In the figure above, the radius of circle with center O is 1 : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 19 Jan 2017, 10:24

### 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 above, the radius of circle with center O is 1

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 09 Aug 2009
Posts: 20
Followers: 0

Kudos [?]: 14 [3] , given: 1

In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

27 Aug 2009, 11:03
3
KUDOS
10
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

74% (01:35) correct 26% (00:37) wrong based on 642 sessions

### HideShow timer Statistics

Attachment:

sample.JPG [ 2.87 KiB | Viewed 41487 times ]
In the figure above, the radius of the circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

(A) √2/2
(B) √3/2
(C) 1
(D) √2
(E) √3

My way
[Reveal] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B
[Reveal] Spoiler: OA
Manager
Affiliations: CFA Level 2 Candidate
Joined: 29 Jun 2009
Posts: 221
Schools: RD 2: Darden Class of 2012
Followers: 3

Kudos [?]: 210 [7] , given: 2

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

27 Aug 2009, 12:26
7
KUDOS
2
This post was
BOOKMARKED
You could use Pythagorean Theorum Here to solve AB.

AC = Diameter or Rx2 = 2
BC = 1

$$x^2+1^2=2^2$$
$$x^2=3$$
$$x=sqrt3$$

Now we have $$(b*h)/2 = (sqrt3*1)/2$$

B!
Manager
Joined: 28 Jul 2009
Posts: 124
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Followers: 6

Kudos [?]: 75 [4] , given: 12

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

28 Aug 2009, 00:06
4
KUDOS
Quote:
My way

[Obscure] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

You should try and avoid assuming any detail in GMAT. Though, you can use a similar way for POE.

Here, the solution is linked to the height of the triangle.
Given, BC = 1, Radius = 1. ie OB = OC = 1.
Therefore, we can say that /\ OBC is an equilateral triangle.
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.
Ans : B.

Hope this helps.
_________________

GMAT offended me. Now, its my turn!
Will do anything for Kudos! Please feel free to give one.

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1123
Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8
Followers: 181

Kudos [?]: 1965 [4] , given: 219

Re: In the figure above, the radius of the circle with center O [#permalink]

### Show Tags

24 Apr 2013, 22:12
4
KUDOS
The triangle is a right triangle because one side is the diameter of the circle.

Calculate the third side as $$\sqrt{2^2-1^2}=\sqrt{3}$$

I created a rectangular by translating the triangle (see picture), and the area will be half of the rectangular's.
$$AreaRec=b*h=1*\sqrt{3}$$
$$AreaTri=\frac{AreaRec}{2}=\sqrt{3}/2$$
B

Let me know if this helps
Attachments

sample.JPG [ 16.43 KiB | Viewed 34687 times ]

_________________

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]

Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7079

Kudos [?]: 93184 [3] , given: 10553

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

03 Jul 2010, 13:53
3
KUDOS
Expert's post
2
This post was
BOOKMARKED
abhas59 wrote:
In the figure above, the radius of circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

Hi guys,
Just one doubt, how can we say that triangle ABC is a right angled triangle given the radius is one & other side BC is one.
I mean to say that is there any property for circle that if radius is equal to any one side of the triangle, then the inscribed triangle so formed will be right-angled one?

We know that AC is a diameter. There is a property of a right triangle inscribed in circle:

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 side, then that triangle is a right triangle and diameter is hypotenuse.
Attachment:

Math_Tri_inscribed.png [ 6.47 KiB | Viewed 39746 times ]

Hope it helps.
_________________
Manager
Joined: 28 Jul 2009
Posts: 124
Location: India
Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business
Followers: 6

Kudos [?]: 75 [1] , given: 12

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

28 Aug 2009, 21:00
1
KUDOS
defoue wrote:
bhanushalinikhil wrote:
Quote:
My way

[Obscure] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

You should try and avoid assuming any detail in GMAT. Though, you can use a similar way for POE.

Here, the solution is linked to the height of the triangle.
Given, BC = 1, Radius = 1. ie OB = OC = 1.
Therefore, we can say that /\ OBC is an equilateral triangle.
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.
Ans : B.

Hope this helps.

Hi Buddy, I like your approach because I did not want to assume it was a rect triangle.
But, could you pls explain this : not sure I got it
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.

Thx

Sure thing. But what part did you not understand?
>Height of equilateral triangle is SQRT(3)/2 * (side)^2 - Its a generic formula. you just have to remember it.
>Therefore, height = SQRT(3)/2.
>Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2. (Area of triangle formula)
>Area = SQRT(3)/2.

I am sorry if I missed something. Please tell me which part would you like to know and I will try my best to help you.
_________________

GMAT offended me. Now, its my turn!
Will do anything for Kudos! Please feel free to give one.

