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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]  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
[Reveal] Spoiler: OA
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Re: Geometry Problem Gmat Prep (hard one) [#permalink]  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: Geometry Problem Gmat Prep (hard one) [#permalink]  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: Geometry Problem Gmat Prep (hard one) [#permalink]  28 Aug 2009, 00:06
4
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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: Geometry Problem Gmat Prep (hard one) [#permalink]  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: Geometry Problem Gmat Prep (hard one) [#permalink]  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: Geometry Problem Gmat Prep (hard one) [#permalink]  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: Geometry Problem Gmat Prep (hard one) [#permalink]  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.
Attachment:

Math_Tri_inscribed.png [ 6.47 KiB | Viewed 19403 times ]

Hope it helps.
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Re: Geometry Problem Gmat Prep (hard one) [#permalink]  07 Jul 2010, 08:54
Thanks bunuel for explanation
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Re: Geometry Problem Gmat Prep (hard one) [#permalink]  07 Jul 2010, 09:18
Thanks for the explanation as well!
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Re: Geometry Problem Gmat Prep (hard one) [#permalink]  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 the circle with center O [#permalink]  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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sample.JPG [ 16.43 KiB | Viewed 14365 times ]

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Re: In the figure above, the radius of the circle with center O [#permalink]  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
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]  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
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Re: In the figure above, the radius of circle with center O is 1 [#permalink]  10 May 2014, 08:57
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Re: In the figure above, the radius of circle with center O is 1   [#permalink] 10 May 2014, 08:57
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