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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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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:
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[Reveal] Spoiler: OA

Last edited by Bunuel on 26 May 2017, 09:33, 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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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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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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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
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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

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

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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.

Image

Hope it helps.

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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...
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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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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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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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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.


To help you visualize, I will leave you with a diagram.
Attachment:
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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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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rishit1080 wrote:
How do you know that BC is the base and AC is the height? Is it because on right angle triangle the 90 degree angle is never the base or the height? Bunuel


The altitude (height) of a triangle is the perpendicular from the base to the opposite vertex (the base may need to be extended). Since there are three possible bases, there are also three possible altitudes. In case of a right triangle both legs are altitudes as well as perpendicular from right angle to hypotenuse.
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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New post 11 Sep 2008, 06:36
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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New post 27 Aug 2009, 12:38
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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New post 28 Aug 2009, 05:34
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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New post 03 Jul 2010, 14:18
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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New post 16 Jul 2010, 02:49
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
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]

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New post 07 Jun 2011, 09:55
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] 07 Jun 2011, 09:55

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