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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3078
AB and CD are two parallel chords of a circle, with center O .........  [#permalink]

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1
2 00:00

Difficulty:   25% (medium)

Question Stats: 77% (02:15) correct 23% (02:47) wrong based on 93 sessions

HideShow timer Statistics 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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NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1021
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
AB and CD are two parallel chords of a circle, with center O .........  [#permalink]

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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.
Attachments ps1.png [ 7.86 KiB | Viewed 1232 times ]

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3078
AB and CD are two parallel chords of a circle, with center O .........  [#permalink]

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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.

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Senior Manager  G
Joined: 13 Feb 2018
Posts: 450
GMAT 1: 640 Q48 V28 Re: AB and CD are two parallel chords of a circle, with center O .........  [#permalink]

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EgmatQuantExpert

Please explain a bit more why AE=EB

Regards
L
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3078
Re: AB and CD are two parallel chords of a circle, with center O .........  [#permalink]

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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

Regards,
_________________ Re: AB and CD are two parallel chords of a circle, with center O .........   [#permalink] 22 Jan 2019, 23:38
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