Circle O is inscribed in equilateral triangle ABC, which is

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."

Once you understand these relations, you will jump to (D) immediately.

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!!!

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Sorry Hussain15, it's just what I was thinking when took a look at the problem. If it doesn't work for you, just leave it.

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.

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.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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 123 √ .

In the original condition, the there is only one variable (radius), and we need one equation to solve for the question.
2 equations are given from the 2 conditions, so there is high chance (D) will be our answer; in fact, (D) is our answer.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

Thanks for the question and the intuitive approach to solve it! I'll try to practice this approach on similar questions.

Answer D.

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.

so answer D

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.

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

Edited the question. Thank you.

how do you solve, in theory, to come to the conclusion that radius of the circumscribed circle is a*sqrt(3)/3?
is it because the diameter will be 1/2 time of the side?
in this case, the diameter will be a/2, and R is a/4, where a is the side of the equilateral triangle.

knowing that the triangle is inscribed in a circle, we can draw 2 radii which will connect with one side of the triangle, creating a 30-30-120 triangle. Then, we can draw a perpendicular, and get 2 triangles of 30-60-90, in which the longest leg will be a/2, where a is the side of the equilateral triangle.

is my way of thinking right?
in case we know area of the small circle, we can find the side of the equilateral triangle, and thus, can find the radius of the big circle.
in case we know the area of the equilateral triangle, we can deduct that A=[S^2 sqrt(3)]/4. Now, we can find the side of the equilateral triangle, and hence, find the radius of the big circle.

I believe this is more a 700 level question

I wouldn't say I did many calculates for this.

St1 - Asc= 4 = πr^2 ---> r =2
If we were to draw the radius connecting the centre of the smaller circle to its bottom, then we end up with 2 30-60-90 triangles.

r = 2 is effectively the height of the triangle and we can use this relationship: x:x√3:2x ---> 2x = 2 ---> x = 1 to determine the base and hence hypotenuse. The hypotenuse is the radius of the larger circle.

Sufficient.

St2 - A(ABC) = 12√3 = s^2√3/4 ---> s = ...
So half of s gives you the bottom leg of the 30-60-90 triangle. Again you can determine the hypotenuse.

Sufficient.

D

Edit: I made a mistake above. Finding out one side length is not the same as finding the hypotenuse of the larger circumscribed circle. But, once you get that side length you can use the formula Bunuel mentioned to get the area of the larger circle:

s = side

s x √3/3

Bunuel I just want to make sure I got this right.

For the radius of the inscribed circle:
• The radius of the inscribed circle is \(r=a*\frac{\sqrt{3}}{6}\);

If we are given the area of the inscribed circle then obviously we can get the radius, which is equivalent to the height of the 30-60-90 triangle that you can form.

To be clear, if you didn't have the area, then the formula above allows you to get the radius if you have the side length of the equilateral triangle. i.e. going the opposite direction

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