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Q1. The length of chord which is at a distance of 12 cm from centre of circle of radius 13 cm is:

• 1) 13 cm
• 2) 10 cm
• 3) 5 cm
• 4) 12 cm

Solution

Using Pythagoras theorem: OA² = OM² + AM² AM² = OA² - OM² AM² = 13² - 12² AM² = 169 - 144 AM² = 25 AM = 5 cm AB = 2 AM AB = 2 5 cm = 10 cm

Q2. If AB = 12 cm, BC = 16 cm and AB BC, then radius of the circle passing through A,B and C is

• 1) 12 cm
• 2) 8 cm
• 3) 10 cm
• 4) 6 cm

Solution

Since the vertices of the triangle ABC form a right angled triangle, right angled at B, ABCOA is a semicircle, where O is the centre of the circle. Thus, AC is the diameter of the circle passing through the points A, B and C. Using Pythagoras theorem in ABC, AC² = AB² + BC² = (12 cm)² + (16 cm)² = 144 cm² + 256 cm² = 400 cm² AC = 20 cm Radius of the circle = =10 cm

Q3.

• 1) 80^o, 100^o
• 2) 40^o, 120^o
• 3) 50^o, 100^o
• 4) 60^o, 100^o

Solution

Q4. Prove that the perpendicular from the centre of the circle to a chord bisects the chord.

Solution

Q5. The two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection. Prove that the chords are equal.

Solution

Draw OM AB and ON CD MOP = 180^o - 90^o - MPO = 90^o - MPO = 90^o - NPO = NOP In OMP and ONP, OMP = ONP = 90^o (by construction) MOP = NOP (shown above) OP = OP (common) Chords AB and CD are equidistant from centre O. Therefore, AB = CD (Chords which are equidistant from the centre are equal)

Q6.

Solution

Q7. In the given figure, if OAB = 40^o then ACB is equal to:

• 1) 70^o
• 2) 50^o
• 3) 40^o
• 4) 60^o

Solution

OA = OB = radius of the circle. Therefore, OAB = OBA = 40^o In AOB + OBA + AOB = 180⁰ 40⁰ + 40⁰ + AOB = 180⁰ AOB = 100⁰ Therefore, ACB = 50^o Since, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Q8. O is the centre of the circle and PQ is a diameter. If ROS is equal to 40^o find RTQ.

Solution

Q9. In the given figure, M is 75^o then O is equal to.

• 1) 100^o
• 2) 110^o
• 3) 105^o
• 4) 85^o

Solution

The sum of a pair of opposite angles of a cyclic quadrilateral is 180º Therefore, _{M + O = 75º + O = O =} 105^o

Q10. For what value of x in the figure, points A, B, C and D are concyclic?

• 1) 11^o
• 2) 9^o
• 3) 10^o
• 4) 12^o

Solution

Sum of opposite angles of a cyclic quadrilateral is 180.

Q11. In a circle of radius 5 cm, NG and AG are two chords such that NG = AG = 6 m. Find the length of the chord AN.

Solution

Q12. AD is a diameter of a circle and AB is chord. If AB = 24 cm, AD = 30 cm, the distance of AB from the centre of the circle is :

• 1) 8 cm
• 2) 15 cm,
• 3) 9 cm
• 4) 17 cm

Solution

The distance of AB from the centre of the circle is 9 cm.

Q13. The perpendicular distance of a chord 8 cm long from the centre of a circle of radius 5 cm is

• 1) 9 cm
• 2) 4 cm
• 3) 3 cm
• 4) 2 cm

Solution

Let OP be perpendicular to the chord AB. We know that perpendicular drawn from the centre of a circle to a chord bisects the chord. So, AP = BP = = 4 cm We know that triangle OPB is right-angled at P, so we can use Pythagoras theorem: OP² + BP² = OB² OP² + 4² = 5² OP² + 16 = 25 OP² = 25 - 16 OP² = 9 OP = 3 cm

Q14. In the figure, two circles intersect at B and C. Through B two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that ACP = _DCQ.

Solution

_{segments AD and PQ intersect at B} PBA = DBQ ….(i) (Vertically opposite angles) _{In major circle, angle subtended by minor arc AP: ....(ii) In minor circle, angle subtended by minor arc DQ: ….(iii) from (i) ,(ii) and (iii)}

Q15. The region bounded by an arc and a chord of a circle is called a ______.

