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

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

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11 Sep 2008, 04:49
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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

[Reveal] Spoiler:
Attachment:

tmp.JPG [ 16.95 KiB | Viewed 54669 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 May 2017, 08:33, edited 3 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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

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11 Sep 2008, 05:26
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greatchap wrote:
Hi Everyone,

I encountered the following question in GMAT Prep Exam and was unable to solve it. The answer that is selected (image below) is correct. Though I did select the right answer but it was a fluke.

Q-1) In the figure above (below here) , the radius of circle with center O is 1 and BC = 1. What is the area of triangular region ABC ?

(a) sqrt(2)/2
(b) sqrt(3)/2
(c) 1
(d) sqrt(2)
(e) sqrt(3)

Image below shows diagram and ques.

Can anyone help me out??

Thanks,

Cheers,
GR

All triangles inscribed in a circle with the hypotenuse as the diameter of the circle are right triangles.

Therefore

x^2 + 1^2 = 2^2
x = sqrt(3)
Area = x*1*(1/2) = sqrt(3)/2

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

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11 Sep 2008, 05:36
Given,
OA=OC=1
BC=1

Angle ABC = 90
=> ab^2+bc^2 = ac^2
=> ab^2 = ac^2 - bc^2
=> ab^2 = 2^2 - 1^2 = sqrt 3

Area of a traingle = b*h/2
=> 1*sqrt 3/2

B

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

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27 Aug 2009, 11:03
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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

Attachment:

sample.JPG [ 2.87 KiB | Viewed 87946 times ]

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

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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

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

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27 Aug 2009, 12:26
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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!

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

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28 Aug 2009, 00:06
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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.
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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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

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

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28 Aug 2009, 21:00
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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.
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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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?

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

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03 Jul 2010, 13:53
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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.

Hope it helps.

[Reveal] Spoiler:
Attachment:

Math_Tri_inscribed.png [ 6.47 KiB | Viewed 81198 times ]

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

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16 Jul 2010, 01:49

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

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

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07 Jun 2011, 08:55
I did not apply pythagoras.

I solved this pretty quickly, as I interpreted this as a right triangle inscribed in a circle as 30-60-90 triangle.

so longer leg = 1/2*hypotenuse*root3.

area = 1/2 *root 3 * 1

hope I am correct ?
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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07 Jun 2011, 10:28
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bblast wrote:
I did not apply pythagoras.

I solved this pretty quickly, as I interpreted this as a right triangle inscribed in a circle as 30-60-90 triangle.

so longer leg = 1/2*hypotenuse*root3.

area = 1/2 *root 3 * 1

hope I am correct ?

Yes, you are. There are multiple ways of arriving at the value of AB.
You see that Cos C = 1/2 so C must be 60 degrees
$$Sin 60 = \sqrt{3}/2$$ so $$AB = \sqrt{3}$$
(Let me point out here that you are not expected to know trigonometry in GMAT.)

Or since the sides are 1 and 2, the third side must be $$\sqrt{3}$$ by Pythagorean theorem.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18128 [2], given: 236 Manager Affiliations: The Earth organization, India Joined: 25 Dec 2010 Posts: 190 Kudos [?]: 15 [0], given: 12 WE 1: SAP consultant-IT 2 years WE 2: Entrepreneur-family business 2 years Re: In the figure above, the radius of circle with center O is 1 [#permalink] ### Show Tags 08 Jun 2011, 07:11 thanks karishma so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ? _________________ Cheers !! Quant 47-Striving for 50 Verbal 34-Striving for 40 Kudos [?]: 15 [0], given: 12 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7796 Kudos [?]: 18128 [3], given: 236 Location: Pune, India Re: In the figure above, the radius of circle with center O is 1 [#permalink] ### Show Tags 08 Jun 2011, 07:26 3 This post received KUDOS Expert's post bblast wrote: thanks karishma so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ? Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle. GMAT generally questions you on one of 30-60-90 and 45-45-90... _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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08 Jun 2011, 07:36
Thank u . +1 Kudos

But it makes sense that if one triangle-circle holds this relation. Any triangle having the diameter as the hypotenuse should. As the angles for it in a circle will not change. Only the sides will grow or shrink. But ya maybe the ratio 1:\sqrt{3} :2 occurs at this particular size of the 2 geometric figures !!

So I will put ur words in my flashcard.
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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08 Jun 2011, 18:06
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These type of question can easily be solved by applying a simple formula.

A question in which triangle is given within a circle and we find out the area of triangle,then we can apply the formula given below

A= abc/4R
where a,b,c are the side of a triangle and R is the circum radius.

here,
a = AC = 2
b = BC = 1
c = AB = √3 ( AB²+BC²=AC² )
R = 1

Put these values on the formula, we get
A= (2*1*√3)/(4*1)
A= √3/2 Ans.

For such formula's, click the link below
http://gmatclub.com/forum/fundas-of-geometry-part-i-114507.html

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

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08 Jun 2011, 18:17
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bblast wrote:
Thank u . +1 Kudos

But it makes sense that if one triangle-circle holds this relation. Any triangle having the diameter as the hypotenuse should. As the angles for it in a circle will not change. Only the sides will grow or shrink. But ya maybe the ratio 1:\sqrt{3} :2 occurs at this particular size of the 2 geometric figures !!

So I will put ur words in my flashcard.

Attachment:

Ques2.jpg [ 10.94 KiB | Viewed 52946 times ]

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

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20 Nov 2011, 21:45
VeritasPrepKarishma wrote:
bblast wrote:
thanks karishma

so whenever I see a triangle in a circle and one arm of triangle as diameter if circle, I can quickly solve it using 30-60-90 formulas ?

Actually no. Here we know that the diameter is 2 units and one side is 1 unit so the triangle is 30-60-90. It may not always be the case. Say, if the diameter is 2 and one side is given as root 2, it will be a 45-45-90 triangle.
GMAT generally questions you on one of 30-60-90 and 45-45-90...

Hi Karishma

I wanted to clarify my misunderstanding. In 45:45:90 the multiples will be x:x: $$\sqrt{2x}$$
So if the diameter is 2 the other 2 sides should be 2 and 2$$\sqrt{2x}$$?

Thanks

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Re: In the figure above, the radius of circle with center O is 1   [#permalink] 20 Nov 2011, 21:45

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