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05 May 2025, 11:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
100% (02:20) correct
0% (00:00) wrong based on 1 sessions
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Date
Time
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Not Attempted Yet
O is the center of the circle of radius 6. OXYZ is a square.
Quantity A
The area of the shaded region.
Quantity B
12
Attachment:
GMAT-Club-Forum-3z283y26.png [ 37.43 KiB | Viewed 229 times ]
09 May 2025, 08:12
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↧↧↧ Detailed Solution to the Problem Series ↧↧↧
O is the center of the circle of radius 6.
=> OB = OY = 6
Let side of square OXYB = a (as shown in the figure below)
GP-6 Image.jpg [ 17.79 KiB | Viewed 185 times ]
Quantity A
Area of the shaded region = Area of the Quarter circle OAB - Area of Square OXYZ = \(\frac{1}{4} * \pi * r^2\) - a*a = \(\frac{1}{4} * \pi * 6^2 - a^2 \) = \(9 * \pi - a^2 \)
Now, Diagonal of a square with side a = a\(\sqrt{2}\)
=> a\(\sqrt{2}\) = 6
=> a = \(\frac{6}{\sqrt{2}}\)
=> Area of Shaded region = \(9 * \pi - (\frac{6}{\sqrt{2}})^2\)
= \(9 \pi - \frac{36}{2}\) = \(9 \pi - 18\)
= 9 * \((\pi - 2)\) = 9 * (3.14 - 2) = 9 * 1.14 = 10.26
Clearly, Quantity B(12) > Quantity A(10.26)
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Circles
O is the center of the circle of radius 6.
=> OB = OY = 6
Let side of square OXYB = a (as shown in the figure below)
Attachment:
GP-6 Image.jpg [ 17.79 KiB | Viewed 185 times ]
Quantity A
Area of the shaded region = Area of the Quarter circle OAB - Area of Square OXYZ = \(\frac{1}{4} * \pi * r^2\) - a*a = \(\frac{1}{4} * \pi * 6^2 - a^2 \) = \(9 * \pi - a^2 \)
Now, Diagonal of a square with side a = a\(\sqrt{2}\)
=> a\(\sqrt{2}\) = 6
=> a = \(\frac{6}{\sqrt{2}}\)
=> Area of Shaded region = \(9 * \pi - (\frac{6}{\sqrt{2}})^2\)
= \(9 \pi - \frac{36}{2}\) = \(9 \pi - 18\)
= 9 * \((\pi - 2)\) = 9 * (3.14 - 2) = 9 * 1.14 = 10.26
Clearly, Quantity B(12) > Quantity A(10.26)
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Circles
18 May 2025, 00:27
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OE
The area of the shaded region is the area of quarter-circle AOB minus the area of the square. Since r = OA = 6, the area of the quartercircle is \(\frac{1}{4}π r^2\)\(=\frac{1}{4}36π=9π\). OY, the diagonal of the square, is 6 (since it is a radius of the circle), so OZ, the side of the square is \(\frac{6}{\sqrt{2}}\). So the area of the square is \([\frac{6}{\sqrt{2}}]^2\)=\(\frac{36}{2}=18.\)
Finally, the area of the shaded region is \(9π – 18\), which is approximately 10.
*The solution above requires several steps. If you can’t reason through this, you still should be able to answer this question correctly. The shaded region has a definite area, which is either 12, more than 12, or less than 12. Eliminate D. Also, the area of a curved region almost always involves π, so assume the area isn’t exactly 12.
Eliminate Choice C. You can now guess between Choices A and B, but if you trust the diagram and know a little bit you can improve your guess. If you know that the area of the circle is 36π, so that the area of the quarter-circle is 9π or about 28, you can estimate the shaded region. It’s well less than half of the quarter-circle, so less than 14 and probably less than 12. Guess Choice B.
Answer: B
The area of the shaded region is the area of quarter-circle AOB minus the area of the square. Since r = OA = 6, the area of the quartercircle is \(\frac{1}{4}π r^2\)\(=\frac{1}{4}36π=9π\). OY, the diagonal of the square, is 6 (since it is a radius of the circle), so OZ, the side of the square is \(\frac{6}{\sqrt{2}}\). So the area of the square is \([\frac{6}{\sqrt{2}}]^2\)=\(\frac{36}{2}=18.\)
Finally, the area of the shaded region is \(9π – 18\), which is approximately 10.
*The solution above requires several steps. If you can’t reason through this, you still should be able to answer this question correctly. The shaded region has a definite area, which is either 12, more than 12, or less than 12. Eliminate D. Also, the area of a curved region almost always involves π, so assume the area isn’t exactly 12.
Eliminate Choice C. You can now guess between Choices A and B, but if you trust the diagram and know a little bit you can improve your guess. If you know that the area of the circle is 36π, so that the area of the quarter-circle is 9π or about 28, you can estimate the shaded region. It’s well less than half of the quarter-circle, so less than 14 and probably less than 12. Guess Choice B.
Answer: B
