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

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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, 23:12
5
1
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 24 Apr 2013, 23:30
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]

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New post 25 Apr 2013, 08:28
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
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 15 May 2015, 05:49
The properties that will come in handy here are:
(i) In an inscribed triangle, if one side of the triangle is the diagonal of the circle, then that triangle is right angled. So, here, ABC is right angled.
(ii) Pythagoras theorem.

So, Area of ABC = (1/2)*BC*AB

From Pythagoras theorem, AB^2 = AC^2 - BC^2 = 2^2 - 1^2 = 3
=> AB = root(3)

So, Area of ABC = (1/2)*BC*AB = (1/2)*1*root(3) = root(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, 11:57
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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New post 03 Jan 2016, 11:13
reza52520 wrote:
Guyz why the height is not 1. isn't it obvious that the height is also a radius of the circle !!


The height is the perpendicular from a vertex to the opposite side. So, both AB and CB are heights as well as the perpendicular from B to AC. Which one turns to be a radius?
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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 06 Jan 2016, 14:20
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
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


To solve this one you don't even need to use pythagoras formula. It's a right triangle (bcs. Hypotenus=Diameter of the circle) with a hypotenuse of 2 and one side equal to 1. Here you have a 90-60-30 triangle, hence the third side is equal \(\sqrt{3}\) --> \(Area = \sqrt{3}*1/2\)
Answer 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 06 Jun 2016, 08:40
Always remember: A circle inscribed in a semicircle with always be a right angle triangle.
In this case,
∠ABC = 90

Hence in the triangle ABC, we have base = BC = 1
Hypotenuse = CA = 2
Height = AB^2 = 2^2 + 1^2
Hence AB = √3

Therefore, area = 1/2*√3*1 = √3/2

Correct option: 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 18 Sep 2016, 15:06
I have two ways for this:

According to theory since one of triangle's side is the diameter the triangle is a right triangle. In case you have any reservations about the hypotenuse Recall the rule which states: The larger side is opposite to to the larger angle. Clearly the larger side AC hence the angle B has to be 90 degrees. BUT this also means that since the larger side is AC is also the hypotenuse.

By using Pythagoras theorem AC^2 = BC^2 + AB^2 and given the info from the question you have 2^2 - 1 = AB^2 hence AB = sqrt(3). Also AB is the HEIGHT of the right triangle and thats why we bother to find it. Thus area = (base * height ) / 2 = (1 * sqrt(3) ) / 2


Another method

By drawing the MEDIAN BO we divide the triangle into two triangles with EQUAL areas (by definition). The MEDIAN BO splits the side AC into two EQUAL length lines hence AO = 1 and OC = 1 (By definition). Therefore the triangle BCO is a equilateral triangle with side r and the are of an equilateral is given by ( r^2 * sqrt(3) ) / 4 = ( 1 * sqrt(3) ) / 4

According to the MEDIAN properties the MEDIAN divides the triangle into TWO triangle with EQUAL areas hence the total are of ABC = 2*Area(BCO) hence sqrt(3) / 2 .

I think the second method is faster....
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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 Feb 2017, 23:03
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.


The sides of the triangle are 1, SQRT(3) and 2. Hence the area is (1/2)*1*SQRT(3) = SQRT(3)/2

Answer: 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 Feb 2017, 01:54
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
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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 Feb 2017, 03:03
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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 Feb 2017, 11:40
Plenty of good methods are already here. Am just going to demonstrate yet another way to do this. I agree this is a bit long.

Join OB. OB is 1 (radius).
We need to find area of triangle ABC = Ar(ABO) + Ar(OBC). let's say this is equation 1

OBC is an equilateral triangle with side=1.

So Ar (OBC) = root3/4.

Now need to find Ar(ABO). Note that angle ABO=30 degrees (because angle ABC =90 and angle OBC =60).
This means angle OAB is also 30 (because OA=OB and angles opposite to equal sides are equal).
This leaves angle AOB = 120 (sum of angles of a triangle is 180. So 180 - (30+30) = 120).

