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If B and D are centers of the circles shown in the figure above, and

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If B and D are centers of the circles shown in the figure above, and  [#permalink]

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New post 21 Nov 2017, 22:11
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

64% (02:17) correct 36% (02:45) wrong based on 41 sessions

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Image
If B and D are centers of the circles shown in the figure above, and if BD = 12, BC = 2 and DE = 3, then AB =

(A) 12
(B) 14
(C) 21
(D) 22
(E) 24

Attachment:
2017-11-21_1032.png
2017-11-21_1032.png [ 8.73 KiB | Viewed 695 times ]

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Re: If B and D are centers of the circles shown in the figure above, and  [#permalink]

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New post 21 Nov 2017, 22:55
1
Bunuel wrote:
Image
If B and D are centers of the circles shown in the figure above, and if BD = 12, BC = 2 and DE = 3, then AB =

(A) 12
(B) 14
(C) 21
(D) 22
(E) 24

Attachment:
2017-11-21_1032.png


Let AB = x..Consider two triangles
A and a Diameter passing through B
A and a Diameter passing through D

We know Diameter of smaller circle = 4 and Diameter of larger circle = 6
Hence
\(AB/AD = 4/6\)
\(x/x+12 = 4/6\)
x= 24
E
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If B and D are centers of the circles shown in the figure above, and  [#permalink]

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New post 22 Nov 2017, 00:18
1
Answer is E as follows.

Attachment:
File comment: Image
1966360.jpg
1966360.jpg [ 9.26 KiB | Viewed 622 times ]


We have to use properties of similar triangle here.

Triangle ACB is similar to AED
\(\frac{AD}{AB} = \frac{DE}{BC}\) => \(\frac{x+12}{x} = \frac{3}{2}\) => 2x+24 = 3x => x=24

Hence E is the answer
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Re: If B and D are centers of the circles shown in the figure above, and  [#permalink]

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New post 24 Jul 2018, 05:20
The problem can be solved using the similar triangle theorem.
where Triangle ABC and Triangle ADE are similar
Thus BC/DE=AB/AD
2/3=AB/(AB+BD)
Hence we get AB=24 by solving the above equation.
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Re: If B and D are centers of the circles shown in the figure above, and &nbs [#permalink] 24 Jul 2018, 05:20
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If B and D are centers of the circles shown in the figure above, and

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