = 20 what is the value of angle +BAC ? (a) 120 (c) 110 (b) 150 (d) 140 Sol : Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Thus reflex+AOC = 2+ABC = 2 # 110 = 220 +AOC = 360 - 220 x = 140 Thus (d) is correct option. (a) 30 (c) 60 (b) 70 (d) 50 Sol : We have redrawn the given figure as below CH 16 : ANGLE PROPERTIES OF CIRCLE The angle in a semi-circle is always a right angle. +ABC = 90 Angles in the same segment are always equal. +ADB = +ACB = 20 Sum of all angles of triangle is always 180c. Now, +BOC = 360 - (120 - 80 ) = 160 Since arc BPC makes +BOC at the centre and +BAC at a point on the remaining part of the circle, we have +ABC + +BCA + +CAB = 180c +BOC = 2+BAC 90c + 20c + +CAB = 180c +BAC = 12 +BOC 110c + +CAB = 180c = 12 # 160 +CAB = 70 Thus (b) is correct option. = 80 Thus (d) is correct option. QUESTION 17. QUESTION 18. 17. In given figure, A, B, C are three points on a circle such that the angles subtended by the chords AB and AC at the centre O are 80 and 120 respectively. What is the value of angle +BAC ? 18. C is a point on the minor arc AB of the circle O. Given +ACB = x and +AOB = y express y in terms of x. What is value of angle x in if ACBO is a parallelogram? (a) 120 (b) 150 (c) 110 (d) 140 Sol : As per given in question we have shown the figure below. (a) 30 (c) 40 (b) 60 (d) 80 Sol : We have redrawn the figure as below. Arc ACB makes Reflex+AOB at the centre of the circle and +ACB at a point on the remaining part of the circle. Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference. ICSE CHAPTERWISE PYQ CLASS 10 MATHS Thus PAGE 361 Reflex +AOB = 2+ACB 360 - y = 2x y = 360 - 2x Opposite angles of a parallelogram are equal. If ACBO is a parallelogram, we have x =y x = 360c - 2x 3x = 360c x = 120c Thus (a) is correct option. Sol : We have redrawn the given figure as below. Here we have joined OA. QUESTION 19. 19. In the given figure, O is the centre of a circle. If the length of chord PQ is equal to the radius of the circle, the +PRQ is (a) 60 (b) 45 (c) 30 (d) 15 Sol : We have redrawn the given figure as below. Due to angle in same segment we have +ADB = +ACB = 2x Arc AB subtends +AOB at the centre and +ADB at the remaining part of the circle. Thus +AOB = 2+ADB = 2 # 2x Here OQ and OP are radius of circle. PQ is equal to the radius of the circle. Thus DOPQ is an equilateral triangle. Thus +POQ = 60c Arc PQ subtends +POQ at the centre and +PRQ at the remaining part of the circle +POQ = +PRQ +PRQ = 12 +POQ = 12 # 60c = 30c = 4x Due to the radius of same circle, OA = OB Angle opposite to equal sides of a triangle are equal. Thus in DOAB , +OAB = +OBA = 3x Sum of all internal angle of a triangle is always 180 . Thus in DOAB , +OAB + +ABO + +AOB = 180 3x + 3x + 4x = 180 Thus (c) is correct option. 10x = 180 QUESTION 20. 20. In the given figure, if O is centre of the circle then the value of x is (a) 18 (b) 20 (c) 24 (d) 36 x = 18c Thus (a) is correct option. QUESTION 21. 21. In the given figure A, B, and C are three points on a circle such that the angles subtended by the chords AB and AC at the centre O are 90 and 110 , respectively. What is the value of angle CH 16 : ANGLE PROPERTIES OF CIRCLE +BAC ? (a) 30 (c) 40 (b) 75 (d) 80 Sol : In the given circle (a) 80 (c) 40 (b) 60 (d) 50 +AOB = 50 , +BDC = 80 Arc AB subtends +AOB at the centre and +ACB at the remaining part of the circle. Thus Sol : +AOB = 2+ACB We have redrawn the figure as below. +ACB = 12 +AOB +ACB = 12 # 50 +DCB = 25 Now in DBCD , +BDC + +DCB + +CBD = 180 80 + 25 + x = 180 105 + x = 180 x = 180 - 105 = 75 Here +BOA = 90 and +AOC = 110 Thus x = 75 . Thus (b) is correct option. Sum of all angles around a point is always 360 . +BOC + +BOA + +AOC = 360 THREE MARKS QUESTIONS +BOC + 90 + 110 = 360 +BOC + 200 = 360 +BAC = 160 Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference. Now, QUESTION 23. Comp 2023 23. In the given circle with diameter AB , find the value of x . +BOC = 2+BOC +BAC = 12 +BOC = 12 # 160 = 80 Thus (a) is correct option. QUESTION 22. 