• 1) Sector
• 2) Semicircle
• 3) Segment
• 4) None of the above

Solution

The region bounded by an arc and a chord of a circle is called a segment.

Q16. AOB is a diameter of the circle and C, D, E are any three points on the semicircle. Find the value of ACD + BED.

Solution

Join BC. Then ACB = 90^o [Angle in a semi circle] Since DCBE is a cyclic quad. BCD + BED = 180^o Adding ACB on both sides BCD + BED + ACB = ACB + 180^o. (BCD + ACB) + BED = 90^o + 180^o ACD + BED = 270^o

Q17. If C = D = 50^o then four points A, B, C, D:

• 1) Are collinear
• 2) Are con-cyclic
• 3) Do not lie on same circle
• 4) A,B,D and A,B,C lie on different circles

Solution

If a line segment joining two points subtends equal angles at the two points lying on the same side of the line segment, then the four points are con-cyclic.

Q18.

Solution

Q19. In a cyclic quadrilateral ABCD, if AB||DC and B = 70^o find the measures of the remaining angles of the quadrilateral.

Solution

In cyclic quad. ABCD, if B = 70^o Then D = 180^o - 70^o = 110^o (opp. angles of cyclic quad. are supplementary) Since AB||CD, B + C = 180^o (adj. angles of cyclic quad. are supplementary) And A + D = 180^o (adj. angles of cyclic quad. are supplementary) C = 110^o, A = 70^o

Q20. An equilateral ABC is inscribed in a circle with centre O. The measure of BOC is:

• 1) 120^o
• 2) 110^o
• 3) 130^o
• 4) 100^o

Solution

Since ABC is an equilateral triangle, therefore, _{BAC = 600 We know,} The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Therefore, _{BOC =} 120^o

Q21. Find the length of the chord, which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

Solution

In rightOMA, by Pythagoras theorem, OA² = OM² + AM²     (The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.) AM² = 5² - 3² = 16 AM = 4 cm AB = 2AM = 2 4 cm = 8 cm

Q22. In the figure, AB and CD are two equal chords of a circle with centre O. OP and OQ are perpendiculars on chords AB and CD respectively. If POQ = 150^o, find APQ.

Solution

As AB = CD So, OP = OQ  (equal chords are equidistant from the centre) 1 = 2      (angles oppsite to equal sides are equal) 1 + 2 + POQ = 180^o 1 + 1 + 150 ^o = 180^o 1 = 15^o Since APB is a line segment BPO + 1 + APQ = 180^o 90^o + 15 ^o + APQ = 180^o APQ = 75^o

Q23. In the given figure, AB is a side of a regular hexagon and AC is a side of a regular octagon inscribed in a circle with centre O. Calculate angles AOB and AOC.

Solution

Q24. In the figure, O is the center of the circle. If ACB = 30^o then ABC equal to:

• 1) 60^o
• 2) 15^o
• 3) 30^o
• 4) 45^o

Solution

In the given figure, ACB = 30^o BC is the diameter, therefore, CAB = 90^o(Angle in a semi-circle) Now, in ABC ACB + CAB + ABC = 180^o (sum of angles of a triangle = 180⁰) 30⁰ + 90⁰ + ABC = 180⁰ ABC = 60^o

Q25. The vertices of a parallelogram lie on a circle. Prove that its diagonals are equal.

Solution

B = D (opposite angles of a parallelogram) But B + D = 180^o (opposite angles of a cyclic quad. are supplementary) B = D = 90^o We know that, a parallelogram with one angle as 90^o is a rectangle. So, ABCD is a rectangle. AC = BD (diagonals of a rectangle are equal)

Q26. In the figure, ABCD is a parallelogram. The circle through A, B and C intersects CD produced at E. Prove that AE = AD.

Solution

1 = 2 (Opp. angles of a parallelogram are equal)…(i) 1 + AED = 180^o (Opp. Angles of cyclic Quad.)…(ii) 2 + ADE = 180^o (Linear pair)…….(iii) AED = ADE….from (i),(ii),(iii). AD = AE (in ADE, lines opp. to equal angles are equal)

Q27. In the figure, B and E are points on line segment AC and DF respectively. Prove that AD||CF.

