Question 134

Two parallel chords of a circle, of diameter 20 cm lying on the opposite sides of the centre are of lengths 12 cm and 16 cm. The distance between the chords is

Solution

OB = OD = 10 , AB = 12 , CD = 16

Now, BF = 12/2 = 6 and DE = 16/2 = 8

In $$\triangle$$OBF

=> $$OF = \sqrt{(OB)^2 - (BF)^2}$$

=> $$OF = \sqrt{100 - 36} = \sqrt{64} = 8$$

Similarly, EO = 6

=> EF = EO + OF = 8+6 = 14 cm

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