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If Band D are centers of the circles shown in the figure a b

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If Band D are centers of the circles shown in the figure a b [#permalink] New post 22 May 2020, 10:25
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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
[Reveal] Spoiler: OA

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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 22 May 2020, 18:14
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Hey, Carcass Be=2, is it correct?
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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 22 May 2020, 18:20
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I think it's not Be but BC, am I right?
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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 01:29
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Yes Sir. Fixed. Of course was BC.

When you take the big book stems sometimes this happens because the book in pdf is really old.

Thank you so much. I am sorry of the typo.

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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 03:39
which property is used here
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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 03:57
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triangle similarity or congruent triangles
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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 04:13
thank u so much

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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 04:16
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We have to use properties of similar triangle here.

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

Hence E is the answer

Orrrr

We know Diameter of smaller circle = 4 and Diameter of larger circle = 6
Hence
\(\frac{AB}{AD} = \frac{4}{6}\)

\(\frac{x}{x}+12 = \frac{4}{6}\)
x= 24
E
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Re: If Band D are centers of the circles shown in the figure a b [#permalink] New post 23 May 2020, 04:31
Triangle ABC is similar to ADE hence

\(\frac{AB}{AD}\) = \(\frac{BC}{DE}\)

Lets value of AB = x

Hence \(\frac{x}{(AB + BD)}\) = \(\frac{2}{3}\)

\(\frac{x}{(x+12)}\) = \(\frac{2}{3}\)

3x = 2x + 24

x=24

Hence answer is E
Re: If Band D are centers of the circles shown in the figure a b   [#permalink] 23 May 2020, 04:31
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