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

 It is currently 02 Jul 2015, 04:48

### 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, ABC is an equilateral triangle, and DAB is

Author Message
TAGS:
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 552
Location: United Kingdom
GMAT 1: 730 Q49 V40
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 29

Kudos [?]: 1154 [0], given: 217

In the figure, ABC is an equilateral triangle, and DAB is [#permalink]  03 Mar 2012, 12:06
1
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

47% (02:08) correct 53% (00:58) wrong based on 173 sessions
Attachment:

InscribedTwice1.gif [ 2.27 KiB | Viewed 5433 times ]
In the figure, 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
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Intern
Joined: 28 Sep 2011
Posts: 35
Location: India
WE: Consulting (Computer Software)
Followers: 1

Kudos [?]: 15 [0], given: 18

Re: Inscribed Triangle [#permalink]  03 Mar 2012, 12:57
Yes , Angle ACB = ADB as they inscribe the same arc in the circle.
Check this out....
Hope that helps....

The rest of the trick you know...
_________________

Kudos if you like the post!!!

Math Expert
Joined: 02 Sep 2009
Posts: 28242
Followers: 4461

Kudos [?]: 44999 [4] , given: 6638

Re: Inscribed Triangle [#permalink]  03 Mar 2012, 13:31
4
KUDOS
Expert's post
Attachment:

Inscribed Triangles.gif [ 2.27 KiB | Viewed 5394 times ]
In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

You should know the following properties to solve this question:
• All inscribed angles that subtend the same arc are equal. The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle. Hence, all inscribed angles that subtend the same arc are equal.
• A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
• In a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$. Notice that the smallest side (1) is opposite the smallest angle (30°), and the longest side (2) is opposite the largest angle (90°).
For more check Circles Triangles and chapters of Math Book: math-circles-87957.html and math-triangles-87197.html

So, from above we'll have that as DAB=90 degrees then DB must be a diameter of the circle. Next, as angles ACB and ADB subtend the same arc AB then they must be equal and since ACB=60 (remeber ACB is an equilateral triangle) then ADB=60 too. Thus DAB is 30-60-90 triangle and its sides are in ratio $$1 : \sqrt{3}: 2$$.

(1) DA = 4 --> the side opposite 30 degrees is 4, then hypotenuse DB=diameter=4*2=8 --> radius=4 --> $$area=\pi{r^2}=16\pi$$. Sufficient.

(2) Angle ABD = 30 degrees --> we knew this from the stem, so nothing new. Not sufficient.

Hope it's clear.
_________________
Intern
Joined: 03 Nov 2010
Posts: 13
Followers: 0

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

Re: Geometry: Circle in [#permalink]  03 Apr 2012, 15:46
Bunuel wrote:

Thanks for merging!
I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?
Math Expert
Joined: 02 Sep 2009
Posts: 28242
Followers: 4461

Kudos [?]: 44999 [1] , given: 6638

Re: Geometry: Circle in [#permalink]  04 Apr 2012, 00:15
1
KUDOS
Expert's post
KGG88 wrote:
Bunuel wrote:

Thanks for merging!
I don't get why the Central Angle Theorem applies. In the figure, there is no central angle for that to occur with. So why would ADB be the same as ACB?

All inscribed angles that subtend the same arc are equal. Since angles ACB and ADB subtend the same arc AB then they must be equal. Next, there is no central angle shown on the diagram, I just mentioned Central Angle Theorem to explain why is above property true: angles ACB and ADB have the same central angle AOB (O is the center of the circle) and since they both equal to half of it then they must be equal.

For more check Circles and Triangles chapters of Math Book: math-circles-87957.html and math-triangles-87197.html
_________________
Intern
Joined: 26 Jul 2010
Posts: 25
Followers: 2

Kudos [?]: 20 [0], given: 2

In the figure above, if ABC is an equilateral triangle, and [#permalink]  03 Apr 2013, 12:42
In the figure above, if ABC is an equilateral triangle, and DAB is a right triangle, what is the area of the circumscribed circle?

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

I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

Intern
Joined: 26 Jul 2010
Posts: 25
Followers: 2

Kudos [?]: 20 [0], given: 2

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 12:43
So I chose both statements together are sufficient but here is the explanation provided:
Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.

Why does that hold true? I get that :
1. DAB is 90
2. CAB = ACB = BAC = 60

and based on that all i can say is ABD < 60
so if we arent told ABD's value, cant ABD = 70 and ADB = 20?

In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.

thanks.
VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1125
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 145

Kudos [?]: 1408 [2] , given: 219

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 12:59
2
KUDOS
wown wrote:
So I chose both statements together are sufficient but here is the explanation provided:
Since both angles ACB and ADB are opposite side AB, and angle ACB is part of an equilateral triangle, we can say both ACB and ADB are 60 degrees.

