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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]  22 May 2020, 10:25
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40% (02:21) correct 60% (05:11) wrong based on 5 sessions
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#greprepclub If Band D are centers of the circles shown.jpg [ 22.12 KiB | Viewed 148 times ]

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]  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]  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]  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.

Resrds
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]  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]  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]  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]  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

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]  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

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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