Solution

Join B and E. ABED is a cyclic quadrilateral. ADE = EBC … (1) (exterior angle of a cyclic quadrilateral = Interior opposite angle) Similarly, EFCB is a cyclic quadrilateral. EBC + CFE = 180^o … (2) (opposite angles of a cyclic quadrilateral are supplementary) From (1) and (2), ADE + CFE = 180^o But these are interior angles made by the transversal DF on the same side of the lines AD and CF. Since, these angles are supplementary, therefore, AD||CF

Q28. AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, the distance of AB from the centre of the circle is :

• 1) 17 cm
• 2) 4 cm
• 3) 15 cm
• 4) 8 cm

Solution

Let O be the centre.Draw OC perpendicular to AB. We know that the perpendicular from the centre to a chord bisects the chord. So, CA==15 cm OA =17 cm   Using Pythagoras theorem: OC² = OA² - AC² OC² = (17 17) - (15 15) OC² = 289 - 225 OC² = 64 OC = 8 cm

Q29.

Solution

Q30. AB and CD are two parallel chords of a circle such that AB = 12 cm and CD = 24 cm. The chords are on the opposite sides of the centre and the distance between them is 17 cm. Find the distance between them.

Solution

Q31. 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 the chords.

Solution

Q32.

Solution

Q33. O is the circumcentre of the ABC and D is the mid point of the base BC. Prove that BOD = A.

Solution

OD BC [A line drawn through the centre of a circle bisecting a chord is to the chord] In rt. OBD and OCD OB = OC (radii) OD = OD (common) ODB = ODC [each 90^o] (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.) From (i) and (ii) BOD = A

Q34. Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Solution

Let arc PQ of a circle with centre O, subtends POQ at the centre and PAQ at a point P which lies on the remaining part of the circle. To prove: POQ = 2 PAB Proof: There are 3 cases: (a) Arc PQ is minor (b) Arc PQ is a semi-circle (c) Arc PQ is major In all the three cases, we have: BOQ = OAQ + AQO ...(1) (exterior angle is equal to the sum of int. opp. angles.) Also, in OAQ, OA = OQ ... (Radii of a circle) OAQ = OQA ...(2) From (1) and (2), BOQ = 2OAQ …..(3) Similarly, BOP = 2OAP ……(4) Adding (3) and (4), BOP + BOQ = 2(OAP + OAQ) Or, POQ = 2PAQ ...(5)

Q35. In the figure, O is the centre of the circle and BAC = 60^o. Find the value of x.

Solution

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.) BOC = 2BAC = 2 60^o = 120^o x = 360^o - 120^o = 240^o

Q36.

Solution

      ΔABC and ΔADC are right triangles with AC as common hypotenuse. Two triangles together make a quadrilateral ABCD In quadrilateral ABCD,  ABC +  ADC = 90° + 90° = 180° Now, sum of opposite angles of a quadrilateral is 180°.      (∵ by thm.) ∴ ABCD is a cyclic quadrilateral. Then,  CAD =  CBD                    (∵ angles in the same segment of a circle are equal)

Q37. Two congruent circles interest each other at points A and B. Through point A, line segment PAQ is drawn so that P and Q lie on the two circles. Prove that BP = BQ.

Solution

Join AB. AB is the common chord. We know that equal chords subtends equal angles,and here two circles are congruent.

Q38. Prove that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Solution

Given: A circle with centre O in which PQ is any arc which makes POQ at the centre and PRQ on the circle at R. To Prove: POQ = 2PRQ Construction: Join RO and produce which meet the circle at point S. Proof : In OPR OP = OR (Radii of the same circle) 1 = 3 (Opposite angles of equal sides) Exterior 5 = 1 + 3 (by exterior angle theorem) 5 = 1 + 1 5 = 21 ...(1) Similarly, 6 = 22 ...(2) On adding equation (1) and (2) 5 + 6 = 21 + 22 POQ = 2[1 + 2] POQ = 2PRQ

Q39. In the figure, chord AB of circle with centre O, is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ACD = y and AOD = x, show that x = 3y.

