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gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,266
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Updated on: 07 Mar 2010, 14:56
Originally posted by gurpreetsingh on 07 Mar 2010, 14:25.
Last edited by gurpreetsingh on 07 Mar 2010, 14:56, edited 1 time in total.
Last edited by gurpreetsingh on 07 Mar 2010, 14:56, edited 1 time in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
52% (02:54) correct
48%
(02:51)
wrong
based on 134
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History
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Time
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Not Attempted Yet
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.
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.
07 Mar 2010, 14:41
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gurpreetsingh
Ha.. this one was easy
.... Its BLook 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
Attachments
semicircle.png [ 23.84 KiB | Viewed 29220 times ]
General Discussion
gurpreetsingh
Joined: 12 Oct 2009
Last visit: 15 Jun 2019
Posts: 2,266
Given Kudos: 235
Status:<strong>Nothing comes easy: neither do I want.</strong>
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Products:
07 Mar 2010, 14:44
Kudos
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oh yea this was quite easy....its 3 am here i think i should sleep, my brain is outa senses...
