A circular park of radius 20m is situated in a colony. Three boy

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 Multiple Choice QuestionsShort Answer Type

31.  Prove that the perpendicular from the centre of a circle to a chord, bisects the chord.
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32.

AB and CD are two parallel chords of a circle, on the same side of the centre. If AB = 12 cm, CD = 24 cm and the distance between the chords is 4 cm, find the radius of the circle.

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33. Prove that the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.   
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34.

In the figure, two circles with centres A and B intersect each other at C and D. Prove that ∠ACB = ∠ADB.

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35. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
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36. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
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37.

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

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38.

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig.)

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39.

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

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40.

A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.


Let BD = x m


Let BD = x mThen in right triangle ODB,                   

Then in right triangle ODB,

                     OB2 = OD2 + BD
                                | By Pythagoras Theorem
rightwards double arrow       (20)2 = OD2 + x2
rightwards double arrow       OD2 = 400 - X2

rightwards double arrow       OD = square root of 400 minus straight x squared end root

Area of equilateral triangle

ABC equals fraction numerator square root of 3 over denominator 4 end fraction left parenthesis side right parenthesis squared equals fraction numerator square root of 3 over denominator 4 end fraction BC squared
space space space space space space space equals space fraction numerator square root of 3 over denominator 4 end fraction left parenthesis 2 BD right parenthesis squared space equals space square root of 3 space BD squared space equals space square root of 3 space straight x squared space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Again, area of equilateral triangle ABC = Area of increment OBC + Area of increment OCA + Area of increment OAB

= 3 Area of increment OBC equals 3 fraction numerator left parenthesis BD right parenthesis left parenthesis OD right parenthesis over denominator 2 end fraction

equals space fraction numerator 3 left parenthesis 2 BD right parenthesis left parenthesis OD right parenthesis over denominator 2 end fraction equals 3 left parenthesis BD right parenthesis left parenthesis AD right parenthesis
equals 3 straight x square root of 400 minus straight x squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis

From equations (1) and (2),

3 straight x square root of 400 minus straight x squared end root equals square root of 3 straight x squared
rightwards double arrow space space space space square root of 3 space square root of 400 minus straight x squared end root equals straight x

Squaring both sides,

                 3(400 - x2) = x
rightwards double arrow            1200 - 3x= x2

rightwards double arrow space space space space space 4 straight x squared space equals space 1200 space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space straight x squared equals 300
rightwards double arrow space space space space space straight x equals 10 square root of 3 space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space BD equals 10 square root of 3
rightwards double arrow space space space space space 2 BD space equals space 20 square root of 3 space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space BC equals 20 square root of 3

Hence, the length of string of each phone is 20 square root of 3 m.

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