Solution

In OBC, OB = BC _{BOC = BCO = y (angles opp. to equal sides are equal) ABO is the exterior angle of} BCO So, ABO = 2y (ext. angle is equal to the sum of int. opp. Angles) ABO = BAO (OA and OB are radius of same circle) Similarly, AOD is the exterior angle of AOC. AOD = BAO + ACO = 2y + y = 3y

Q40. In the figure, if AB is the diameter of the circle, then the value of x is:

• 1) 80^o
• 2) 50^o
• 3) 90^o
• 4) 40^o

Solution

AB is the diameter. Therefore, (angle in a semicircle) In ,

Q41.

Solution

Q42. A regular polygon is inscribed in a circle. If a side subtends an angle of 36^o at the centre, find the number of sides of the polygon. Name the polygon.

Solution

Q43. ABCD is a cyclic quadrilateral. Find ADB

Solution

_{In a cyclic quadrilateral, opposite angles are supplementary.} DAB + BCD = 180^o DAB + 120^o = 180^o DAB = 60^o In DAB, DAB + ABD + ADB = 180^o 60^o + 50^o + ADB = 180^o ADB = 180^o - 110^o = 70^o

Q44. In the given figure, O is the centre of the circle. The angle subtended by arc ABC at the centre is 140^o. AB is produced to P. Determine ADC and CBP

Solution

Q45. ABDC is a cyclic quadrilateral and AB = AC. If ACB = 70^o find BDC.

Solution

In AB = AC ABC = ACB = 70^o BAC + 70^o+ + 70^o = 180^o BAC = 180^o - 140^o = 40^o ABCD is a cyclic quadrilateral and opposite angles of a cyclic quadrilateral are supplementary BAC + BDC = 180^o 40^o+ BDC = 180^o BDC = 140^o

Q46.

Solution

Q47. In the given figure, if PQR is 40^o, then the value of PSR is.

• 1) 40^o
• 2) 80^o
• 3) 90^o
• 4) 60^o

Solution

Angles in the same segment are equal. Therefore, PQR = PSR = 40^o

Q48. In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Find the distance between the chords if they are (i) on the same side of the centre. (ii) on the opposite sides of the centre.

Solution

Q49.

Solution

Q50. Draw a circle whose diameter is 4 cm. Mark the points P, Q, and R in the interior, exterior and on the circle respectively.

Solution

Q51. In the figure, if BAC = 60^o, ACB = 20^o find ADC.

Solution

ABC = 180^o - (BAC + BCA) (Angle sum property) = 180^o - 80⁰ = 100^o ABCD is a cyclic quadrilateral, ABC + ADC = 180^o ADC = 180^o - 100 = 80^o

Q52. In the figure, quadrilateral PQRS is cyclic. If P = 80^o, thenR =

• 1) 40^o
• 2) 120^o
• 3) 100^o
• 4) 80^o

Solution

The sum of either pair of opposite angles of a cyclic quadrilateral is 180º. Hence, P + R = 180^o 80^o + R = 180^o R = 100^o

Q53. In the figure, if O is the centre of the circle and BOA = 120^o, then the value of x is

• 1) 30^o
• 2) 60^o
• 3) 90^o
• 4) 120^o

Solution

_{OA = OC (radii of same circle) OAC = OCA(angles opposite to equal sides)} BOA = OAC + OCA (Exterior angle of a triangle is equal to sum of interior angles) 120⁰ = 2OAC = 2x x = 60⁰

Q54. AB and AC are two equal chords of a circle whose centre is O. If OD is perpendicular to AB and OE is perpendicular to AC, prove that ADE is an isosceles triangle.

Solution

   

Q55. In the figure A, B and C are three points on a circle such that the angle subtended by the chords AB and AC at the centre are 120^o and 80^o respectively. Determine angle BAC.

Solution

Q56. Draw a line segment AB of length 4.8 cm. Construct two semi-circles on opposite sides of AB such that half the length of AB is the diameter of each of the semi-circles.

Solution

Half the length of AB = 2.4. Let O be the mid point of AB. Construct two semi-circles on opposite sides of AB, with AO and OB as diameters.

Q57. In the figure, O is the centre of the circle and .

Solution

Since, angle at the centres is twice the angle at the remaining part of circumference.

Q58. In the figure, O is the centre of a circle. Prove that x + y = z.

Solution

Q59.

Solution

Q60. In the figure, O is the centre of the circle, OM BC, OL AB, ON AC and OM = ON = OL. Is ABC equilateral? Given reasons.

