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Re: Interesting Geometry Problem: Veritas [#permalink]
26 Jun 2010, 08:01

5

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

Hussain15 wrote:

Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?

(1) The area of circle O is \(4\)pie.

(2) The area of triangle ABC is \(12\sqrt{2}\).

For equilateral triangle: • The radius of the circumscribed circle is \(R=a*\frac{\sqrt{3}}{3}\), (where \(a\) is the side of equilateral triangle); • The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\); • The area of equilateral triangle is \(A=a^2*\frac{\sqrt{3}}{4}\).

We are asked to calculate area of bigger circle P - \(area_P=\pi{R^2}\). Note that knowing any of the following: the side of equilateral triangle \(a\), radius of the smaller circle O (as it gives \(a\)) or the radius of P itself is sufficient to calculate area of P.

(1) The area of circle O is \(4\pi\) --> we can find \(r\) --> we can find \(a\) --> we can find \(R\). Sufficient.

(2) The area of triangle ABC is \(12\sqrt{3}\) --> we can find \(a\) --> we can find \(R\). Sufficient.

Re: Interesting Geometry Problem: Veritas [#permalink]
26 Jun 2010, 08:31

1

This post received KUDOS

Expert's post

if you don't have enough time to calculate or don't remember formulas, here is fast "intuitive" approach:

Let's imagine this highly fixed structure. If you change any linear size or area, the structure just scales. We can't change any part of the system without proportionally changing all others parts. Once you get this "intuitive" idea, any linear size or area of any part of the structure defines all other linear sizes and areas of the system. For instance, if we know the height of the triangle, it's enough to find all other parameters in the system. Both statements give us information about one of the parts of the system. So, it's D.

P.S. It's a lot of text but it took 10-20sec. _________________

Re: Interesting Geometry Problem: Veritas [#permalink]
26 Jun 2010, 09:41

1

This post received KUDOS

I wish I could understand the system approach. Perhaps its the thinking of a MBA student, which I am unable to get. I try to go through it again. Let's see!!!

Re: Interesting Geometry Problem: Veritas [#permalink]
26 Jun 2010, 15:56

D it is.

If a circle is inscribed in an equilateral triangle , you can find radius if the side if a triangle /height of the triangle is given or you can find side of a triangle if radius of the inscrbed circle is given

Even if you dont remember formulas as spelled out by Bunnel..you just need to remember the above fact.

Re: Interesting Geometry Problem: Veritas [#permalink]
28 Jun 2010, 06:34

1

This post received KUDOS

PriyaRai wrote:

D it is.

If a circle is inscribed in an equilateral triangle , you can find radius if the side if a triangle /height of the triangle is given or you can find side of a triangle if radius of the inscrbed circle is given

Even if you dont remember formulas as spelled out by Bunnel..you just need to remember the above fact.

Even I don't remember all the formulas used above, i was able to get the answer as D with little logic and knowledge.

Who wants to know all the formulas, some time you can do without that.

there's a saying:

Who wants to know the price of everything and value of nothing.

_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

Re: Interesting Geometry Problem: Veritas [#permalink]
10 Nov 2010, 17:51

2

This post received KUDOS

Expert's post

walker wrote:

if you don't have enough time to calculate or don't remember formulas, here is fast "intuitive" approach:

Let's imagine this highly fixed structure. If you change any linear size or area, the structure just scales. We can't change any part of the system without proportionally changing all others parts. Once you get this "intuitive" idea, any linear size or area of any part of the structure defines all other linear sizes and areas of the system. For instance, if we know the height of the triangle, it's enough to find all other parameters in the system. Both statements give us information about one of the parts of the system. So, it's D.

P.S. It's a lot of text but it took 10-20sec.

I am myself a proponent of exactly this thinking. It makes perfect sense and takes a few seconds. And, you get very good at it with practice. Something akin to this for the intuitively inclined: "If there is only one way in which you can draw a geometry diagram with certain specifications, you will be able to find all other sides and angles." _________________

A) if the area of circle is given. you can (r1)of the inscribed circle and from that the sides of the triangle. Sides of triangle can give you the radius (r2) of the outer circle, enough to answer the question

B) area of triangle will give you the side and also the radius (r2) or circumscribed circle.

Re: Circle O is inscribed in equilateral triangle ABC, which is [#permalink]
18 Jan 2014, 13:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Interesting Geometry Problem: Veritas [#permalink]
15 May 2014, 01:06

1

This post received KUDOS

Bunuel wrote:

Hussain15 wrote:

Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?

(1) The area of circle O is \(4\)pie.

(2) The area of triangle ABC is \(12\sqrt{2}\).

For equilateral triangle: • The radius of the circumscribed circle is \(R=a*\frac{\sqrt{3}}{3}\), (where \(a\) is the side of equilateral triangle); • The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\); • The area of equilateral triangle is \(A=a^2*\frac{\sqrt{3}}{4}\).

We are asked to calculate area of bigger circle P - \(area_P=\pi{R^2}\). Note that knowing any of the following: the side of equilateral triangle \(a\), radius of the smaller circle O (as it gives \(a\)) or the radius of P itself is sufficient to calculate area of P.

(1) The area of circle O is \(4\pi\) --> we can find \(r\) --> we can find \(a\) --> we can find \(R\). Sufficient.

(2) The area of triangle ABC is \(12\sqrt{2}\) --> we can find \(a\) --> we can find \(R\). Sufficient.

Answer: D.

Dear Members,

Has anyone noticed that both the statements contradict each other

From statement 1 , we get \(a = 4\sqrt3\) or \(a^2 = 48\)

from statement 2 we get \(12\sqrt2 = a^2 *\frac{\sqrt3}{4}\)

or \(a^2 = 48*\frac{\sqrt2}{sqrt3}\)

both the statements should give the same value for a and \(a^2\) ,( side of the triangle).