22. In the following figure O is the centre of circle. What is the value of x ? Sol : (2003) We redraw the figure as below. ICSE CHAPTERWISE PYQ CLASS 10 MATHS PAGE 363 +ADC = 62c (ii) +BAC Here AB is diameter. Since angle in a semi-circle is always 90c, thus +ACB = 90c Using angle sum property in triangle TABC we have +ACB + +ABC + +BAC = 180c Due to angle in the same segment. +ABD = +ACD = 30c Since angle in a semi-circle is always 90c, +BAC + 62c + 90c = 180c +BAC + 152c + 90c = 180c +ADB = 90c Since sum of all angles in a triangles is 180c, in TABD , we have +DAB + +ABD + +ADB = 180c = 28c QUESTION 25. Comp 2022 25. In given figure, O is the centre of the circle. The angle subtended by the arc BCD at the centre is 140 . BC is produced to P. (i) Find +BAD (ii) Find +DCB x + 90c + 30c = 180c x + 120c = 180c x = 180c - 120c = 60c Hence, the value of x is 60c. QUESTION 24. +BAC = 180c - 152c Main 2022 Sem II 24. In the given figure A, B, C and D are points on the circle with centre O. Given +ABC = 62c. Find : (i) +ADC (ii) +BAC Sol : We have redrawn the figure as below. Sol : We have redrawn the figure as below. (i) +ADC Here +ABC = 62c Due to angle in the same segment we have +ADC = +ABC Since the arc BCD makes +BOD at the centre of circle and +BAD at a point on the remaining part of the circle, we have +BOD = 2+BAD CH 16 : ANGLE PROPERTIES OF CIRCLE +BAD = 12 +BOD any point on the major (or minor) arcs of the circle. Here AC = BD, thus = # 140 1 2 = 70 Again, arc DAB makes angle reflex +BOD = ^360 - 140 h = 220 at the centre and +BCD at a point C on the circumference. Thus +BAD = +ABC = 36c (iii) +ABD Angle in semi-circle is a right angle, thus +ADB = 90c Using angle sum property in triangle TABD we have Reflex+BOD = 2+BCD 220 = 2+BCD +BCD = 12 # 220 +ABD + +ADB + +BAD = 180c = 110 QUESTION 26. +ABD + 90c + 36c = 180c Main 2020 +ABD + 126c = 180c 26. In the figure below, O is the centre of the circle and AB is diameter. If AC = BD and +AOC = 72 . Find +ABD = 180c - 126c = 54c QUESTION 27. Main 2019 27. In the given circle, chord QN is perpendicular to chord MP and +MNQ = 20 . Find +NQP . (i) +ABC (ii) +BAD (iii) +ABD Sol : We have redrawn the figure as below. Here O is the centre of the circle and AB is a diameter. Sol : We have redrawn the given figure as below. Here AC = BD and +AOC = 72c (i) +ABC Angle subtended by an arc at the remaining circumference is half the angle subtended at the centre. Thus +ABC = 12 +AOC = 12 # 72c = 36c (ii) +BAD Equal chords of a circle subtend equal angles at In right DMNE, we have +NEM = 90 +MNE = 20 Sum of two acute angles in the right angle is always 90 . ICSE CHAPTERWISE PYQ CLASS 10 MATHS Thus +EMN + +MNE = 90 PAGE 365 to 180 . +QSR + +SQR + +SRQ = 180 +PMN + 20c = 90 70 + +SQR + 30 = 180 +PMN = 70 Angles in the same segment are always equal. +SQR + 100 = 180 +NQP = +PMN +SQR = 80 = 70 QUESTION 29. Thus +NQP = 70 QUESTION 28. Comp 2019 28. In the given circle with centre O, +POQ = 60 and +QSR = 70 . (i) Find +PRQ . (ii) Find +SQR Main 2018 29. In the figure given below, O is the centre of the circle. If QR = OP and +QRP = 20c. Find the value of 'x' giving reasons. Sol : (2018) We have redrawn the figure as below. Sol : We have redrawn the given figure as below Given QR = OP ...(1) Due to raddi of circle OP = OQ From Eqs. (1) and (2), we get ...(2) OP = OQ = QR Since angle opposite to equal sides are equal, (i) +PRQ . Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Arc PQ subtends +POQ at the centre and +PRQ at any point on the circumference of the circle. +POQ = 2+PRQ +PRQ = 12 +POQ = 12 # 60 = 30 (ii) +SQR The sum of angles of a linear pair is always equal +1 = +ORP +1 = 20c Due to external angle +2 = +1 + +ORP = 20c + 20c = 40c Due to raddi of circle, OP = OQ , thus +3 = +2 = 40c Sum of all angles in a triangles is 180c, +2 + +3 + +4 = 180c 40c + 40c + +4 = 180c +4 = 100c CH 16 : ANGLE PROPERTIES OF CIRCLE PC = PD PB PC Due to straight angle, x + +1 + +4 = 180c x + 20c + 100c = 180c 1 = PD PB x + 120c = 180c PB = PD x = 60c QUESTION 30. AP + PB = AP + PD AP + PB = CP + PB Comp 2017 30. In given figure, AB and CD are two chords of a circle, intersecting each other at P such that AP = CP . Show that AB = CD . AB = CD [AP = CP] Hence Proved QUESTION 31. Comp 2016 31. In the given figure, ABC is a triangle inscribed in a circle. Given that BD bisects +ABC . Prove AB BE that BD . = BC Sol : We have redrawn the figure as below. Here we have joined AD and BC . In 3 PAD and 3 PCB : Due to angles in the same segment of arc BD, +PAD = +PCB Due to vertically opposite angles, +APD = +CPB So, by AAA criterion of similarity, we have DPAD + PCB Corresponding sides of similar triangles are in the same ratio. PA = PD Thus PB PC Due to radius of same circle PA = PC , thus Sol : We have redrawn the figure as below. Here DABC is inscribed. Bisector of +ABC meets AC at E and the circle at D. We have joined DC. In DABE and DBCD : Due to angle in the same segment, +BAE = +BDC Since BD is the bisector of +B , +ABE = +DBC By AA Axiom we get DABE + DBCD Corresponding side of similar triangles are in the same ratio. Therefore AB = BE Hence proved. BD BC ICSE CHAPTERWISE PYQ CLASS 10 MATHS QUESTION 32. PAGE 367 Main 2015 32. AB and CD are two chords of a circle intersecting at P . Prove that AP # PB = CP # PD Due to the radius of same circle, OA = OB Angle opposite to equal sides of a triangle are equal. Thus in DABO , Sol : (2015) We have redrawn the figure as below. Here we have joined AC and BD . +OAB = +OBA Thus DAOB is right angle triangle. Sum of two acute angles in the right angle triangle is always 90c. +OAB + +OBA = 90c +OAB + +OAB = 90c 2+OAB = 90c +OAB = 12 # 90c In TACP and TDBP , due to vertically opposite angles = 45c Arc AB subtends +AOB at the centre and +ACB at the remaining part of circle. +APC = +BPD Due to angles is same segment, Thus +ACB = 12 +AOB +CAP = +PDB By AA similarity criterion, = 12 # 90c TACP + TDBP Corresponding sides of similar triangles are always proportional. Thus PA = PC PD PB = 45c Sum of all internal angle of a triangle is always 180 . Thus in DACB +ACB + +CBA + +BAC = 180 45c + 30c + +BAC = 180 PA $ PB = PC $ PD QUESTION 33. +AOB = 2+ACB 75c + +BAC = 180 Comp 2015 33. In the given figure, O is the centre of the circle. If +AOB = 90c and +ABC = 30c, then find +CAO . +BAC = 105c Now +CAO + +OAB = +BAC +CAO + 45c = 105c +CAO = 60c QUESTION 34. SQP 2015 34. Two circles intersect at A and B. Given that AC and AD are respectively the diameters of the circles. Prove that C, B, D are collinear. Sol : Sol : We have redrawn the given figure as below. As per given in question we have shown the figure below. Here we have joined CB , BD and AB. CH 16 : ANGLE PROPERTIES OF CIRCLE AM = MB = AB = 24 = 12 cm 2 2 (i) Radius of the cirlce. Let r be the radius of circle. In right TAMO , by Pythagoras theorem, AO2 = AM2 + OM2 r2 = (12) 2 + (5) 2 Here AC is a diameter of the circle with centre at O and AD is a diameter of the circle with centre at O l . Since the angle in a semi-circle is a right angle, we have +ABC = 90 (i) and +ABD = 90 Adding (i) and (ii), we get (ii) Therefore = 144 + 25 = 169 2 r = (13) 2 r = 13 cm (ii) Length of chord CD . In right TCNO , by Pythagoras theorem CO2 = ON2 + CN2 r2 = 122 + CN2 +ABC + +ABD = 90 + 90 2 13 = 12 + CN = 180 Therefore CBD is a straight line and C, B, D are collinear. QUESTION 35. [r = OC] 2 CN2 = 169 - 144 = 25 Main 2014 35. In the given figure, 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 . If AB = 24 cm, OM = 5 cm, ON = 12 cm, find the (i) radius of the cirlce. (ii) length of chord CD . 