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luyennguyen
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)

:-)

Why is the triangle less than 4 units ?

For another explanation of this question please click the link below

https://gmatclub.com/forum/an-isosceles ... l#p1534951
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I would say this is the best answer I saw. However the black area needs a bit edit. The center portion is not swept. Am I understanding it correctly?

Shobhit7
Approx calculations as per image

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The area we are looking for can be described as an area of a portion of the circle minus areas of two triangles.
Since choice A,B, and C only have pi term, these choices can be ruled out.
Then, we have D and E left.
If you see the drawings that bunch of other people had drawn, the angle of the portion of the circle is substantially larger than 180 degrees.
Since the radius is 2, the area of the whole circle is 4 pi. From this, we can tell that choice D is saying the angle of the portion of the circle we are looking for is 120 degrees, and that choice E is saying the angle is much larger than 180 degrees(we don't need to calculate this since this is the only choice we have left).
Since the angle is much larger than 180 degrees, E is the answer.
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sindhugclub
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 = 4-1.7 = 2.3
E. 25π/12 - √3 = 25 (3/12) - 1.7
= 25(1/4) -1/7
= 6.2-1.7 =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!


How did you get root 6 as the height? sindhugclub
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Once I found out ABO is an equilateral triangle, finding answer became easy.

1) As 30-60-90 triangle has 1-2-\(\sqrt{3} \) proportion and we incidentally have "2" as a hypotenuse(!),
we now know all the length of triangles, the radius of small quarter-circle inside big circle.
2) We need to subtract two triangles with the area of \(\sqrt{3}/2\) from big circle, ---> Eliminate answer c. 1, 2, 3
3) We're left with the answer choices 4, 5.
The area of the big circle is obviously \(4π\), and the shaded area looks like about quarter~half of the big circle (about \(1.5π\) ~ \(2 π\) )
It seems answer choice 4 is too small for that. (about \(1π - \sqrt{3}\))
----> choose answer choice E.

(of course it is totally possible to calculate figures, but in the spirit of GMAT, elimination seems a better strategy.)
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File comment: Most difficult part was to find out that ABO is an equilateral triangle.
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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.


The target here is NOT to solve this perfectly but solve it fast enough and with good enough accuracy to be able to mark the correct option.
So, we can make some approximations. The working is shown in the diagram below:

Attachment:
1.jpg
1.jpg [ 81.85 KiB | Viewed 1004 times ]

Answer E
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KarishmaB Bunuel IanStewart - Your thoughts on this question?
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Sonia0106
KarishmaB Bunuel IanStewart - Your thoughts on this question?

This is more suitable to IIT, not even CAT, forget GMAT. No point discussing even for intellectual purposes. To make it GMAT relevant I would have to put a vertex at the centre and talk about the area of the sector, not triangle.
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