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AB and CD are two parallel chords of a circle, with center O .........

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AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   05 Dec 2018, 01:11
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A
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Difficulty:

45% (medium)

Question Stats:

73% (02:26) correct 27% (02:37) wrong based on 343 sessions
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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

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B
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AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   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.
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AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   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

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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   22 Jan 2019, 05:34
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EgmatQuantExpert

Please explain a bit more why AE=EB

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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   22 Jan 2019, 23:38
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LevanKhukhunashvili wrote:
EgmatQuantExpert

Please explain a bit more why AE=EB

Regards
L


Hi,

If you see the two triangles AOE and EOB, both are congruent triangles.
    AO = OB = radius,
    OE is a common side, and
    angle OEA = angle OEB = 90 degrees

Thus, the length of the third sides must be equal, that is, AE = EB

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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   22 Nov 2019, 01:29
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It wasn't mentioned that OE & OF bisect the chords right? Should we always assume that the line drawn from the center bisects every chord?
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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   25 May 2020, 23:07
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Hi EgmatQuantExpert,

In your solution, you assumed that the line drawn from centre 'O' to the two chords (AB & CD) is perpendicular. However, no information is given in the question to to make that inference. Further, when you prove congruency of the two triangles AOe and EOB, same assumption is made i.e. angle OEA = angle OEB - 90 degrees.

Could you please help me understand how have you arrived at the conclusion that line 'O' from centre is a perpendicular (bisector) to the two chords?
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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   20 Jun 2020, 02:25
Question has not been framed correctly.
It is assumed the line drawn from centre 'O' to the two chords (AB & CD) is perpendicular.
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Re: AB and CD are two parallel chords of a circle, with center O ......... [#permalink] New post   04 May 2023, 21:23
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