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05 Dec 2018, 01:11
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:26) correct
27%
(02:37)
wrong
based on 343
sessions
History
Date
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Not Attempted Yet
AB and CD are two parallel chords of a circle, with center O, such that AB = 12 and CD = 6. If the diameter of the circle is 6√5, then find the distance between the two chords?
- A. √3
B. 3
C. √5
D. 2√3
E. 6
05 Dec 2018, 02:08
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OA = radius = \(6\sqrt{5}/2\) = \(3\sqrt{5}\) = OD
From the attached below image. AE = 6.
CF = 3.
Then OE as per Pythagorean theorem gives \(\sqrt{45-36}\) = 3.
OF = \(\sqrt{45-9}\)= 6.
OF = OE + EF
EF = 6-3 = 3.
B is the answer.
From the attached below image. AE = 6.
CF = 3.
Then OE as per Pythagorean theorem gives \(\sqrt{45-36}\) = 3.
OF = \(\sqrt{45-9}\)= 6.
OF = OE + EF
EF = 6-3 = 3.
B is the answer.
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10 Dec 2018, 02:01
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Solution
Given:
We are given that,
- • AB and CD are two parallel chords of a circle
• O is the center of the circle
• AB = 12 and CD = 6
• Diameter of the circle = 6√5
To find:
- • The distance between the two chords, AB and CD
Approach and Working:
From the above figure, we can infer that we need to find the value of EF
In triangle, OAE, we know that,
- • \(AE = \frac{AB}{2} = \frac{12}{2} = 6\)
• OA = radius of the circle =\(\frac{6√5}{2} = 3√5\)
• \(∠OEA = 90^o\)
So, we can write, \(OA^2 = OE^2 + AE^2\)
- • Implies, \((3√5)^2 = OE^2 + 6^2\)
• \(OE^2 = 45 – 36 = 9\)
• Thus, OE = √9 = 3
Similarly, in triangle DOF, we ca say that
- • \(OD^2 = OF^2 + FD^2\)
- o \((3√5)^2 = OF^2 + (\frac{6}{2})^2\)
o \(OF^2 = 45 – 9 = 36\)
• Thus, OF = √36 = 6
Therefore, EF = OF = OE = 6 – 3 = 3
Hence the correct answer is Option B.
Answer: B
