Question : The distance between two parallel chords of length 6 cm each, in a circle of diameter 10 cm is:
Option 1: 12 cm
Option 2: 8 cm
Option 3: 6 cm
Option 4: 4 cm
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Correct Answer: 8 cm
Solution : A line passing through the centre perpendicularly bisects the chord into two equal parts. Equal chords are at equal distances from the centre. AB = CD = 6 cm, R = $\frac{10}{2}$ = 5 cm We have to find the length of EF. In $\triangle$BEO, $\angle$E is 90°, OB = 5 cm, BE = $\frac{6}{2}$ = 3 cm, From Pythagoras theorem in $\triangle$OBE, EO 2 = (OB 2 - BE 2 ) ⇒ EO 2 = (5 2 – 3 2 ) ⇒ EO 2 = (25 – 9) ⇒ EO 2 = 16 ⇒ EO = 4 cm Now, the length of the EF = 2 × OE = 2 × 4 = 8 cm ∴ The required distance between the chords is 8 cm. Hence, the correct answer is 8 cm.
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