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In the figure above, the radius of the circle with center O

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In the figure above, the radius of the circle with center O [#permalink] New post 11 Sep 2008, 04:49
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A
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D
E

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Question Stats:

71% (01:34) correct 29% (00:23) wrong based on 79 sessions
Image
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

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figure-above-the-radius-of-circle-with-center-o-is-83121.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 20 Oct 2013, 03:28, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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Re: Geometry ques...tricky [#permalink] New post 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: Geometry ques...tricky [#permalink] New post 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

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Re: Geometry ques...tricky [#permalink] New post 11 Sep 2008, 07:04
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Thanks a lot :) guys for the explanation. I got it. Cheers...;-)
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Re: Geometry ques...tricky [#permalink] 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: Geometry ques...tricky [#permalink] New post 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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Re: Geometry ques...tricky [#permalink] 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: Geometry ques...tricky [#permalink] New post 08 Jun 2011, 07:26
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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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Re: Geometry ques...tricky [#permalink] New post 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: Geometry ques...tricky [#permalink] 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: Geometry ques...tricky [#permalink] New post 08 Jun 2011, 18:17
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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:
Ques2.jpg
Ques2.jpg [ 10.94 KiB | Viewed 10283 times ]

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Re: Geometry ques...tricky [#permalink] New post 08 Jun 2011, 20:06
Yup, the lights are finally switched on inside my cranium, :| +1 again for clearing this,
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Re: Geometry ques...tricky [#permalink] New post 09 Jun 2011, 20:48
B...using Pythagoras theorem and using the property that a triangle with the hypotenuse as the diameter of a circle subtends a 90 degree anywhere on the circle.
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Re: Geometry ques...tricky [#permalink] 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}?

Thanks
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Re: Geometry ques...tricky [#permalink] New post 21 Nov 2011, 20:35
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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: Geometry ques...tricky [#permalink] New post 28 Nov 2011, 22:59
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...



thanks- ur ex helped a lot
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Re: Geometry ques...tricky [#permalink] New post 19 Oct 2013, 11:30
Can someone explain why sq.root 3 is the height in this question
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Re: Geometry ques...tricky [#permalink] New post 20 Oct 2013, 03:29
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Re: Geometry ques...tricky   [#permalink] 20 Oct 2013, 03:29
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