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06 May 2021, 08:00
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C
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Difficulty:
65%
(hard)
Question Stats:
58% (02:28) correct
42%
(02:34)
wrong
based on 45
sessions
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Two circular regions of equal radii are centered at adjacent corners C and D of square ABCD, as shown above. If shaded regions, X and Y, have equal areas, and if DE=CF=√18 , what is the length of AB?
A. \(\frac{3}{√3}\)
B. \((\frac{3}{√2})*(√π)\)
C. \(3√π\)
D. \(3√2√π\)
E. \(3π\)
Attachment:
1.png [ 8.23 KiB | Viewed 2181 times ]
08 May 2021, 06:15
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Bunuel
Look at the image below:
If we add quarter of the circle centered at C and quarter of the circle centered at D, we get the area of red portion, the area of blue portion and the area of Y twice (because Y is included in both quarters). Since the area of X = the area of Y, then quarter of the circle centered at C and quarter of the circle centered at D, gives Red + Blue + X + Y = Square.
Let the side of the square be x, then \(x^2 =\frac{\pi{r^2}}{4}+\frac{\pi{r^2}}{4} =2*\frac{\pi{r^2}}{4}\) --> \(x^2=2*\frac{\pi{18}}{4}\) --> \(x^2=9\pi\) --> \(x=3\sqrt{\pi}\).
Answer: C.
Attachment:
Untitled.png [ 6.86 KiB | Viewed 2039 times ]
General Discussion
06 May 2021, 11:31
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Bunuel
Set length of square as \(s\). The area of one of the quarter circles is \((\sqrt{18})^2*\pi / 4 = \frac{9}{2}*\pi\)
Label one of the blank spaces Z. Since we have X = Y, then we also have
X + Z = Y + Z
\(s^2 - \frac{9}{2}*\pi = \frac{9}{2}*\pi\)
\(s^2 = 9*\pi\)
\(s = 3*\sqrt{\pi}\)
Ans: C
Attachments
Square.png [ 14.49 KiB | Viewed 2014 times ]
