An isosceles triangle with one angle of 120° is inscribed in

mau5 san,

These GMAT challenges are damn tough..Why did you categorize this problem 600-700.....it hurts when u don't even know where to start.
Any hints where to begin??

I think if you see this level problem, more likely you would have achieved your dream score so you can just chillax and go easy.....

Wow...this is brutal. I am completely stumped.

The area of the circle is a given, but I cannot for the life of me figure out how to determine the size of the triangle. We know that It's largest measure is 120 and naturally, the two smaller measures are 30 a piece. The triangle has to be some kind of defined size, right? The base of the triangle is obviously smaller than the diameter because this is NOT a right triangle. The area of the triangle is therefore less than a=1/2 (4*2) = 4 which is what it would be if it were a right triangle with the base on diameter 4.

So, the area of the circle is 4pi and the triangle is smaller than 4. The triangle rotates around the center - to make things easy, start with the hypotenuse of the triangle parallel to what the y axis would be. A 90 degree rotation would have the triangle cover roughly 3/4ths of the area of the triangle but because this triangle is smaller than a right triangle (it's hypotenuse, if drawn vertically parallel to the y axis, would be somewhere left of the origin) it doesn't cover quite 3/4ths of the triangle - let's say it covers 3/5ths. So, it's covered 3/5ths of the 4pi triangle which mean's that it covered 4pi*.6 = 7.53 units of the triangle. Apparently, my very rough calculation isn't even close to the correct answer of E which is about 4.8.

Yup, totally lost!

I got E, but here how will i make figure?

I can make it on my rough sheet though. lol

Are there any experts willing to take a crack at this?
Looks like a really tough one.

Can anyone explain this with figure?

Hi all,
Its a tough Q to be able to come up with exact answer in exactly 2 mins untill you visualize it. thereafter it becomes easy...
Trial : i think a bit of trial and we can be close to the correct answer..
Given that the iso tri has one angle as 120. so we can safely assume this triangle is inscribed on one side of the diameter.
Total area of circle=4pi.. the triangle rotates through 90 degree... so traversed area should be between pi and 2 pi approximately..
only E fits in...
for the exact answer, a sketch is reqd visualizing the triangle making a circle of radius 1 as it rotates and i'll explain after sometime as office time...

Bunuel, what is the better approach to resolve this question?

GMATinsight gave a detailed solution.
But I suppose we do not have much time, while taking GMAT.
So I would use a different approach - checking the answers (maybe with not too many approximations as sindhugclub did, but still).

Look at GMATinsight painting.
We need to know the yellow area. As we can see, it is equal to sum of 2 big sectors (2*ABC sector) minus their common part minus small sector and minus some triangle areas.
So what can we see.
The area of all circle is 4Pi.
The area of ABC sector is (120/360)*4Pi=4Pi/3.
The area of 2 such sectors is 8Pi/3
1. So we know that the area will be less than 8Pi/3 ------> we can eliminate options B and C
2. But we know that this area will not be less or even very close to area of one sector ABC, so it will be bigger than 4Pi/3------> thus we eliminate A and D.

And that leaves us with E.

Maybe we took too much by statement 2. Ok. Lets do more calculations.
Look at triangle AOC. It can be easily found that AC=2*sqrt(3), XO=1, so Area of triangle AOC = sqrt(3).
Ok. We have 8Pi/3 - sqrt(3) - something else. But all this will not be even close to 4Pi/3 ---------------> option E is an answer.

This is not very accurate approach, but we can use it

thanks for your explanation, how much time it took to solve this ? I didnt even able to start

A Lot of time... so don't worry because this Question doesn't represent the actual GMAT question's difficulty level. Most questions are doable in less than 2 mins with an average speed.

A. approximately 4
B. approx. 8
C. Approx. 12
D. Approx. 2
E. Approx. 5
S the triangle is less than 4. if it rotates 180 degree, max total S=8. If it rotates 90 degree, the S= 2/3*8 = approx.5 ( E)

Hi sindhugclub,

You cannot assume O to be the centre and the base of the isosceles triangle. If you do so, you are saying that Angle ABC is 90 as AC is the diameter, which is not possible as one of the angles must be 120. I may be wrong here and I have no intention of pointing out mistakes, but I feel that luyennguyen answer's uses the option reduction technique that you have mentioned here in a better way.

square has sides with 90 deg adjacent to each other.. going by the diagram, i dont think its correct approach.

hello everyone.
I find the question is bit ambiguous. It does not clearly defines how the triangle is inscribed in the circle. As described in the the detailed answer given, the Triangle is taken as ABC which is above center of the circle. Can somebody please explain me why can't we take triangle AOC as our triangle which is supposed to be rotated by 90 deg. if we take AOC as our triangle. The problem becomes simpler and can be easily solved by Area of sector, which yields the result 4pi/3. Please give your views on this approach.

Thanks

Approx calculations as per image

@ Luyenguen. Please hiw is the triangle less than 4? do you mean area? kindly expantiate your explanation

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Gmat Insight team, How did you arrive at area of Arc XOY? (1/4 * π * (1)^2), could you please explain the logic here?

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