Manager
Joined: 10 Jul 2009
Posts: 169
Followers: 1

Kudos [?]: 82 [0], given: 8

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

27 Aug 2009, 11:38
Given AC = 2, BC = 1.

Since AC is the diameter and B is a point on the circle, triangle ABC is a right angled triangle.

Area of the triangle = 1/2 * base* height

In the figure, base and height are BC & AB.
AB = \sqrt{3}

So area of the triangle = 1/2 *\sqrt{3}*1 = \sqrt{3}/2
Hence B
Intern
Joined: 30 Jun 2009
Posts: 48
Followers: 1

Kudos [?]: 11 [0], given: 2

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

28 Aug 2009, 04:34
bhanushalinikhil wrote:
Quote:
My way

[Obscure] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

You should try and avoid assuming any detail in GMAT. Though, you can use a similar way for POE.

Here, the solution is linked to the height of the triangle.
Given, BC = 1, Radius = 1. ie OB = OC = 1.
Therefore, we can say that /\ OBC is an equilateral triangle.
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.
Ans : B.

Hope this helps.

Hi Buddy, I like your approach because I did not want to assume it was a rect triangle.
But, could you pls explain this : not sure I got it
Height of equilateral triangle is SQRT(3)/2 * (side)^2.
Therefore, height = SQRT(3)/2.
Therefore, area of /\ ABC = (1/2)*2 (Given) * SQRT(3)/2.
Area = SQRT(3)/2.

Thx
Manager
Joined: 06 Mar 2010
Posts: 107
Followers: 2

Kudos [?]: 5 [0], given: 11

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

03 Jul 2010, 13:18
Hi guys,
Just one doubt, how can we say that triangle ABC is a right angled triangle given the radius is one & other side BC is one.
I mean to say that is there any property for circle that if radius is equal to any one side of the triangle, then the inscribed triangle so formed will be right-angled one?
Manager
Joined: 06 Mar 2010
Posts: 107
Followers: 2

Kudos [?]: 5 [0], given: 11

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

07 Jul 2010, 08:54
Thanks bunuel for explanation
Manager
Joined: 04 Feb 2010
Posts: 200
Followers: 1

Kudos [?]: 41 [0], given: 8

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

07 Jul 2010, 09:18
Thanks for the explanation as well!
Manager
Joined: 12 Jun 2007
Posts: 128
Followers: 6

Kudos [?]: 71 [0], given: 2

Re: Geometry Problem Gmat Prep (hard one) [#permalink]

### Show Tags

16 Jul 2010, 01:49
answer should be b

BC=1, AC would become diameter so 2

AB = square root of 3

Area of triangle = 1/2 * base * height
= 1/2 * square root of 3 * 1
= 1/2 * square root of 3
Current Student
Joined: 04 Mar 2013
Posts: 69
Location: India
Concentration: Strategy, Operations
Schools: Booth '17 (M)
GMAT 1: 770 Q50 V44
GPA: 3.66
WE: Operations (Manufacturing)
Followers: 5

Kudos [?]: 52 [0], given: 27

Re: In the figure above, the radius of the circle with center O [#permalink]

### Show Tags

24 Apr 2013, 22:30
Zarrolou wrote:
The triangle is a right triangle because one side is the diameter of the circle.

Calculate the third side as $$\sqrt{2^2-1^2}=\sqrt{3}$$

I created a rectangular by translating the triangle (see picture), and the area will be half of the rectangular's.
$$AreaRec=b*h=1*\sqrt{3}$$
$$AreaTri=\frac{AreaRec}{2}=\sqrt{3}/2$$
B

Let me know if this helps

You could just use the formula for the area of a triangle which is 1/2xBxH where AB is the height and BC is the base as /_ABC is 90
_________________

When you feel like giving up, remember why you held on for so long in the first place.

Manager
Status: Trying.... & desperate for success.
Joined: 17 May 2012
Posts: 78
Location: India
Schools: NUS '15
GMAT 1: Q33 V27
GPA: 2.92
WE: Analyst (Computer Software)
Followers: 0

Kudos [?]: 76 [0], given: 61

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

25 Apr 2013, 07:28
over2u wrote:
Attachment:
sample.JPG
In the figure above, the radius of the circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

(A) √2/2
(B) √3/2
(C) 1
(D) √2
(E) √3

My way
[Reveal] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

When a triangle that is right angled as 30, 60 & 90 then the sides opposite to it will be in ratio of 1:\sqrt{3} : 2.

Angle ABC is 90.
Triangle OBC is equilateral triangle. Hence, Angle OCB is 60.

