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# On a semicircle with diameter AD , chord BC is parallel to AD. Further

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CEO
Status: Nothing comes easy: neither do I want.
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07 Mar 2010, 13:25
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Difficulty:

65% (hard)

Question Stats:

59% (02:31) correct 41% (02:37) wrong based on 82 sessions

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On a semicircle with diameter AD , chord BC is parallel to AD. Further each of the chords AB and CD has length 2.
Given AD = 8 , what is BC?

a) 7.5
b) 7
c) 7.75
d) 7.25
e) None of the above.

Do not have OA, open for discussion.
[Reveal] Spoiler: OA

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Last edited by gurpreetsingh on 07 Mar 2010, 13:56, edited 1 time in total.
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07 Mar 2010, 13:41
3
KUDOS
gurpreetsingh wrote:
On a semicircle with diameter AD , chord BC is parallel to AD. Further each of the chords AB and CD has length 2.
Given AD = 8 , what is BC?

a) 7.5
b) 7
c) 7.75
d) 7.25
e) None of the above.

Do not have OA, open for discussion.

Ha.. this one was easy .... Its B

Look at the figure:

$$2^2 - x^2 = 4^2 - (4-x)^2$$
$$4 - x^2 = 16 - (16+x^2-8x)$$
$$x = \frac{1}{2}$$

Therefore BC = AD - 1 = 7
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CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2750
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35

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07 Mar 2010, 13:44
oh yea this was quite easy....its 3 am here i think i should sleep, my brain is outa senses...
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07 Dec 2015, 19:37
Hello from the GMAT Club BumpBot!

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22 Jan 2016, 12:56
Coordinate Geometry Approach:

BC = two times the x coordinate of the intersection of 2 circles y^2 = 16 - x^2 and y^2 = 4- (x-4)^2
Therefore solve for x: 16 - x^2 = 4 - (x-4)^2
x=7/2
BC = 2* (7/2) = 7
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10 Sep 2017, 05:51
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: On a semicircle with diameter AD , chord BC is parallel to AD. Further   [#permalink] 10 Sep 2017, 05:51
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