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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:

66% (01:24) correct 33% (01:55) wrong based on 12 sessions
Attachment:
figure 14.jpg
figure 14.jpg [ 21.06 KiB | Viewed 1809 times ]



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

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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?
2 KUDOS received
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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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GRE Math Challenge #23 equilateral triangle ABC is inscribed

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