In fig., l and m are two parallel tangents to a circle with centr

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 Multiple Choice QuestionsShort Answer Type

331.

 In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.Find: (i) the length of the arc (ii) area of the sector formed by the arc.
  Use π = 227 


 Multiple Choice QuestionsMultiple Choice Questions

332.

In fig., a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively, If AB = 29 cm, AD = 23 cm, B = 90o and DS = 5 cm, then the radius of thecircle (in cm) is 

  • 11

  • 18

  • 6

  • 15


333.

In fig., PA and PB are two tangents drawn from an external point P to a circlewith centre C and radius 4 cm. If PA PB, then the length of each tangent is

  • 3

  • 4

  • 5

  • 6


 Multiple Choice QuestionsShort Answer Type

334.

Prove that the parallelogram circumscribing a circle is a rhombus.


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335.

Two circular pieces of equal radii and maximum area, touching each otherare cut out from a rectangular card board of dimensions 14 cm x 7 cm. Find the area of the remaining card board.

[ Use π = 22/7 ]


336.

In fig., a circle is inscribed in triangle ABC touches its sides AB, BC and ACat points D, E and F respectively. If AB = 12 cm, BC = 8 cm and AC = 10cm, then find the length of AD, BE and CF.


 Multiple Choice QuestionsLong Answer Type

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337.

In fig., l and m are two parallel tangents to a circle with centre O,touching the circle at A and B respectively. Another tangent at C intersects the line l at D and m at E. Prove that   DOE = 900


Given:  l and m are two parallel tangents to the circle with centre O touching the circle at A and B respectively. DE is a tangent at the point C, which intersect  l at  D  and  m at  E.To prove:  DOE = 90°Contruction: Join OC.Proof:                     

In ODA and ODC,OA = OC     ( Radii of the same circle )AD = DC      (Length of tangents drawn from an external point to a circle                            are equal )DO = OD       ( Commmon side ) ODA  ODC,         (SSS congruence criterion) DOA = COD          ..........(1)Similarly, OEB OECEOB = COE           ...........(2)Now, AOB is a diameter of the circle. Hence, it is a straight line.DOA +COD + COE + EOB =180°From (1) and (2), we have:2COD + 2COE = 180°COD + COE = 90°DOE = 90°Hence, proved.


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338.

Prove that the tangent at any point of a circle is perpendicular to theradius through the point of contact.


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 Multiple Choice QuestionsMultiple Choice Questions

339.

Two circles touch each other externally at P. AB is a common tangent to the circlestouching them at A and B. The value of ∠APB is

  • 30°

  • 45°

  • 60°

  • 90°


340.

A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

  • 52

  • 102

  • 52

  • 103


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