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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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New post Updated on: 19 Jul 2018, 20:18
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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

Attachment:
tmp.JPG
tmp.JPG [ 16.95 KiB | Viewed 129911 times ]

Originally posted by greatchap on 11 Sep 2008, 04:49.
Last edited by Bunuel on 19 Jul 2018, 20:18, edited 4 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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New post 03 Jul 2010, 13:53
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abhas59 wrote:
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

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.

Attachment:
Math_Tri_inscribed.png
Math_Tri_inscribed.png [ 6.47 KiB | Viewed 163898 times ]

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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, 12:26
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6
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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New post 11 Sep 2008, 05:26
6
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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New post 27 Aug 2009, 11:38
1
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, 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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New post 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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New post 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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New post 07 Jun 2011, 10:28
2
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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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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New post 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 ?
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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 08 Jun 2011, 07:26
4
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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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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New post 08 Jun 2011, 07:36
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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.
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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 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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New post 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.


To help you visualize, I will leave you with a diagram.
Attachment:
Ques2.jpg
Ques2.jpg [ 10.94 KiB | Viewed 119824 times ]

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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 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\(\sqrt{2x}\)?

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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 21 Nov 2011, 20:35
melguy wrote:

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


In 45-45-90, the sides will be in the ratio 1:1:\(\sqrt{2}\) or you can say they will be x, x and \(\sqrt{2}x\) (Mind you, the root is only on 2, not on x). \(\sqrt{2}x\), the longest side, is the hypotenuse.

Here, the diameter is the hypotenuse i.e. the longest side. The right angle is opposite to the diameter. We know that the side opposite to the largest angle is the longest side. Hence the diameter has to be the longest side i.e. \(\sqrt{2}x\).
If \(\sqrt{2}x\) = 2
x = \(\sqrt{2}\)
So the other two equal sides will be \(\sqrt{2}\) each.
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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 24 Apr 2013, 22:12
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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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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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New post 25 Apr 2013, 07:28
over2u wrote:
Attachment:
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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
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
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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 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 !!
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Re: In the figure above, the radius of circle with center O is 1  [#permalink]

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Re: In the figure above, the radius of circle with center O is 1   [#permalink] 03 Jan 2016, 10:13

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