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GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink]
Expert's post 00:00

Question Stats: 71% (01:40) correct 28% (01:55) wrong based on 14 sessions
Attachment: figure 14.jpg [ 21.06 KiB | Viewed 2056 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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Sandy
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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? Intern Joined: 25 May 2018
Posts: 7
Followers: 0

Kudos [?]: 2  , given: 1

Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed [#permalink]
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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. Intern Joined: 08 Apr 2018
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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.
Intern Joined: 25 May 2018
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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. Re: GRE Math Challenge #23 equilateral triangle ABC is inscribed   [#permalink] 16 Jul 2018, 00:09
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