2 = 52 CN = 5 cm Since, ON = CD , so N is the mid-point of CD . CD = 2CN = 2 # 5 = 10 cm Hence, radius of circle is 13 cm and length of chord is 10 cm. QUESTION 36. Comp 2014 36. D is a point on the circumcircle of DABC in which AB = AC such that B and OM on the opposite side of line AC. If CD is produced to a point E such that CE = BD , prove that AD = AE . Sol : Sol : (2014) As per given in question we have shown the figure below. We have redrawn the figure as below. Here AB = 24 cm, OM = 5 cm, ON = 12 cm OM = AB and ON = CD Since perpendicular drawn from center to the chord bisects the chord. Thus M is the mid-point of AB . We have, AB = AC and CE = BD ICSE CHAPTERWISE PYQ CLASS 10 MATHS In DABD and DACE , +ACE = 25 But, +DEC and +ACE are alternate angles, because AC | | DE . Given AB = AC Due to angles in the same segment, Thus +ABD = +ACE +DEC = +ACE = 25 Given BD = CE So, by SAS congruence criterion, we have QUESTION 38. DABD , DACE AD = AE QUESTION 37. PAGE 369 SQP 2013 Main 2013 38. In the given figure, +BAD = 65c, +ABD = 70c , +BDC = 45c. (i) Prove that AC is a diameter of the circle. (ii) Find +ACB . 37. In given figure, chord ED is parallel to the diameter AC of the circle. Given +CBE = 65 , find +DEC . Sol : (2013) We have redrawn the figure as below. Sol : We have redrawn the figure as below. Here +BAD = 65c, +ABD = 70c, and +BDC = 45c (i) Prove that AC is a diameter of the circle. Since sum of all angles in a triangles is 180c, in TABD , we can write +BAD + +ABD + +ADB = 180c 65c + 70c + +ADB = 180c 135c + +ADB = 180c +ADB = 180c - 135c Consider the arc CDE. Angles +CBE and +CAE are in the same segment of arc CDE. Thus +CAE = +CBE +CAE = 65 Since AC is the diameter of the circle and the angle in a semi-circle is a right angle, we have = 45c Now +ADB + +BDC = 45c + 45c = 90c So, angle in a semi-cricle is 90c. Hence, AC is a diameter of a circle. (ii) +ACB Since angles in a same segment are equal, from (1) we have +AEC = 90 Now, in DACE , we have +ACE + +AEC + +CAE = 180 +ACE + 90 + 65 = 180 ...(1) +ACB = +ADB = 45c QUESTION 39. SQP 2014, Comp 2005 39. In the given figure O is the centre of the circle. Given that CM = OA and CN = OB . If CH 16 : ANGLE PROPERTIES OF CIRCLE ! CM = CN , prove that C is the midpoint of AB . Angle in a semi-circle is a right angle, thus +ACO = 90 Sol : +ADB = 90 and We have redrawn the figure as below. So we have +ACO = +ADB Thus, in DABD , C and O are points on AD and AB respectively such that CO | | DB . Therefore, by basic proportionality theorem, we have BD = AB OC AO BD = 2AO OC AO BD = 2OC Here in the given circle, O is the centre, OA and OB are the two radii of the circle. From a point C on the circle CM = OA and CN = OB and CM = CN . We have joined OC. In right DOMC and DONC , Common OC = OC Given CM = CN QUESTION 41. Main 2011 41. ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not down to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles. Due to RHS axiom we get DOMC , DONC Now due to CPCT we have +COM = +CON or +COA = +COB Equal arcs subtends equal angles at the centre. ! ! Thus AC = CB ! Hence C is the midpoint of AB . QUESTION 40. Main 2012 40. A circle with centre O, diameter AB and a chord AD are drawn. Another circle is drawn with AO as diameter to cut AD at C. Prove that BD = 2OC . Sol : As per given in question we have shown the figure below. Sol : (2011) We have redrawn the figure as below. ICSE CHAPTERWISE PYQ CLASS 10 MATHS In DAOB : Due to the radius of same circle, Let the radii of three circles be BR = BP = x , AO = PA = y and CR = OC = z Sides of a triangle are AB = 10 cm, BC = 8 cm and AC = 6 cm. x + y = 10 ...(1) x+z = 8 ...(2) and y+z = 6 Adding Eqs. (1), (2) and (3), we get ...(3) 2 (x + y + z) = 10 + 8 + 6 x + y + z = 24 2 ...(4) x + y + z = 12 Subtracting Eq. (1) from Eq. (4), we obtain z = 12 - 10 = 2 cm Subtracting Eq. (2) from Eq. (4), we obtain y y = 12 - 8 = 4 cm Subtracting Eq. (3) from Eq. (4), we obtain x = 12 - 6 = 6 cm QUESTION 42. PAGE 371 SQP 2011 42. In the given figure calculate the measure of DAOC . OA = OB Angle opposite to equal sides of a triangle are equal. Thus +OBA = +OAB +OBA = 30 In DBOC : Due to the radius of same circle, ...