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

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Math Expert
Joined: 02 Sep 2009
Posts: 47948
If B and D are centers of the circles shown in the figure above, and  [#permalink]

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21 Nov 2017, 23:11
00:00

Difficulty:

45% (medium)

Question Stats:

67% (01:41) correct 33% (02:30) wrong based on 33 sessions

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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 [ 8.73 KiB | Viewed 588 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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21 Nov 2017, 23:55
1
Bunuel wrote:

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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22 Nov 2017, 01:18
1

Attachment:
File comment: Image

1966360.jpg [ 9.26 KiB | Viewed 518 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

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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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24 Jul 2018, 06:20
The problem can be solved using the similar triangle theorem.
where Triangle ABC and Triangle ADE are similar
2/3=AB/(AB+BD)
Hence we get AB=24 by solving the above equation.
Re: If B and D are centers of the circles shown in the figure above, and &nbs [#permalink] 24 Jul 2018, 06:20
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