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

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

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New post Updated on: 07 Mar 2010, 14:56
1
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

48% (02:43) correct 52% (02:59) wrong based on 79 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.

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

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New post 07 Mar 2010, 14:41
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1
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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Re: On a semicircle with diameter AD , chord BC is parallel to AD. Further  [#permalink]

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New post 07 Mar 2010, 14:44
oh yea this was quite easy....its 3 am here i think i should sleep, my brain is outa senses...
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On a semicircle with diameter AD , chord BC is parallel to AD. Further  [#permalink]

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New post 22 Jan 2016, 13:56
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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Re: On a semicircle with diameter AD , chord BC is parallel to AD. Further  [#permalink]

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New post 26 Mar 2019, 23:43
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Re: On a semicircle with diameter AD , chord BC is parallel to AD. Further   [#permalink] 26 Mar 2019, 23:43
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