(i) OB = OC Angle opposite to equal sides of a triangle are equal. Thus +OBC = +OCB +OBC = 40 (ii) Using eq. (i) and (ii), we have +ABC = +OBA + +OBC = 30 + 40 = 70 Angle subtended by an arc of a circle at the circle is double the angle subtended by the arc on the circumference. Therefore +AOC = 2+ABC = 2 # 70 = 140 QUESTION 43. Main 2010 43. In the given figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm, respectively. Sol : We have joined BO and redrawn the figure as below. Sol : (2010) We have redrawn the given figure as below CH 16 : ANGLE PROPERTIES OF CIRCLE We have OA = OC = 15 cm, AB = 24 cm, CD = 18 cm A perpendicular drawn to a chord from the centre, always bisects the chord. Here AB and CD are chord, OM and ON are perpendicular drawn from center. Thus AM = MB = AB = 24 = 12 cm 2 2 and CN = ND = CD = 18 = 9 cm 2 2 In the figure, Reflex +AOC = 360 - 144 In right TAMO , using Pythagoras theorem, AO2 = OM2 + MA2 152 = OM2 + 122 OM2 = 225 - 144 = 81 OM = 9 cm Similarly, in TCNO , using Pythagoras theorem, = 216 Major arc AC subtends +AOC at the centre and +ABC at the remaining part of the circle. Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. +ABC = Reflex +AOC = 12 # 216 OC2 = ON2 + CN2 = 108 152 = ON2 + 92 Due to linear pair we have ON2 = 225 - 81 = 144 = (12) 1 2 +ABC + +ABD = 180 2 108 + +ABD = 180 ON = 12 cm +ABD = 72 MN = OM + ON In DABD, AB = AD Same side make equal angle in any triangle. = 9 + 12 = 21 cm. Hence, the value of MN is 21 cm. QUESTION 44. +ADB = +ABD SQP 2010 44. In the circle with centre O, +AOC = 144 , If AB = AD , find +ADB = 72 QUESTION 45. Main 2008 45. In the given figure, AE and BC intersect each other at point D . If +CDE = 90c, AB = 5 cm, BD = 4 cm and CD = 9 cm, find DE . Sol : We have redrawn the given figure as below Sol : (2008) We have redrawn the given figure as below ICSE CHAPTERWISE PYQ CLASS 10 MATHS Here we have +CDE = 90c, AB = 5 cm, PAGE 373 have +APB + +PBA + +BAP = 180c BD = 4 cm, CD = 9 cm In right TADB , Due to vertically opposite anlges 90c + 45c + +BAP = 180c +BAP = 180c - 135c +ADB = +CDE = 90c Using Pythagoras theorem, = 45c Since angles in a same segment are equal, AB2 = AD2 + BD2 2 2 (5) = AD + (4) +PQB = +BAP = 45c Hence, the value of +PQB is 45c. 2 AD2 = 25 - 16 = 9 QUESTION 47. AD = 3 cm If two chords intersect internally or externally, then the product of the lengths of their segment is equal. Thus Main 2005 47. In the alongside figure, O is the centre of the circle and +AOC = 160c. Prove that 3+y - 2+x = 140c. DA # DE = DB # DC 3 # DE = 4 # 9 DE = 4 # 9 = 12 cm 3 Hence, the value of DE is 12 cm. QUESTION 46. Main 2007 46. In the given figure, O is the centre of the circle and +PBA = 45c. Calculate the value of +PQB . Sol : (2007) We have redrawn the given figure as below Sol : (2005) We have redrawn the given figure as below Angle that an arc of a circle, subtend at the centre is twice that angle, which subtend at any point on the remaining part of a circle. Thus +AOC = 2 # +ADC +AOC = 2x But given that +AOC = 160c, thus we have 160c = 2x x = 160c = 80c 2 Here +PBA = 45c In this give figure, AOB is a straight line, thus Similarly using same concept we have 2y = 360c - 160c y = 360c - 160c 2 +AOB = 180c Since, AB is a diameter of circle, so TAPB is in semi-circle. Therefore +APB = 90c Using angle sum property in triangle TAPB we = 200c = 100c 2 Now, 3+y - 2+x = 2 # 100c - 2 # 80c CH 16 : ANGLE PROPERTIES OF CIRCLE = 300c - 160c = 140c Hence proved. QUESTION 48. Main 2000 48. In the given figure, chords AB and CD , when extended meet at X . Given AB = 4 cm, BX = 6 cm, XD = 5 cm. Calculate the length of CD . (i) Prove that TPAB - TPCD . (ii) Find the length of CD. (iii) Find area of TPAB : Area of TPCD . Sol : We have redrawn the figure as below. Sol : (2000) We have AB = 4 cm, BX = 6 cm, XD = 5 cm We have redrawn the given figure as