Let me know if I am misinterpreting anything.

Although the answer is still D, both the statements shouldn't give different values for a.

Re: Interesting Geometry Problem: Veritas [#permalink]
20 May 2014, 02:37

qlx wrote:

Bunuel wrote:

Hussain15 wrote:

Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?

(1) The area of circle O is \(4\)pie.

(2) The area of triangle ABC is \(12\sqrt{2}\).

For equilateral triangle: • The radius of the circumscribed circle is \(R=a*\frac{\sqrt{3}}{3}\), (where \(a\) is the side of equilateral triangle); • The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\); • The area of equilateral triangle is \(A=a^2*\frac{\sqrt{3}}{4}\).

We are asked to calculate area of bigger circle P - \(area_P=\pi{R^2}\). Note that knowing any of the following: the side of equilateral triangle \(a\), radius of the smaller circle O (as it gives \(a\)) or the radius of P itself is sufficient to calculate area of P.

(1) The area of circle O is \(4\pi\) --> we can find \(r\) --> we can find \(a\) --> we can find \(R\). Sufficient.

(2) The area of triangle ABC is \(12\sqrt{2}\) --> we can find \(a\) --> we can find \(R\). Sufficient.

Answer: D.

Dear Members,

Has anyone noticed that both the statements contradict each other

From statement 1 , we get \(a = 4\sqrt3\) or \(a^2 = 48\)

from statement 2 we get \(12\sqrt2 = a^2 *\frac{\sqrt3}{4}\)

or \(a^2 = 48*\frac{\sqrt2}{sqrt3}\)

both the statements should give the same value for a and \(a^2\) ,( side of the triangle).

Let me know if I am misinterpreting anything.

Although the answer is still D, both the statements shouldn't give different values for a.

The DS question asks for data sufficiency and not the final answer. Two statements may give two different answers or same answers is not of any merit in these questions.

Re: Interesting Geometry Problem: Veritas [#permalink]
20 May 2014, 03:27

Expert's post

mittalg wrote:

qlx wrote:

Bunuel wrote:

For equilateral triangle: • The radius of the circumscribed circle is \(R=a*\frac{\sqrt{3}}{3}\), (where \(a\) is the side of equilateral triangle); • The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\); • The area of equilateral triangle is \(A=a^2*\frac{\sqrt{3}}{4}\).

We are asked to calculate area of bigger circle P - \(area_P=\pi{R^2}\). Note that knowing any of the following: the side of equilateral triangle \(a\), radius of the smaller circle O (as it gives \(a\)) or the radius of P itself is sufficient to calculate area of P.

(1) The area of circle O is \(4\pi\) --> we can find \(r\) --> we can find \(a\) --> we can find \(R\). Sufficient.

(2) The area of triangle ABC is \(12\sqrt{2}\) --> we can find \(a\) --> we can find \(R\). Sufficient.

Answer: D.

Dear Members,

Has anyone noticed that both the statements contradict each other

From statement 1 , we get \(a = 4\sqrt3\) or \(a^2 = 48\)

from statement 2 we get \(12\sqrt2 = a^2 *\frac{\sqrt3}{4}\)

or \(a^2 = 48*\frac{\sqrt2}{sqrt3}\)

both the statements should give the same value for a and \(a^2\) ,( side of the triangle).

Let me know if I am misinterpreting anything.

Although the answer is still D, both the statements shouldn't give different values for a.

The DS question asks for data sufficiency and not the final answer. Two statements may give two different answers or same answers is not of any merit in these questions.

Don't fall for such traps.

Cheers!!!

That's not true. On the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other or the stem. _________________

Re: Interesting Geometry Problem: Veritas [#permalink]
20 May 2014, 03:28

Expert's post

qlx wrote:

Bunuel wrote:

Hussain15 wrote:

Circle O is inscribed in equilateral triangle ABC, which is itself inscribed in circle P. What is the area of circle P?

(1) The area of circle O is \(4\)pie.

(2) The area of triangle ABC is \(12\sqrt{2}\).

For equilateral triangle: • The radius of the circumscribed circle is \(R=a*\frac{\sqrt{3}}{3}\), (where \(a\) is the side of equilateral triangle); • The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\); • The area of equilateral triangle is \(A=a^2*\frac{\sqrt{3}}{4}\).

We are asked to calculate area of bigger circle P - \(area_P=\pi{R^2}\). Note that knowing any of the following: the side of equilateral triangle \(a\), radius of the smaller circle O (as it gives \(a\)) or the radius of P itself is sufficient to calculate area of P.

(1) The area of circle O is \(4\pi\) --> we can find \(r\) --> we can find \(a\) --> we can find \(R\). Sufficient.

(2) The area of triangle ABC is \(12\sqrt{2}\) --> we can find \(a\) --> we can find \(R\). Sufficient.

Answer: D.

Dear Members,

Has anyone noticed that both the statements contradict each other

From statement 1 , we get \(a = 4\sqrt3\) or \(a^2 = 48\)

from statement 2 we get \(12\sqrt2 = a^2 *\frac{\sqrt3}{4}\)

or \(a^2 = 48*\frac{\sqrt2}{sqrt3}\)

both the statements should give the same value for a and \(a^2\) ,( side of the triangle).

Let me know if I am misinterpreting anything.

Although the answer is still D, both the statements shouldn't give different values for a.

You are right. I guess the second statement should read: he area of triangle ABC is \(12\sqrt{3}\).

Re: Circle O is inscribed in equilateral triangle ABC, which is [#permalink]
08 Sep 2015, 18:49

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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