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Lines and Angles

Short Answer Type

71. In figure, find the value of x.

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72. In figure, if QT ⊥ PR, ∠TQR = 40° and ∠SPR = 30°, find the value of x and y.

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73. In figure, if line segment AB intersects CD at O such that ∠OAD = 80°, ∠ODA = 50° and ∠OCB = 40°, then find ∠OBC.

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74. In figure, AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD. Prove that 2 ∠APD = ∠B + ∠C.

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75. In figure, AB II DE, ∠ABC = 75°, ∠CDE = 145°, find ∠BCD.

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76. In figure, the side BC of ∆ABC is produced to D. The bisector of ∠A meets BC in L. Prove that ∠ABC + ∠ACD = 2 ∠ALC

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Long Answer Type

77. Prove that the sum of the angles of a hexagon is 720°.

Given: A hexagon ABCDEF
To Prove:
∠ A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°
Construction: Join AD, BE and FC so as to intersect at O.
Proof: In ∆OAB,
∠1 + ∠7 + ∠9 = 180° ...(1)
| Angle sum property of a triangle

Given: A hexagon ABCDEFTo Prove:∠ A + ∠B + ∠C + ∠D + ∠E + ?

In ∆AOBC,
∠2 + ∠10 + ∠11 = 180°    ...(2)
| Angle sum property of a triangle
In ∆OCD,
∠3 + ∠12 + ∠13 = 180° ...(3)
| Angle sum property of a triangle
In ∆CDE,
∠4 + ∠14 + ∠15 = 180°    ...(4)
| Angle sum property of a triangle
In ∆OEF,
∠5 + ∠16 + ∠17 = 180°    ...(5)
| Angle sum property of a triangle
In ∆OFA,
∠6 + ∠18 + ∠8 = 180°    ...(6)
| Angle sum property of a triangle
Adding (1), (2), (3), (4), (5) and (6), we get
(∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6)
+ (∠7 + ∠8) + (∠9 + ∠10)
+ (∠11 + ∠12) + (∠13 + ∠14)
+ (∠15 + ∠16) + (∠17 + ∠18)
=1080°
⇒ 360° + ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080°
| ∵ Sum of all the angles round a point is equal to 360°
⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080° - 360° = 720°