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Math Expert V
Joined: 02 Sep 2009
Posts: 58428
AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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2
4 00:00

Difficulty:   65% (hard)

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

HideShow timer Statistics 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 [ 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.
NUS School Moderator V
Joined: 18 Jul 2018
Posts: 1020
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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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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Intern  B
Joined: 18 Oct 2018
Posts: 16
Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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2
Bunuel wrote: 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.
Director  V
Joined: 27 May 2012
Posts: 903
Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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Zoom96 wrote:
Bunuel wrote: 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.
_________________
- Stne
Intern  B
Joined: 18 Oct 2018
Posts: 16
Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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stne wrote:
Zoom96 wrote:
Bunuel wrote: 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.
Intern  B
Joined: 20 Jan 2019
Posts: 12
Re: AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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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...
Manager  G
Joined: 10 Apr 2018
Posts: 245
Location: India
Concentration: Entrepreneurship, Strategy
GMAT 1: 680 Q48 V34 GPA: 3.3
AB and CD are chords of the circle, and E and F are the midpoints of  [#permalink]

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Bunuel wrote: 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. 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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