Solution

OL AB, OM BC and ON AC OM = ON = OL Perpendicular distance of chords from the centre of a circle AB = BC = AC [chords equidistant from the centre of a circle are equal.] ABC is an equilateral triangle.

Q61. Two parallel chords of a circle whose diameter is 13 cm are respectively 5 cm and 12 cm. Find the distance between them if they lie on opposite sides of centre.

Solution

Let OM = x cm and ON = y cm _{.....(OM is perpendicular from centre to the chord)} ....(ON is perpendicular from centre to the chord) Distance MN

Q62. Prove that quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.

Solution

AH, BF, CF and DH are bisectors of A, B, C and D respectively which form quadrilateral EFGH. FEH = AEB = 180^o - EAB - EBA (angles of AEB) = 180^o - (A + B) FGH = DGC = 180^o - GDC - GCD (angles of DGC) = 180^o - (D + C) FEH + FGH = 180^o - EFGH is a cyclic quadrilateral (If the sum of a pair of opposite angles of a quadrilateral is 180º, then the quadrilateral is cyclic)

Q63. O is the centre of the circle as shown in the figure. Find CBD

Solution

Take a point E on the circle, join AE and CE. AEC + ABC = 180^o (Opposite Angles of a cyclic quadrilaterals) ABC = 130^o ABC + CBD = 180^o (linear pair) 130^o + CBD = 180^o CBD = 50^o

Q64. If two intersecting chord of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Solution

Let AB and CD be the two chords. Construct OM AB and ON CD In OME and ONE, 1 = 2        (given) OME = ONE = 90^o OE = OE          (common) By AAS congruence rule, OM = ON        (c.p.c.t) Hence, AB = CD (Since, chords equidistant from the centre are equal)

Q65. Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Solution

Q66. AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters of the circle and (ii) ABCD is a rectangle.

Solution

Q67. Two circles intersect at two points A and B. AD and AC respectively are diameters to the two circles. Prove that B lies on the line segment DC.

Solution

Join AB. ABD = 90^o (angle in a semicircle) ABC = 90^o (angle in a semicircle) ABD + ABC = 90^o + 90^o = 180^o DBC is a straight line and hence, B lies on line DC

Q68.

Solution

Q69.

Solution

Q70.

Solution

Q71. In the given figure, equal chords AB and CD of a circle C(O, r) cut at right angles at E. If M and N are the mid points of AB and CD respectively, prove that OMEN is a square.

Solution

Q72. In the given figure, AB = BC = CD and angle ABC is 132^o. Calculate angle COD.

Solution

  

Q73. Two circles of radii 10 cm and 8 cm intersect each other at two points, and the length of the common chord is 12 cm. Find the distance between their centers.

Solution

Let O and O' be the centers of the circle of radii 10 cm and 8 cm respectively. Let PQ be their common chord. We have, OP = 10 cm O'P = 8 cm PQ = 12 cm _{(Perpendicular from the center of the circle to a chord bisects the chord).} In right OLP, we have OP² = OL² + PL² In right O'LP we have (O'P)² = (PL)² + (O'L)² OO' = 8 + 5.29 = 13.29 cm

Q74. In the given figure, O is the centre of the circle. If OAC = 35^o and OBC = 40^o, find the value of x.

Solution

Join OC Since OA = OC ACO = OAC = 35^o Similarly, OB = OC and OCB = OBC = 40^o ACB = 35^o + 40^o = 75^o _{AOB = 2 750 = 1500 (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.)}

Q75. Prove that the opposite angles of a cyclic quadrilateral are supplementary.

Solution

 

Q76. Prove that equal chords of congruent circles subtend equal angles at their centres

Solution

Given: Two congruent circles C(O, r) and C(P, r) such that AB and CD are equal chords of C(O, r) and C(P, r) respectively.

Q77. The given figure shows a circle with centre O. Also PQ = QR = RS and angle PTS = 75^o. Calculate: .

Solution

Q78. Two equal chords AB and CD of a circle when produced, intersect at point P. Prove that PB = PD.

Solution

Const: Join centre O to P. Draw Proof: We have, AB = CD (given) OR = OS (equal chords are equidistant from the centre) In ORP and OSP, OR = OS (each 90⁰) and, OP = OP (common) ORP OSP (RHS criterion) RP = SP …..(i) Now, AB = CD BR = DS ……(ii) Subtracting (ii) from (i) RP - BR = SP - DS PB = PD

Q79. In the figure, ABC = 45^o, Prove that OA OC.

