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11 Jan 2014, 08:16
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (01:59) correct
36%
(02:01)
wrong
based on 486
sessions
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Attachment:
File comment: Geometry
ABCD.png [ 20.95 KiB | Viewed 19295 times ]
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?ABCD.png [ 20.95 KiB | Viewed 19295 times ]
(1) The radius of the circle with center O is 4.
(2) The area of triangle ADC is 32.
OE
To find area of shaded region, you need to find area of triangle ACD, and subtract area of unshaded region inside this triangle from it. To find area of unshaded region, you need to first draw in a line perpendicular to diameter AB of circle from center O to diagonal AC. Call this point E. This point is also one of endpoints of arc AFE.
(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.
(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient
Attachment:
File comment: ABCD2
ABCD2.png [ 19.07 KiB | Viewed 16341 times ]
Arc AFE and radii OA and OE bound a sector of circle O: this sector is one quarter of entire area of circle. Now, if you take area of small triangle AEO, and subtract this area from area of sector OAFE, then you have area of unshaded portion of triangle ADC. So, to solve this question, you need to find area of triangle ADC, one quarter of area of circle O, and area of small triangle AEO. To find these three areas, you need radius of circle, or length of one side of square.ABCD2.png [ 19.07 KiB | Viewed 16341 times ]
(1): This is exactly what we need; once we have radius of circle we can find diameter of circle which is same as a side of square and also same as each leg of right triangle ADC. So you can, with this information, solve for different areas necessary, and then solve for area of shaded region. Sufficient.
(2): able to find length of height and width (which are equal to each other) of triangle ADC.
Let's say that both base and height of triangle ADC = x.
32 = (½)x^2 → x = 8
You have leg lengths (which are equal) of isosceles right triangle ADC; this means that you also have diameter of circle → radius.
Sufficient
11 Jan 2014, 08:26
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goodyear2013
What is the area of the shaded region above, if ABCD is a square and line segment AB is a diameter of the circle with center O?
Actually there is no need for any calculations. From both statements we have everything "fixed", only one case possible where we know everything possible.
(1) The radius of the circle with center O is 4 --> diameter = side = 8. We have "fixed" circle and square. We can find anything there. Sufficient.
(2) The area of triangle ADC is 32 --> 1/2*AD^2=32 --> AD=8 --> AD = diameter = 8. The same info as above. Sufficient.
Answer: D.
General Discussion
11 Jan 2014, 16:55
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Bunuel
Bunuel, Can you please explain why if you only have the radius you can find anything. How would you get the AF part?
I knew that Statement 1 and 2 gave the same information, but couldn't figure out how to find the white portion in TriangleADC
Thanks
)