Why does that hold true? I get that :
1. DAB is 90
2. CAB = ACB = BAC = 60

and based on that all i can say is ABD < 60
so if we arent told ABD's value, cant ABD = 70 and ADB = 20?

In any case, if we can deduce ABD from just statement 2, then the rest is easy. if ABD = 30 degrees (as told by statement 1), then DB = diameter. we can figure out the length of DB using the 30-60-90 relationship and solve for area.

thanks.

There is a thing you are not considering: in any right triangle inscribed in a circle, the hypotenuse coincides with a diameter of the circle.
So we know that DB is the diameter, that the bisectors of the ACB triangle intersect at the center (ancient math theory if I remember...).
ABC=60° is divided in two 30°angles, DAB=90° and so ADB=60°
And now as you said "we can figure out the length of DB using the 30-60-90 relationship and solve for area."
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Kant , Critique of Pure Reason

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

Manager
Status: Looking to improve
Joined: 15 Jan 2013
Posts: 177
GMAT 1: 530 Q43 V20
GMAT 2: 560 Q42 V25
GMAT 3: 650 Q48 V31
Followers: 1

Kudos [?]: 49 [1] , given: 65

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 13:28
1
KUDOS
Wown,

This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.

That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.

central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.

From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120

Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.

Thanks for sharing this problem...

//Kudos please, if the above explanation is good.
_________________

KUDOS is a way to say Thank You

Intern
Joined: 26 Jul 2010
Posts: 25
Followers: 2

Kudos [?]: 20 [0], given: 2

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 13:37
Oh, did not consider the hypotenuse-circle relationship. thanks for the explains.
Intern
Joined: 26 Jul 2010
Posts: 25
Followers: 2

Kudos [?]: 20 [0], given: 2

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 13:43
nt2010 wrote:
Wown,

This is very good trap C DS questions and if running out of time on exam you want to guess anything but C.

That said, the problem is using the relationship between an inscribed angle and central angle. central angle is the angle formed by an arc at the center and inscribed angle is the angle formed by the same arc at the circle.

central angle = 2 * inscribed angle - Here <ACB and <ADB are inscribed angles formed by arc AB.

From the stem, it is given that <ACB = 60 so the central angle for arch AB is 120. Make O as the center of the circle, so <AOB = 120

Now <ADB is the inscribed angle for <AOB so it will be 60 degrees. Triangle ADB is a 30-60-90 type right angle triangle and stmt 2 gives DA = 8 using which you can calculate the diameter of the circle as 16 and circumference as 16pi. Hence stmt B is sufficient.

Thanks for sharing this problem...

//Kudos please, if the above explanation is good.

Also did not consider this central angle-inscribed angle relationship... very interesting
Intern
Joined: 18 Nov 2011
Posts: 37
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98
Followers: 0

Kudos [?]: 9 [0], given: 0

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 16:05
Did the original question show a 90 degree mark on the illustration?

I was attempting to answer the question, but without the mark I didn't want to assume it was 90 degrees
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
Followers: 57

Kudos [?]: 721 [0], given: 135

Re: Why is this angle 60degrees? [#permalink]  03 Apr 2013, 20:40
Expert's post
wown wrote:
I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

We know that angle subtended by the same arc, on the circumference are always equal.

From F.S 1, the data given is of no value. It is so because Angle ACB = Angle ADB = 60 degrees. Thus, angle ABD has to be (90-60) = 30 degrees.Insufficient.

From F.S 2, we know the length of AD. Thus, in the case of a 60-30-90 triangle, we can easily calculate the hypotenuse = the diameter =BD and hence the area.Sufficient.

B.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 28242
Followers: 4461

Kudos [?]: 44999 [0], given: 6638

Re: In the figure above, if ABC is an equilateral triangle, and [#permalink]  04 Apr 2013, 02:23
Expert's post
wown wrote:
In the figure above, if ABC is an equilateral triangle, and DAB is a right triangle, what is the area of the circumscribed circle?

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

I came across a DS question whose explanation provides reasoning I do not follow. What am i missing?

Merging similar topics. Please refer to the solutions above.

_________________
Re: In the figure above, if ABC is an equilateral triangle, and   [#permalink] 04 Apr 2013, 02:23
Similar topics Replies Last post
Similar
Topics:
5 What is the area of equilateral triangle ABC? 10 23 Jan 2015, 06:37
An equilateral triangle ABC is inscribed in the circle. If 2 23 Apr 2011, 14:32
12 In the figure, ABC is an equilateral triangle, and DAB is a 5 01 Dec 2010, 05:29
19 Circle O is inscribed in equilateral triangle ABC, which is 14 26 Jun 2010, 06:52
In the figure (attached), ABC is an equilateral triangle, 10 08 Aug 2009, 06:41
Display posts from previous: Sort by