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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]
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Expert Reply
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.

Regards,
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]
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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]
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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 :)

A board clean is a board efficient
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Re: In the figure below, equilateral triangle ABC is inscribed i [#permalink]
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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]
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]
2
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]
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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]
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.
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink]
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Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink]
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