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81. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

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82. Let ABC be a right angle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.

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83. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

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84. Draw a line segment of length 7.6 cm and divide it in the ratio 3 : 5.

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85. Divide the given line segment AB of length 6 cm internally in the ratio 3 : 4.

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Long Answer Type

86. Divide a line segment 11 cm in the ratio of 2 : 5 (a) internally (b) externally.

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Short Answer Type

87. Construct a triangle ABC in which AB = 5 cm, ∠B = 60° and altitude CD = 3 cm. Construct a triangle AQR similar to ΔABC, such that each side of ΔAQR is 1.5 times that of the corresponding side of ΔABC.

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88. Construct a ΔPBC, similar to a given ΔPQR, with QR = 6 cm, PR = PQ = 5 cm, such that each of its side is 5/7th of the corresponding sides of the ΔPQR.

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89. Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm. Draw a triangle similar to ΔABC with its sides equal to 3/4th of the corresponding sides of ΔABC.

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90. Construct a ΔABC in which BC = 6 cm, m∠BAC = 60° and median through A is 4.5 cm. Construct a ΔA ‘BC’ similar to ΔABC with BC’ = 8 cm.

Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) Draw a ray BX making an angle 60° with BC.

Steps of Construction:(i) Draw a line segment BC = 6 cm.(ii) ?

(iii)    Through B, draw MB ⊥ BX.
(iv)    Draw perpendicular bisector of BC which intersects BC at D and MB at O.
(v) With O as centre and OB as radius draw a circle.
(vi)    With D as centre and 4.5 cm as radius, draw an arc which intersects the circle at A.
(vii)    Join AB and AC. ΔABC is the required triangle.
(viii)    Produce BC to C’, such that BC’ = 8 cm.
(ix) Through C’, draw C’A || CA which meets BA produced at A’.
Thus A’ BC’ is the required triangle similar to triangle ABC.

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