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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