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An isosceles triangle with one angle of 120° is inscribed in
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02 Dec 2013, 01:42
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An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D. \(\frac{4\pi}{3}\)\(\sqrt{3}\) E. \(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week.
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Re: An isosceles triangle with one angle of 120° is inscribed in
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05 Jun 2015, 23:43
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. Answer: Option A Detailed solution is as mentioned below
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Re: An isosceles triangle with one angle of 120° is inscribed in
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08 Apr 2017, 07:05
the below figure will be one of the possible figures formed as per the given question. Considering it to be a square and O to be the centre and diagonals BD,AC. We need the area of the entire square ABCD  Area of triangle OAD Area of square = (√8)^2 = 8 P.S. side √8 is deduced from Pythagorean theorem, √(4+4) = √8 Area of triangle = 1/2 * √8 * √6 = 2√3 = 3.40(approx considering √3 value is 1.7) So required area = area of the entire square ABCD  Area of triangle OAD = 8  3.4 = 4.6 Check for all options by considering π value as approx 3. A. 4π/3 = 4*(3/3) = 4 B. 8π/3 = 8*(3/3) = 8 C. 4π = 12 D. 4π/3√3 = 41.7 = 2.3 E. 25π/12  √3 = 25 (3/12)  1.7 = 25(1/4) 1/7 = 6.21.7 =4.5 E is the closest answer compared to other answers, Hence E is the answer.. (NOTE: All the values are approximated hence chose the approx close answer) Might look lengthy but should be done within 2 minutes if done properly. Cheers!
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Re: An isosceles triangle with one angle of 120° is inscribed in
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03 Dec 2013, 04:21
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. mau5 san, These GMAT challenges are damn tough..Why did you categorize this problem 600700.....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.....
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Re: An isosceles triangle with one angle of 120° is inscribed in
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11 Dec 2013, 22:07
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!



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Re: An isosceles triangle with one angle of 120° is inscribed in
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14 Dec 2013, 23:28
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. I got E, but here how will i make figure? I can make it on my rough sheet though. lol



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Re: An isosceles triangle with one angle of 120° is inscribed in
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13 Sep 2014, 12:45
Are there any experts willing to take a crack at this? Looks like a really tough one.



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Re: An isosceles triangle with one angle of 120° is inscribed in
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24 May 2015, 04:03
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. Can anyone explain this with figure?



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Re: An isosceles triangle with one angle of 120° is inscribed in
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05 Jun 2015, 22:18
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. 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...
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Re: An isosceles triangle with one angle of 120° is inscribed in
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17 Jan 2017, 16:15
mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D. \(\frac{4\pi}{3}\)\(\sqrt{3}\) E. \(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. Bunuel, what is the better approach to resolve this question?
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Re: An isosceles triangle with one angle of 120° is inscribed in
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10 Aug 2017, 13:52
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



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Re: An isosceles triangle with one angle of 120° is inscribed in
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12 Aug 2017, 04:25
GMATinsight wrote: mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. Answer: Option A Detailed solution is as mentioned belowthanks for your explanation, how much time it took to solve this ? I didnt even able to start



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Re: An isosceles triangle with one angle of 120° is inscribed in
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12 Aug 2017, 06:01
jokschmer wrote: GMATinsight wrote: mau5 wrote: An isosceles triangle with one angle of 120° is inscribed in a circle of radius 2. This triangle is rotated 90° about the center of the circle. What is the total area covered by the triangle throughout this movement, from starting point to final resting point? A. \(\frac{4\pi}{3}\) B. \(\frac{8\pi}{3}\) C. \(4\pi\) D.\(\frac{4\pi}{3}\)\(\sqrt{3}\) E.\(\frac{25\pi}{12}\)\(\sqrt{3}\) Manhattan GMAT challenge of the week. Answer: Option A Detailed solution is as mentioned belowthanks 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.
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An isosceles triangle with one angle of 120° is inscribed in
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31 May 2018, 19:39
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)



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Re: An isosceles triangle with one angle of 120° is inscribed in
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23 Jun 2018, 15:31
sindhugclub wrote: the below figure will be one of the possible figures formed as per the given question.
Considering it to be a square and O to be the centre and diagonals BD,AC. We need the area of the entire square ABCD  Area of triangle OAD
Area of square = (√8)^2 = 8 P.S. side √8 is deduced from Pythagorean theorem, √(4+4) = √8
Area of triangle = 1/2 * √8 * √6 = 2√3 = 3.40(approx considering √3 value is 1.7)
So required area = area of the entire square ABCD  Area of triangle OAD = 8  3.4 = 4.6
Check for all options by considering π value as approx 3.
A. 4π/3 = 4*(3/3) = 4 B. 8π/3 = 8*(3/3) = 8 C. 4π = 12 D. 4π/3√3 = 41.7 = 2.3 E. 25π/12  √3 = 25 (3/12)  1.7 = 25(1/4) 1/7 = 6.21.7 =4.5 E is the closest answer compared to other answers, Hence E is the answer.. (NOTE: All the values are approximated hence chose the approx close answer) Might look lengthy but should be done within 2 minutes if done properly. Cheers! 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.



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Re: An isosceles triangle with one angle of 120° is inscribed in
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19 Jan 2019, 22:30
sindhugclub wrote: the below figure will be one of the possible figures formed as per the given question.
Considering it to be a square and O to be the centre and diagonals BD,AC. We need the area of the entire square ABCD  Area of triangle OAD
Area of square = (√8)^2 = 8 P.S. side √8 is deduced from Pythagorean theorem, √(4+4) = √8
Area of triangle = 1/2 * √8 * √6 = 2√3 = 3.40(approx considering √3 value is 1.7)
So required area = area of the entire square ABCD  Area of triangle OAD = 8  3.4 = 4.6
Check for all options by considering π value as approx 3.
A. 4π/3 = 4*(3/3) = 4 B. 8π/3 = 8*(3/3) = 8 C. 4π = 12 D. 4π/3√3 = 41.7 = 2.3 E. 25π/12  √3 = 25 (3/12)  1.7 = 25(1/4) 1/7 = 6.21.7 =4.5 E is the closest answer compared to other answers, Hence E is the answer.. (NOTE: All the values are approximated hence chose the approx close answer) Might look lengthy but should be done within 2 minutes if done properly. Cheers! square has sides with 90 deg adjacent to each other.. going by the diagram, i dont think its correct approach.



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Re: An isosceles triangle with one angle of 120° is inscribed in
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19 Jan 2019, 22:49
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.
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Re: An isosceles triangle with one angle of 120° is inscribed in
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26 Jan 2019, 15:06
Approx calculations as per image
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Re: An isosceles triangle with one angle of 120° is inscribed in
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29 Mar 2019, 14:49
@ Luyenguen. Please hiw is the triangle less than 4? do you mean area? kindly expantiate your explanation
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Re: An isosceles triangle with one angle of 120° is inscribed in
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