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AB and CD are chords of the circle, and E and F are the midpoints of

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AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post Updated on: 22 Jul 2019, 06:15
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Question Stats:

56% (02:27) correct 44% (02:43) wrong based on 55 sessions

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AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?

(A) 10

(B) 12

(C) 13

(D) 15

(E) 25

Source: Nova GMAT
Difficulty Level: 700

Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg
#GREpracticequestion AB and CD are chords of the circle.jpg [ 17.99 KiB | Viewed 1147 times ]

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Originally posted by Bunuel on 02 Apr 2019, 23:42.
Last edited by SajjadAhmad on 22 Jul 2019, 06:15, edited 1 time in total.
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Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post 03 Apr 2019, 11:46
1
From the figure
AE = BE = 5
CF = FD = 12
EF = 17
OE = x, Then OF = 17-x

OAE is a right angle triangle, Apply Pythagorean theorem,
OA = \(\sqrt{25+x^2}\) = radius
Similarly, OCF is a right angle triangle, Apply Pythagorean theorem,
OC = \(\sqrt{(17-x)^2+144}\) = radius

Equating both equations as both are the radii, we get
\(25+x^2 = 289+x^2-34x+144\)
34x = 408
x = 12 = OE, then OF = 5
OA = OC = \(\sqrt{25+144}\) = 13

C is the answer.
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Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post Updated on: 11 Apr 2019, 12:39
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Bunuel wrote:
Image
AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?

(A) 10

(B) 12

(C) 13

(D) 15

(E) 25


Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg


The hypotenuses of AEO and OCF must be the same (the radius) \(AO = OC\). \(EF = 17\); one can quickly see that \(OE=12\) and \(OF = 5\).
Pythagoras Theorem: \(25 + 144 = 169\) --> Answer: C.
Realizing that \(AO = OC\) and that the specific values for OE and OF can be determined quickly without any math saves a lot of time on this question.

Originally posted by Zoom96 on 07 Apr 2019, 15:28.
Last edited by Zoom96 on 11 Apr 2019, 12:39, edited 3 times in total.
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Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post 11 Apr 2019, 05:58
Zoom96 wrote:
Bunuel wrote:
Image
AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?

(A) 10

(B) 12

(C) 13

(D) 15

(E) 25


Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg


The hypotenuses of AEO and OCF must be the same (the radius) \(AO = CD\). \(EF = 17\); one can quickly see that \(OE=12\) and \(OF = 5\).
Pythagoras Theorem: \(25 + 144 = 169\) --> Answer: C.
Realizing that \(AO =CD\) and that the specific values for OE and OF can be determined quickly without any math saves a lot of time on this question.


Hi Zoom96,
Can you please elaborate how AO=CD? Thank you.
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Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post 11 Apr 2019, 12:41
stne wrote:
Zoom96 wrote:
Bunuel wrote:
Image
AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?

(A) 10

(B) 12

(C) 13

(D) 15

(E) 25


Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg


The hypotenuses of AEO and OCF must be the same (the radius) \(AO = CD\). \(EF = 17\); one can quickly see that \(OE=12\) and \(OF = 5\).
Pythagoras Theorem: \(25 + 144 = 169\) --> Answer: C.
Realizing that \(AO =CD\) and that the specific values for OE and OF can be determined quickly without any math saves a lot of time on this question.


Hi Zoom96,
Can you please elaborate how AO=CD? Thank you.


Oh yeah my bad, I of course meant \(AO = OC\) (both are the radius). I just fixed it. Thanks.
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Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post 17 Apr 2019, 10:47
Can also be solved by estimating the value by range

The the vertical lines of 5 and 12 tell us that the radius is more than 12, but just slightly...
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AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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New post 13 Oct 2019, 10:46
Bunuel wrote:
Image
AB and CD are chords of the circle, and E and F are the midpoints of the chords, respectively. The line EF passes through the center O of the circle. If EF = 17, then what is radius of the circle?

(A) 10

(B) 12

(C) 13

(D) 15

(E) 25

Source: Nova GMAT
Difficulty Level: 700

Attachment:
#GREpracticequestion AB and CD are chords of the circle.jpg


Let OE = x, so OF = 17-x
As OAE is a right angle triangle, we get,
OA=\((25+x^2)^{1/2}\)
Similarly, as OCF is a right angle triangle, we get,
OC = \(((17−x)^2+144)^{1/2}\)
As OA=OC=Radius, we get,
\((25+x^2)^{1/2}\)=\(((17−x)^2+144)^{1/2}\)
=> x=12=OE,
So, from triangle OAE, we get that, OA=13=Radius

Therefore, the answer is option C.
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AB and CD are chords of the circle, and E and F are the midpoints of   [#permalink] 13 Oct 2019, 10:46
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