AB and CD are two chords of equal length of the cirlc e with centre at O. The distance of the chord AB form the centre Ois 4 cm. Then the distance of chord CD form the center O is

Options

A. 2 cm
B. 4 cm
C. 6 cm
D. 8 cm

Answer

B

Explanation

Let AB and CD are two chords of equal length of the circle with centre at O. OP and OQ are two perpendiculars to the chords AB and CD form O .
`therefore` distance of AB from O = OP and that of CD = OQ.
Now , OP is perpendicular to AB.
`therefore` P is the mid - point of AB.
`therefore" " BP = (1)/(2) xx AB ...........(1)`
Again OQ, is perpendicular to CD.
`therefore` Q is the mid -point of CD
`therefore" ' CQ = (1)/(2) xx CD............(2)`
From (1) and (2) we get `BP = CQ , [because AB = CD]`
Now in right - angled trinagles OBP and OCQ.
BP = CQ and `angleOPB = angle OQC [ because " each is right angle "]`
`therefore Delta OBP ~= Delta OCQ`
`therefore OP = OQ or , 4 = OQ [ because OP = 4 cm]`
`therefore OQ = 4 cm ` . Hence the required distance `= 4 cm rArr (b)` is correct.