Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

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.

313 Views

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?

191 Views

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.

264 Views

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.

395 Views

Long Answer Type

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

499 Views

Short Answer Type

56.

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

394 Views

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.

425 Views

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.

∠CDB = ∠BAC
| Angles in the same segment of a circle are equal
= 30°    ...(1)
∠DBC = 70°    ...(2)
In ∆BCD,

∠CDB = ∠BAC| Angles in the same segment of a circle are equal= 30

∠BCD + ∠DBC + ∠CDB = 180°
| Sum of all the angles of a triangle is 180°
⇒ ∠BCD + 70° + 30° = 180°
| Using (1) and (2)
⇒    ∠BCD + 100° = 180°
⇒    ∠BCD = 180° - 100°
⇒    ∠BCD = 80°    ...(3)
In ∆ABC,
AB = BC
∴ ∠BCA = ∠BAC
| Angles opposite to equal sides of a triangle are equal
= 30°    ...(4)
| ∵ ∠BAC = 30° (given)
Now, ∠BCD = 80°    | From (3)
⇒ ∠BCA + ∠ECD = 80°
⇒ 30° + ∠ECD = 80°
⇒    ∠ECD = 80° - 30°
⇒    ∠ECD = 50°.

668 Views

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

250 Views

60.

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

415 Views