below Here From the given figure, we have XB $ XA = XD $ XC If two chords intersects internally or externally, then the product of the lengths of the segments are equal. Therefore XB $ (XB + BA) = XD $ (XD + DC) PC = 5 cm (i) TPAB - TPCD . In TPAB and TPCD Angle in the same segment of a circle are equal, thus +PBA = +PDC Due to vertically opposite angles we have +APB = +CPD By AA similarity criterion, we have 6 $ (6 + 4) = 5 $ (5 + DC) TPAB + TPCD (ii) Length of CD Corresponding sides of similar triangles are always proportional. Since, TPAB + TPCD we have AP = AB CP CD 6 # 10 = 25 + 5DC 60 = 25 + 5DC 5DC = 60 - 25 DC = 35 = 7 cm 5 Hence, the value of DC is 7 cm. 7.5 = 9 5 CD CD = 5 # 9 = 6 cm 7.5 FOUR MARK QUESTIONS QUESTION 49. AB = 9 cm, PA = 7.5 cm and Main 2020 49. In the given figure AB = 9 cm, PA = 7.5 cm and PC = 5 cm. Chords AD and BC intersect at P. (iii) Area of TPAB : Area of TPCD . The ratio of areas of two similar triangles in equal to the square of the ratio of their corresponding sides area of TPAB = AB 2 area of TPCD CD2 ICSE CHAPTERWISE PYQ CLASS 10 MATHS In DBOD , we have 2 = 92 = 9 4 6 area of TPAB : area of TPCD = 9 : 4 QUESTION 50. PAGE 375 +OBD + +ODB = +AOB 40c + +ADB = 70c SQP 2020 50. In the given circle with centre O, +AOB = 70 and+BEC = 50 . (i) Find +AEB (ii) Find +ADB (iii) Find +DAE . +ADB = 30c (iii) +DAE . The sum of angles of a linear pair is always equal to 180 . +AOE + +AOB = 180 +AOE + 70c = 180 +AOE = 110 Due to radii of same circle, OA = OE Same side make equal angle in any triangle. Thus +OAE = +OEA The sum of all angles of a triangle is always equal to 180 . Thus +OAE + +OEA + +AOE = 180 +OAE + +OAE + 110c = 180 Sol : 2+OAE + 110c = 180 We have redrawn the given figure as below 2+OAE = 70 +OAE = 12 # 70 +DAE = 35 QUESTION 51. SQP 2019 51. In the given figure O is the centre of the circle and AB is a diameter. If AC = BD and +AOC = 72 , find : (i) +ABC (ii) +BAD (iii) +ABD (i) +AEB Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. +AOB = +AEB +AEB = 12 +AOB = 35 (ii) +ADB The angle in a semi-circle is always a right angle. +ECB = 90 Due to angle sum property of a triangle, +BEC + +ECB + +CBE = 180 50c + 90c + +CBE = 180 140c + +CBE = 180 +CBE = 40 Sol : We have redrawn the given figure as below. CH 16 : ANGLE PROPERTIES OF CIRCLE Since +ADB is angle in semi-circle +ADB = 90 (i) +ABC Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Sol : We have redrawn the figure as below. +AOC = 2+ABC Thus +ABC = 12 +AOC = 12 # 72c = 36 (ii) +BAD Given AC = BD Equal arcs make equal angle. Thus +BAD = +ABC = 36 In the given circle DABC is inscribed in it Here ! E is midpoint of BC , EOD is diameter. (iii) +ABD In DABD +ABC = 72 , Here +ACB = 48 +ABD + +BAD + +ADB = 180 +ABD + 36 + 90 = 180 (i) +BAE Due to angles in the same segment, +ABD + 126 = 180 +BAE = +BCE +ABD = 54 QUESTION 52. SQP 2018 52. DABC has been inscribed in the circle with centre ! O. E is the midpoint of BC . If +ABC = 72 and +ACB = 48 , find (i) +BAE (ii) +CBE (iii) +DAC (iv) +AEB In DABC , +ABC = 72 and +ACB = 48 Now in DABC , +BAC + +ABC + +ACB = 180 +BAC + 72 + 48 = 180 +BAC + 120 = 180 +BAC = 60 Due to angles subtended by equal arc, +BAE = +CAE But +BAE + +CAE = 60 +BAE + +BAE = 60 +BAE = 12 # 60 = 30 (ii) +CBE ICSE CHAPTERWISE PYQ CLASS 10 MATHS PAGE 377 AN = NB = AB = 24 = 12 cm 2 2 Due to angles in the same segment, +CBE = +CAE and = 30 (iii) +DAC CM = MD = CD = 10 = 5 cm 2 2 In right TANO , using Pythagoras theorem, OA2 = ON2 + AN2 Due to angle in a semi-circle, 132 = ON2 + 122 +DAE = 90 ON2 = 169 - 144 = 25 +DAC + +CAE = 90 ON = 5 cm Similarly, in TOMC , using Pythagoras theorem, +DAC + 30 = 90 +DAC = 90 - 30 OC2 = OM2 + CM2 = 60 132 = OM2 + 52 (iv) +AEB In DABE , OM2 = 169 - 25 +ABE + +BAE + +AEB = 180 = 144 = 122 (72 + 30 ) + 30 + +AEB = 180 OM = 12 cm 102 + 30 + +AEB = 180 MN = OM + ON 132 + +AEB = 180 = 5 + 12 = 17 cm. Hence, the value of MN is 17 cm. +AEB = 180 - 132 = 48 QUESTION 53. QUESTION 54. Main 2017 53. AB and CD are two parallel chords of a circle such that AB = 24 cm and CD = 10 cm. If the radius of the circle is 13 cm, find the distance between the two chords. SQP 2016. Comp 2012 54. AC and BD are chords of a circle that bisect each other. Prove that: (i) AC and BD are diameters (ii) ABCD is a rectangle. Sol : As per given in question we have shown the figure below. Sol : (2017) We have redrawn the given figure as below (i) AC and BD are diameters In DAOB and DCOD : Suppose AC and BD bisect each other at point O. We have OA = OC = 13 cm, AB = 24 cm, CD = 10 cm A perpendicular to a chord drawn from the centre, always bisects the chord. Here AB and CD are chord, OM and ON are perpendicular drawn from center. Thus Then OA = OC and OB = OD Due to vertically opposite angles, +AOB = +COD So, by SAS congruence criterion, we have (i) ...(ii) CH 16 : ANGLE PROPERTIES OF CIRCLE DAOB , DCOD Corresponding side of similar triangles are in the same ratio. Therefore AB = CD ! ! AB , CD ...(iii) Similarly, we have ! ! ...(iv) BC , DA Adding (iii) and (iv), we have ! ! ! ! AB + BC = CD + DA ! ! ABC = CDA ! Thus AC divides the circle into two equal parts. ! Hence AC is a diameter of the circle. Similarly, we can prove that BD is also a diameter of the circle. (ii) ABCD is a rectangle. Since AC and BD are diameters of the circle and +ABC and +ADC are angle in semi-circle. Therefore, Here +TRS = 65c Here we have joined SQ and SP is a tangent. Since tangents at any point of a circle and radius are perpendicular to each other. Therefore +OSR = 90c Using angle sum property of a triangle TSTR we have +STR + +TRS + +TSR = 180c x + 65c + 90c = 180c x = 180c - 155c +ABC = 90 = 25c Angle made by a chord at centre is double the angle made by it at remaining part. Thus = +ADC Also, +BCD = +BAD = 90 From (iii) and (iv), we have y = 2x AB = CD = 2 # 25c = 50c and BC = DA . Hence, ABCD is a rectangle. QUESTION 55. Comp 2011 55. In the figure below, O is the centre of the cirlce and SP is a tangent. If +SRT = 65c, find the values of x , y and z . y = 2x & y - = 50c Using angle sum property of a triangle TOPS we have +SOP + +OPS + +OSP = 180c 50c + z + 90c = 180c z = 180c - 140c = 40c QUESTION 56. Sol : (2015) We have redrawn the given figure as below Comp 2010 56. In given figure ABCD is a cyclic quadrilateral in the circle with centre O. +ADC = 50 and +AOB = 68 . (i) Find +ACB (ii) Find +OBA . (iii) Find +BOC . ICSE CHAPTERWISE PYQ CLASS 10 MATHS PAGE 379 = 100 - 68 = 32 QUESTION 57. Comp 2009 57. In the given figure BC is a chord with centre O. A is a point on an arc BC as shown figure. Sol : We have redrawn the given figure as below. (i) Prove that +BAC + +OBC = 90 , if A is the point on the major arc. (ii) Prove that +BAC - +OBC = 90 , if A is the point on the minor arc. Sol : We have redrawn the figure as below. (i) +ACB Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. +AOB = +ACB +ACB = 12 +AOB = 34 (ii) +OBA Due to radii of same circle, OA = OB Same side make equal angle in any triangle. In +OAB , +OBA = +OAB = 180 - 68 2 = 112 = 56 2 (iii) +BOC Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. +AOC = 2+ADC = 2 # 50 = 100 Now +BOC + +AOB = +AOC +BOC = +AOC - +AOB The arc BC makes +BOC = z at the centre and +BAC = x at a point on the circumference. In both case z = 2x (i) In DOBC Sum of all internal angle of a triangle is always 180 . Thus +OBC + +OCB + +BOC = 180 y + z + y = 180 z + 2y = 180 2x + 2y = 180 x + y = 90 +BAC + +OBC = 90 Hence Proved (ii) In DOBC , Sum of all internal angle of a triangle is always 180 . Thus +OBC + +OCB + +BOC = 180 y + y + t = 180 CH 16 : ANGLE PROPERTIES OF CIRCLE t = 180 - 2y Now, Since AF is the bisector of +A , z = 360 - t +DAE = +BAE z = 360 - ^180 - 2y h Since AD | | BC , due to corresponding angles, (i) z = 180 + 2y Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference. Thus z = 2x (ii) From (i) and (ii) we have +CEF = +DAE From (i) and (ii) we get (ii) +CEF = +BAE Due to angles in the same segment, (iii) +BCF = +BAF +ECF = +BAE From (iii) and (iv) we get 2x = 180 + 2y 2x - 2y = 180 (iv) +ECF = +CEF Equal opposite side in triangle make equal angles. x - y = 90 +BAC - +OBC = 90 (i) Hence Proved QUESTION 58. Comp 2008 58. The adjoining figure shows a quadrilateral ABCD in which AD | | BC inscribed in a circle. The bisector of +BAD cuts BC at E and the circle at F. Prove that (i) EF = FC (ii) BF = DF Thus EF = FC (ii) BF = DF . Due to angles in the same segment, and +DAF = +DBF (v) +BAF = +BDF (vi) Since AF is the bisector of +A , +DAF = +BAF From (iii) and (iv) we get (vii) +DBF = +BDF Equal opposite side in triangle make equal angles. BF = DF QUESTION 59. Hence proved. Main 2006 59. In the given circle arc RS subtends 42 angle at the centre O in the given figure. PR and QS extended meet at a point T. (i) Find +RQS (ii) Find +RTQ . Sol : We have redrawn the figure as below. Sol : We have redrawn the given figure as below. Here ABCD is a quadrilateral inscribed in the circle and AD | | BC . The bisector of +BAD , intersects BC at E and the circle at F. We have joined BD. (i) EF = FC ICSE CHAPTERWISE PYQ CLASS 10 MATHS (i) +RQS Angle subtend by an arc at the centre is twice the angle at the remaining part of the circle. Thus PAGE 381 (i) +ABD Since +ADB is angle in a semi-circle, +ADB = 90 Due to angles in the same segment, +ROS = 2+RQS +RQS = +ROS 1 2 +DAB = +DCB = 12 # 42 = 22 Sum of two acute angles in the right angle is always 90 . In 3 ADB , = 21 (ii) +RTQ Angle in a semi-circle is always 90 . +DAB + +ABD = 90 +PRQ = 90 Due to linear pair we have 22c + +ABD = 90 +QRT = 90 Thus 90 is right angle triangle. The sum of acute angles of a right triangle is always equal 90 . Thus +ABD = 68 (ii) +BDE . Due to vertically opposite angle we have +RQT + +RTQ = 90 +BED = +BEC +RQS + +RTQ = 90 = 40 Now in DEBD , 21c + +RTQ = 90 +DEB + +EBD + +BDE = 180 +RTQ = 69 QUESTION 60. 40 + 68 + +BDE = 180 Main 2007 60. In the given figure AB is a diameter of the circle. Given that +AEC = 40 and +BCE = 22 . (i) Find +ABD (ii) Find +BDE . 108 + +BDE = 180 +BDE = 72 Hence +ABD = 68 and +BDE = 72 QUESTION 61. Main 2005 61. A circle with centre O , diameter AB and a chord AD si drawn. Another circle is drawn with AO as diameter to cut AD at C . Prove that BD = 20C . Sol : (2005) As per given in question, we have drawn the given figure as below Sol : We have redrawn the given figure as below. Here two circles intersect each other at A. AB CH 16 : ANGLE PROPERTIES OF CIRCLE and AD are diameter and chord of bigger circle respectively. Also, AO is diameter of smaller circle. We have joined CO and DB . Now in TOCA and TBDA , Since angle in semi-circle is right angle, we have +OCA = +BDA = 90c Due to common angle we have Pythagoras theorem PQ2 = PR2 + QR2 (17) 2 = (9 + x) 2 + (2 + x) 2 289 = 81 + x2 + 18x + x2 + 4x + 4 289 = 2x2 + 22x + 85 2x2 + 22x - 204 = 0 +OAC = +BAD By AA axiom of similarity, we get x2 + 11x - 102 = 0 TOCA + TBDA Since in similar triangle sides are proportional, thus OA = OC BA BD OA = OC 2OA BD [BA = 2AO] BD = 20C Hence proved. x2 + 17x - 6x - 102 = 0 (x - 6) (x + 17) = 0 x = - 17 , 6 Thus x = 6 cm since radius cannot be negative. FIVE MARK QUESTIONS QUESTION 62. Main 2004 62. P and Q are centres of circles of radii 9 cm and 2 cm, respectively, PQ = 17 cm. R is the centre of a circle of radius x cm, which touches the above circles externally. Given that +PRQ = 90c, write an equation in x and solve it. Sol : (2004) As per given in question, we have drawn the given figure as below Here P and Q are the centre of circles, whose radii are 9 cm and 2 cm, respectively. Also, PQ = 17 cm A circle with centre R and radius x touches above circles. Therefore PR = 9 + x and QR = 2 + x . Since PRQ is a right angled triangle, using END

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