Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 03 Sep 2015, 16:01

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the figure (attached), ABC is an equilateral triangle,

Author Message
Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 313 [0], given: 22

In the figure (attached), ABC is an equilateral triangle, [#permalink]  08 Aug 2009, 06:41
00:00

Difficulty:

(N/A)

Question Stats:

50% (01:34) correct 50% (00:18) wrong based on 37 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees

OA later after discussion
Attachments

clip_image002.jpg [ 2.89 KiB | Viewed 8589 times ]

SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 144 [0], given: 39

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 08:26
[quote="crejoc"]In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees

FROM STEM : DB = DIAMETER

from 1

da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff

from 2
we dont have any side length...insuff

A
Director
Joined: 01 Apr 2008
Posts: 903
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 18

Kudos [?]: 364 [0], given: 18

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 09:17
yezz wrote:
crejoc wrote:
In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees

FROM STEM : DB = DIAMETER

from 1

da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff

from 2
we dont have any side length...insuff

A

I could not get where the stem says DB is the diameter??
For me it is C.
We need to find the length of a side of equilateral triangle in order to find the radius and hence the area of the circle.

Combining 1 and 2 we can get the length of AB.
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 144 [0], given: 39

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 09:30
Economist wrote:
yezz wrote:
crejoc wrote:
In the figure (attached), ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees

FROM STEM : DB = DIAMETER

from 1

da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff

from 2
we dont have any side length...insuff

A

I could not get where the stem says DB is the diameter??
For me it is C.
We need to find the length of a side of equilateral triangle in order to find the radius and hence the area of the circle.

Combining 1 and 2 we can get the length of AB.

try to draw any right angle triangle with 3 verticies on the circle and the right angle is one of these verticies without the hyp being the diameter!
Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 313 [0], given: 22

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 15:24
yezz wrote:

db bisects cba,we can know angles of the right triangle

How can we Know the angles of the right triangle, can you explain that..
Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 313 [0], given: 22

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 19:07
yezz wrote:
da = 4 , , db bisects cba,we can know angles of the right triangle and thus the hyp = diam and thus we can get the radius...suff

A

The OA is A , you are right, but how we can know the angles of the right triangle by knowing that db bisects cba.?? explain please..
Current Student
Joined: 03 Aug 2006
Posts: 116
Followers: 4

Kudos [?]: 162 [1] , given: 3

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 20:27
1
KUDOS

Given:
$$\triangle ABC$$ is Equilateral

$$\Rightarrow \angle ABC = \angle BCA = \angle CAB = 60^\circ$$

Also given

$$\triangle BAD$$ is a right triangle

$$\Rightarrow \angle BAD = 90^\circ$$

Now if you notice

Arc AB is intercepted by inscribed angles $$\angle BAC$$and $$\angle BDA$$

$$\Rightarrow \angle BCA = \angle BDA = 60^\circ$$

For $$\triangle BAD$$ we know

$$\angle BAD = 90^\circ$$

$$\angle BDA = 60^\circ$$

$$\Rightarrow \angle ABD = 30^\circ$$

$$\text{Area of circle} = \pi r^2$$

$$\Rightarrow r=? \text{ or } 2r=d=BD=?$$

Lets look at Statement 1.

$$DA = 4$$

$$\text{then } BD = 4 \times 2$$

$$\text {because } 30^\circ-60^\circ-90^\circ \triangle =\rightarrow 1-sqrt{3}-2 \text { sides}$$

Sufficient

Statement 2

$$\angle ABD = 30^\circ$$

This does not tell us anything about the length of any sides. Hence insufficient.

Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 313 [0], given: 22

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 20:50
nookway wrote:
T

$$\Rightarrow \angle BAC = \angle BDA = 60^\circ$$

I think it is
Rightarrow \angle BCA = \angle BDA = 60^\circ
I too reasoned the problem this way, but i cant understand "yezz's" explanation of
yezz wrote:
db bisects cba,we can know angles of the right triangle.

I thought if there is some we can deduce the angles other than quoted by nookway, it would be also helpful..

Last edited by crejoc on 09 Aug 2009, 00:40, edited 3 times in total.
Current Student
Joined: 03 Aug 2006
Posts: 116
Followers: 4

Kudos [?]: 162 [0], given: 3

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 22:23
crejoc wrote:

I think it is
Rightarrow \angle BCA = \angle BDA = 60^\circ

Thanks for pointing out the typo.
SVP
Joined: 05 Jul 2006
Posts: 1516
Followers: 5

Kudos [?]: 144 [0], given: 39

Re: equilateral triangle circumscribed circle [#permalink]  08 Aug 2009, 23:04
crejoc wrote:
yezz wrote:

db bisects cba,we can know angles of the right triangle

How can we Know the angles of the right triangle, can you explain that..

any equilateral triangle drawn inside a circle (60,60,60 angles) and if you draw the diameter it will bisect one side and one angle.ie: the triangle is exactly in the middle of the circle and the diameter cut it into halves ( similar triangles).

if we draw the 3 perpendicular bisectors of the triangle's 3 angles they will intersect at the center of the circle.
Senior Manager
Joined: 17 Mar 2009
Posts: 310
Followers: 6

Kudos [?]: 313 [0], given: 22

Re: equilateral triangle circumscribed circle [#permalink]  09 Aug 2009, 00:39
yezz wrote:
crejoc wrote:
yezz wrote:

db bisects cba,we can know angles of the right triangle

How can we Know the angles of the right triangle, can you explain that..

any equilateral triangle drawn inside a circle (60,60,60 angles) and if you draw the diameter it will bisect one side and one angle.ie: the triangle is exactly in the middle of the circle and the diameter cut it into halves ( similar triangles).

if we draw the 3 perpendicular bisectors of the triangle's 3 angles they will intersect at the center of the circle.

Got it, that was really nice .. thanks for letting know it..
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 6203
Followers: 346

Kudos [?]: 71 [0], given: 0

Re: In the figure (attached), ABC is an equilateral triangle, [#permalink]  23 Aug 2015, 13:37
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In the figure (attached), ABC is an equilateral triangle,   [#permalink] 23 Aug 2015, 13:37
Display posts from previous: Sort by