Three girls Reshma, Salma and Mandip are playing a game by stand

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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?


Let    KR = x m

ar left parenthesis triangle ORS right parenthesis equals ar left parenthesis triangle ORK right parenthesis plus ar left parenthesis triangle SRK right parenthesis
space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis OK right parenthesis left parenthesis KR right parenthesis over denominator 2 end fraction plus fraction numerator left parenthesis KS right parenthesis left parenthesis KR right parenthesis over denominator 2 end fraction
space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis KR right parenthesis left parenthesis OK plus KS right parenthesis over denominator 2 end fraction equals fraction numerator left parenthesis KR right parenthesis left parenthesis OS right parenthesis over denominator 2 end fraction
space space space space space space space space space space space space space space space equals space fraction numerator left parenthesis straight x right parenthesis left parenthesis 5 right parenthesis over denominator 2 end fraction 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 1 right parenthesis


Let    KR = x mAgain, ar (∆ORS)From equations (1) and (2),⇒ KR

Again, ar (∆ORS)

equals fraction numerator RS cross times OL over denominator 2 end fraction equals fraction numerator 6 cross times OL over denominator 3 end fraction
equals space fraction numerator 6 cross times square root of OR squared minus RL squared end root over denominator 2 end fraction
space space space space space space space space space space space vertical line space By space Pythagoras space Theorem
equals space space fraction numerator 6 cross times square root of left parenthesis 5 right parenthesis squared minus open parentheses begin display style RS over 2 end style close parentheses squared end root over denominator 2 end fraction
equals space fraction numerator 6 cross times square root of left parenthesis 5 right parenthesis squared minus open parentheses begin display style 6 over 2 end style close parentheses squared end root over denominator 2 end fraction
equals space fraction numerator 6 cross times square root of 25 minus 9 end root over denominator 2 end fraction
equals space fraction numerator 6 cross times 4 over denominator 2 end fraction equals space 12 space straight m 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... left parenthesis 2 right parenthesis

From equations (1) and (2),

fraction numerator left parenthesis straight x right parenthesis left parenthesis 5 right parenthesis over denominator 2 end fraction equals 12 rightwards double arrow space space straight x equals fraction numerator 12 cross times 2 over denominator 5 end fraction equals 24 over 5 equals 4.8 space straight m

⇒ KR = 4.8 m
∴ RM = 2KR = 2 × (4.8) = 9.6 m
Hence, the distance between Reshma and Mandip is 9.6 m.

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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.

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