Ar(ABO) = (1/2).AO. BO. sin120 = root3/4 (because AO=BO=1 = radii and sin120 = root3/2)

Put these results in equation 1, Ar(ABC) = root3/4 + root3/4 = root3/2. Choice B.




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
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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New post 22 May 2017, 11:56
Hi sorry but i dont get why 1 and srqt3 are the base and the height. Is for every rectangle triangle a*b/2 (where c si the hypotenuse) the area?

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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 26 May 2017, 09:11
Bunuel wrote:
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


Hope it helps.


Usually it's always a GMAT rule that you should just assume ANYTHING...

The question only states that "the radius of the circle with center O is 1 and BC = 1".

How do I know that AC is the diameter? Yes, I see it on the graph but the graphs aren't accurate, are they? A could also be placed 1mm before the circle's line. It's not stated anywhere that the triangle is inscribed, is it?

Thanks for explaining!
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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 26 May 2017, 09:31
guenthermat wrote:
Bunuel wrote:
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


Hope it helps.


Usually it's always a GMAT rule that you should just assume ANYTHING...

The question only states that "the radius of the circle with center O is 1 and BC = 1".

How do I know that AC is the diameter? Yes, I see it on the graph but the graphs aren't accurate, are they? A could also be placed 1mm before the circle's line. It's not stated anywhere that the triangle is inscribed, is it?

Thanks for explaining!


First of all, if it were the way you are saying, how would you solve the question? It's a PS question, so one of the answers should be correct. How you get it?

Next, GMAT will not use such tricks. Check what is OG saying about the figures.

OFFICIAL GUIDE:

Problem Solving
Figures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Data Sufficiency:
Figures:
• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
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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 Apr 2018, 13:26
Quick way to solve this problem is by noticing that the inscribed triangle will make a right angle at point b.

Point B is 90 degrees. Following the 30 60 90 rule, the corresponding ratios are 1, sqrt3 and 2.

If you rotate the triangle upright, you can make height AB and the base BC. You can solve for the triangle area by setting the base to 1 and the height to sqrt3.

Triangle area = (1*sqrt3)/2

Answer 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 19 Jul 2018, 19:00
Hi all,

Why can't AC be the base, and a vertical line drawn down from B, the height? I keep getting answer A, Sqrt(2)/(2).

The vertical line from B forms a smaller 45-45-90 triangle to the right, with BC as the hypotenuse, and the height of the entire triangle becomes 1/sqrt(2). Since the base is then 2 times radius, or 2, the Area = b*h/2, so the answer is A. Where am I going wrong?
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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 19 Jul 2018, 21:03
TippingPoint93 wrote:
Hi all,

Why can't AC be the base, and a vertical line drawn down from B, the height? I keep getting answer A, Sqrt(2)/(2).

The vertical line from B forms a smaller 45-45-90 triangle to the right, with BC as the hypotenuse, and the height of the entire triangle becomes 1/sqrt(2). Since the base is then 2 times radius, or 2, the Area = b*h/2, so the answer is A. Where am I going wrong?


A triangle has three bases, so yes you can consider AC to be the base. The problem is that the perpendicular from B to AC does not create 45-45-90 triangle. Check the image below:
Image


Attachment:
triangle1.png
triangle1.png [ 8.55 KiB | Viewed 411 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 19 Jul 2018, 21:07
Bunuel wrote:
TippingPoint93 wrote:
Hi all,

Why can't AC be the base, and a vertical line drawn down from B, the height? I keep getting answer A, Sqrt(2)/(2).

The vertical line from B forms a smaller 45-45-90 triangle to the right, with BC as the hypotenuse, and the height of the entire triangle becomes 1/sqrt(2). Since the base is then 2 times radius, or 2, the Area = b*h/2, so the answer is A. Where am I going wrong?


A triangle has three bases, so yes you can consider AC to be the base. The problem is that the perpendicular from B to AC does not create 45-45-90 triangle. Check the image below:
Image


Attachment:
triangle1.png


Ahh, got it. Thanks, Bunuel.


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Re: In the figure above, the radius of circle with center O is 1 &nbs [#permalink] 19 Jul 2018, 21:07

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