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GRE Math Challenge #23 equilateral triangle ABC is inscribed

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GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink] New post 07 Sep 2014, 06:39
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Question Stats:

75% (02:07) correct 25% (01:22) wrong based on 32 sessions
In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?

Attachment:
figure 14.jpg
figure 14.jpg [ 21.06 KiB | Viewed 3369 times ]



A. 1

B. \(\sqrt{3}\)

C. 2

D. 2 \(\sqrt{3}\)

E. 4 \(\sqrt{3}\)
[Reveal] Spoiler: OA

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In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 22 Jan 2017, 09:44
Hi everyone, I'm having struggle to find out how to do this question...
1. Equilateral Triangle inscribed in circle (as the attachment of question.jpg, question 9)
2. According mathopenref.c0m/trianglecentroid.html (please exchange 0 to o to access the urI), one of the Centroid facts mentioned this
"The centroid is exactly two-thirds the way along each median. Put another way, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex. These lengths are shown on the one of the medians in the figure at the top of the page so you can verify this property for yourself."

Image

Since centroid divides each median into two segments with 2:1 ratio.
if A to O is 4 (as the attachment of math.jpg), O to X will be 2?? Am I getting the right understanding??

In the figure below, equilateral triangle ABC is inscribed in the circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE ?

A) 1

B) \(\sqrt{3}\)

C) 2

D) \(2\sqrt{3}\)

E) \(4\sqrt{3}\)

Image


As for question 9 since radius is 4, therefore BO = 4.
With 2:1 ratio, OD will be 2. (If I get the correct understanding from the facts)
OE - OD = DE
therefore 4 - 2 = DE
Answer is C
Am I right??
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 22 Jan 2017, 10:50
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Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Thank you a lot for your collaboration. ASAP our expert sandy will reply to your question.

Regards
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 22 Jan 2017, 11:27
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Hi darkdevil8z,

Yes you are quite correct.

If you dont know this property you might even solve it with trigonometry.

Join Ao; and AO=BO=OE= 4 and triangle AOD is a right triangle with AO bisecting the angle CAB of triangle ABC.

So OD = \(AO \times sine(\frac{60}{2})\) = \(AO \times sine(30)\) =\(AO\times\frac{1}{2}\) =2

DE = OE-OD = 4 -2=2.

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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 23 Jan 2017, 04:30
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Carcass wrote:
Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Thank you a lot for your collaboration. ASAP our expert sandy will reply to your question.

Regards

Hi Carcass, sorry for didn't post properly, as a new user (less than 5 posts), I'm not able to post the urI... (because it restrict new user to post a urI, if he/she less than 5 posts) as for the sub-forum, sorry for that, I didn't know the sub-forum exist.

As for Sandy, thanks for reply. =)
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 23 Jan 2017, 05:24
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darkdevil8z wrote:
Carcass wrote:
Hi,

please format the question properly. Screencast of a question instead of the text are avoidable. Moreover, follow the rules for posting on the board and chose the right sub-forum to post the questions.

Thank you a lot for your collaboration. ASAP our expert sandy will reply to your question.

Regards

Hi Carcass, sorry for didn't post properly, as a new user (less than 5 posts), I'm not able to post the urI... (because it restrict new user to post a urI, if he/she less than 5 posts) as for the sub-forum, sorry for that, I didn't know the sub-forum exist.

As for Sandy, thanks for reply. =)


No sorry Sir. Just point out :)

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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink] New post 23 Jan 2017, 06:09
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After this message, I think I no longer have urI restrict problem, and ill post the sub-forum as you mentioned, thanks for point out. =)
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink] New post 05 Jul 2018, 19:30
sandy wrote:
Attachment:
figure 14.jpg



In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?




A. 1
B. \(\sqrt{3}\)
C. 2
D. 2 \(\sqrt{3}\)
E. 4 \(\sqrt{3}\)

Please solution for this?
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink] New post 05 Jul 2018, 23:14
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Emike56 wrote:
sandy wrote:
Attachment:
figure 14.jpg



In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?




A. 1
B. \(\sqrt{3}\)
C. 2
D. 2 \(\sqrt{3}\)
E. 4 \(\sqrt{3}\)

Please solution for this?


@Emike56,

Join the line AO. Now, we have triangle AOD, which is a 30:60:90 triangle (<DAO = 30, <AOD = 60, <ADO =90). Hence, sides will be in the ratio of \(1:\sqrt{3}:2\).

We know AO = 4 (Radius of the circle).Hence, OD = 2 and AD = \(2\sqrt{3}\).

DE = OE - OD = 4 - 2 = 2. Answer(C).

Hope it helps.
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink] New post 08 Jul 2018, 19:14
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ganand wrote:
Emike56 wrote:
sandy wrote:
Attachment:
figure 14.jpg



In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?




A. 1
B. \(\sqrt{3}\)
C. 2
D. 2 \(\sqrt{3}\)
E. 4 \(\sqrt{3}\)

Please solution for this?


@Emike56,

Join the line AO. Now, we have triangle AOD, which is a 30:60:90 triangle (<DAO = 30, <AOD = 60, <ADO =90). Hence, sides will be in the ratio of \(1:\sqrt{3}:2\).

We know AO = 4 (Radius of the circle).Hence, OD = 2 and AD = \(2\sqrt{3}\).

DE = OE - OD = 4 - 2 = 2. Answer(C).

Hope it helps.

Thanks. I am assuming that we have made O the centre of the circle. This was not stated in the question nor in your solution.
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink] New post 16 Jul 2018, 00:09
Emike56 wrote:

In figure below, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?




A. 1
B. \(\sqrt{3}\)
C. 2
D. 2 \(\sqrt{3}\)
E. 4 \(\sqrt{3}\)
Please solution for this?

Thanks. I am assuming that we have made O the centre of the circle. This was not stated in the question nor in your solution.


@Emike56,
Thank you for the query.
Let me explain.

When we say triangle ABC this implies that A, B, and C are the three vertices of the triangle.

Similarly, the question says "Circle O", which implies that O is the center of the circle.

I hope this helps.
Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed   [#permalink] 16 Jul 2018, 00:09
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