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Founder  Joined: 18 Apr 2015
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If Band D are centers of the circles shown in the figure a b [#permalink]
Expert's post 00:00

Question Stats: 40% (02:21) correct 60% (05:11) wrong based on 5 sessions
Attachment: #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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GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests. Intern Joined: 17 Feb 2020
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
1
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Hey, Carcass Be=2, is it correct? Intern Joined: 17 Feb 2020
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
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I think it's not Be but BC, am I right?
Founder  Joined: 18 Apr 2015
Posts: 11096
Followers: 237

Kudos [?]: 2772 , given: 10521

Re: If Band D are centers of the circles shown in the figure a b [#permalink]
Expert's post
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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Intern Joined: 13 Mar 2020
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
which property is used here Intern Joined: 17 Feb 2020
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
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triangle similarity or congruent triangles
Intern Joined: 13 Mar 2020
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Kudos [?]: 3 , given: 10

Re: If Band D are centers of the circles shown in the figure a b [#permalink]
thank u so much

Posted from my mobile device Founder  Joined: 18 Apr 2015
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
Expert's post
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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Intern Joined: 24 Jan 2020
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Re: If Band D are centers of the circles shown in the figure a b [#permalink]
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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