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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

Re: Geometry Problem Gmat Prep (hard one) [#permalink]
28 Aug 2009, 00:06

4

This post received KUDOS

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. _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

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.

Re: Geometry Problem Gmat Prep (hard one) [#permalink]
28 Aug 2009, 21:00

1

This post received KUDOS

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. _________________

GMAT offended me. Now, its my turn! Will do anything for Kudos! Please feel free to give one.

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?

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 20201 times ]

Re: In the figure above, the radius of the circle with center O [#permalink]
24 Apr 2013, 22:12

3

This post received KUDOS

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

Attachments

sample.JPG [ 16.43 KiB | Viewed 15162 times ]

_________________

It is beyond a doubt that all our knowledge that begins with experience.

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 _________________

When you feel like giving up, remember why you held on for so long in the first place.

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

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]
15 May 2015, 04:49

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

gmatclubot

Re: In the figure above, the radius of circle with center O is 1
[#permalink]
15 May 2015, 04:49

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