In triangle ABC, angle BAC = 180 -90- 60 = 30

Given that AC = 2, BC = 1, hence AB = \sqrt{3}

Therefore area of triangle = 1/2 * 1 * \sqrt{3} = \sqrt{3}/2
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13456
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

10 May 2014, 08:57
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.
_________________
Intern
Joined: 10 Oct 2013
Posts: 34
Followers: 0

Kudos [?]: 1 [0], given: 3

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

15 May 2015, 04:49
The properties that will come in handy here are:
(i) In an inscribed triangle, if one side of the triangle is the diagonal of the circle, then that triangle is right angled. So, here, ABC is right angled.
(ii) Pythagoras theorem.

So, Area of ABC = (1/2)*BC*AB

From Pythagoras theorem, AB^2 = AC^2 - BC^2 = 2^2 - 1^2 = 3
=> AB = root(3)

So, Area of ABC = (1/2)*BC*AB = (1/2)*1*root(3) = root(3)/2
Intern
Joined: 22 Nov 2014
Posts: 6
Concentration: Entrepreneurship, Marketing
GPA: 3.2
Followers: 2

Kudos [?]: 11 [0], given: 48

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

02 Jan 2016, 10:57
Guyz why the height is not 1. isn't it obvious that the height is also a radius of the circle !!
Math Expert
Joined: 02 Sep 2009
Posts: 36566
Followers: 7079

Kudos [?]: 93184 [0], given: 10553

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

03 Jan 2016, 10:13
reza52520 wrote:
Guyz why the height is not 1. isn't it obvious that the height is also a radius of the circle !!

The height is the perpendicular from a vertex to the opposite side. So, both AB and CB are heights as well as the perpendicular from B to AC. Which one turns to be a radius?
_________________
Director
Joined: 10 Mar 2013
Posts: 608
Location: Germany
Concentration: Finance, Entrepreneurship
GMAT 1: 580 Q46 V24
GPA: 3.88
WE: Information Technology (Consulting)
Followers: 15

Kudos [?]: 266 [0], given: 200

In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

06 Jan 2016, 13:20
over2u wrote:
Attachment:
sample.JPG
In the figure above, the radius of the circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

(A) √2/2
(B) √3/2
(C) 1
(D) √2
(E) √3

My way
[Reveal] Spoiler: Description
We should find the area of triangle with this formula: 0.5*h*b. It is quite easy but we don't know the length of heigh. So what I do in this situation: I just imagine that h equal BC, so the area of triangle is 2*0.5*1=1, but in fact its smaller, because heigh is slightly less than BC. So we can look on answers and define the aproriate value for our triangle area. It is definetly less than 1, sqr(2), sqr(3). So we have two choises: sqr(2)/2 which equals to 1.4/2 or 0.7 and sqr(3)/2 which equals to 1.7/2 or 0.85. I pick the biggest because heigh just SLIGHTLY less then BC. So the answer is B

To solve this one you don't even need to use pythagoras formula. It's a right triangle (bcs. Hypotenus=Diameter of the circle) with a hypotenuse of 2 and one side equal to 1. Here you have a 90-60-30 triangle, hence the third side is equal $$\sqrt{3}$$ --> $$Area = \sqrt{3}*1/2$$
_________________

When you’re up, your friends know who you are. When you’re down, you know who your friends are.

Share some Kudos, if my posts help you. Thank you !

800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50
GMAT PREP 670
MGMAT CAT 630
KAPLAN CAT 660

Optimus Prep Instructor
Joined: 06 Nov 2014
Posts: 1775
Followers: 51

Kudos [?]: 393 [0], given: 21

Re: In the figure above, the radius of circle with center O is 1 [#permalink]

### Show Tags

06 Jun 2016, 07:40
Always remember: A circle inscribed in a semicircle with always be a right angle triangle.
In this case,
∠ABC = 90

Hence in the triangle ABC, we have base = BC = 1
Hypotenuse = CA = 2
Height = AB^2 = 2^2 + 1^2
Hence AB = √3

Therefore, area = 1/2*√3*1 = √3/2

Correct option: B
_________________

# Janielle Williams

Customer Support

Special Offer: $80-100/hr. Online Private Tutoring GMAT On Demand Course$299
Free Online Trial Hour

Re: In the figure above, the radius of circle with center O is 1   [#permalink] 06 Jun 2016, 07:40

Go to page    1   2    Next  [ 21 posts ]

Similar topics Replies Last post
Similar
Topics:
6 In the figure above, point O is the center of the circle 7 05 Mar 2013, 10:38
23 In the figure above, showing circle with center O and points 8 28 Feb 2013, 04:23
21 In the figure above, showing circle with center O and points 19 25 Feb 2013, 01:21
150 In the figure above, point O is the center of the circle and 39 16 Jan 2011, 15:56
31 In the figure above, the radius of the circle with center O 17 11 Sep 2008, 04:49
Display posts from previous: Sort by

# In the figure above, the radius of circle with center O is 1

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.