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Circles

Short Answer Type

51. Two circles of radii 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.

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52. A chord 12 cm long is 8 cm away from the centre of the circle. What is the length of a chord which is 6 cm away from the centre?

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53.

In figure A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

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54. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

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Long Answer Type

55. In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

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Short Answer Type

56.

In figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

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57. In figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

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Long Answer Type

58. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

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59. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

In ∆OAB and ∆OCD,
OA = OC | Radii of a circle
OB = OD | Radii of a circle
∠AOB = ∠COD
| Vertically Opposite Angles

In ∆OAB and ∆OCD,OA = OC | Radii of a circleOB = OD | Radii of a

∴ ∆OAB ≅ ∆OCD     | SAS Rule
∴ AB = CD     | C.P.C.T.
⇒ Arc AB = Arc CD    ...(1)
Similarly, we can show that
Arc AD = Arc CB    ...(2)
Adding (1) and (2), we get
Arc AB + Arc AD = Arc CD + Arc CB
⇒     Arc BAD = Arc BCD
⇒ BD divides the circle into two equal parts (each a semicircle)
∴ ∠A = 90°, ∠C = 90°
| Angle of a semi-circle is 90°
Similarly, we can show that
∠B = 90°, ∠D = 90°
∴ ∠A = ∠B = ∠C = ∠D = 90°
∴ ABCD is a rectangle.

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60.

If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Prove that an isosceles trapezium is cyclic.

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