Solution

_{Join AC.} ABC = 45^o AOC AOC = 2 ABC AOC = 2 45^o = 90^o Since, the angle subtended by an arc (here AC)at the centre is double the angle subtended by it at any point on the remaining part of the circle. OA OC.

Q80. Draw a circle with radius 2.1 cm. Draw three chords of the circle such that they form an equilateral triangle.

Solution

ABC is an equilateral triangle. Measure of each side is 3.7cm.

Q81. In fig., O is the center of the circle of radius 5 cm. OP AB, OQ CD, AB||CD. = 6 cm, = 8 cm. Determine PQ.

Solution

Since the perpendicular from the center of the circle to a chord bisects the chord. P and Q are the mid points of AB and CD In right triangles OAP and OCQ, we have OA² = OP² + AP² and OC² = OQ² + CQ² 5² = OP² + 3² and 5² = OQ² + 4² OP² = 5² - 3² and OQ² = 5² - 4² OP² = 16 and OQ² = 9 OP = 4 and OQ = 3 PQ = OP + OQ = 4 + 3 = 7 cm

Q82. Prove that a cyclic parallelogram is a rectangle.

Solution

_{Let ABCD be a cyclic parallelogram.} ... (1) _{(Opp. angles of a cyclic quadrilateral) We know that opposite angles of a parallelogram are equal. From equation (1), Parallelogram ABCD has one of its interior angles as 90°. Therefore, it is a rectangle.}

Q83. Prove that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Solution

Q84. In a circle of radius 5 cm, AB and AC are two chords such that AB = AC = 6 cm. Find length of chord BC.

Solution

Given - AB and AC are two equal words of a circle, therefore the centre of the circle lies on the bisector of BAC. OA is the bisector of BAC. Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. P divides BC in the ratio = 6 : 6 = 1 : 1. P is mid-point of BC. OP BC. In ABP, by Pythagoras theorem, In right OBP, we have Equating (1) and (2), we get Putting AP in (1), we get or,alternate method: Area of AOB by Heron's formula arAOB) = Also arAOB = BM = h = 4.8 cm BC = 2 BM = 9.6 cm.

Q85. The radius of a circle is 5 cm and the length of a chord in the circle is 8 cm. Find the distance of the chord from the centre of the circle.

Solution

Radius of circle (r) = 5 cm AB = 8 cm OM AB AM = MB = 4 cm [ drawn from the centre of the circle bisects the chord] In right OMA, by Pythagoras theorem OA² = OM² + AM² 5² = OM² + (4)² 25 = OM² + 16 OM² = 9 OM = 3 cm

Q86. In the figure, ABCD is a cyclic quadrilateral and ABC = 85^o. Find the measure of ADE.

Solution

ABCD is a cyclic quadrilateral. ABC + ADC = 180^o (opp. s are supplementary) ADC = 180^o - 85^o = 95^o ADE = 180^o - 95^o = 85^o (linear pair)

Q87. Find the length of the chord which is at a distance of 4 cm from the center of a circle whose radius is 5 cm.

Solution

Since ODA is a right angled triangle By Pythagoras Theorem (OA)²= (OD)² + (AD)² (5)² = (4)² + (AD)² 25 - 16= (AD)² AD² = 9 AD = 3 cm Chord AB = 2 AD[ from the center on a chord bisects the chord] = 2 3 = 6 cm

Q88. In the figure, PQ is a diameter of the circle and XY is chord equal to the radius of the circle. PX and QY when extended intersect at point E. Prove that PEQ = 60^o

Solution

PQ is the diameter of the circle, chord XY = r (radius of circle) PX and QY extended intersect at a point E. To prove PEQ = 60^o XY = OX = OY [radii of a circle, Given] XOY is an equilateral triangle XQY = 30^o [Inscribed angle is half of the central angle] PXQ = 90^o [Angle in a semi circle] QXE = 180^o - PXQ = 90⁰ [Linear pair] In XEQ, XEQ = 180^o - (EXQ + EQX) (Angle sum property) XEQ = 180^o - (90⁰ + 30⁰) = 60⁰ PEQ = 60^o

Q89.

Solution

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