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25 May 2017, 09:57
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
85% (01:17) correct
15%
(01:26)
wrong
based on 46
sessions
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Attachment:
7777.jpg [ 9.58 KiB | Viewed 3563 times ]
In the figure above, the circle with center O is inscribed in the square PQRS. The combined area of the shaded regions is
(A) \(36 – 9π\)
(B) \(36 − \frac{9}{2} π\)
(C) \(\frac{36 − 9π}{2}\)
(D) \(18 – 9π\)
(E) \(9 − \frac{9}{4} π\)
Source: Nova GMAT
25 May 2017, 11:00
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The combined area of the shaded regions is = [Area of square PQRS - Area of circle with center O]/2
Area of square PQRS = 6*6 = 36
Area of circle with center 0 = π * 3 * 3 = 9π
The combined area of the shaded regions is = [36 - 9π]/2
Answer is C
Area of square PQRS = 6*6 = 36
Area of circle with center 0 = π * 3 * 3 = 9π
The combined area of the shaded regions is = [36 - 9π]/2
Answer is C
25 May 2017, 11:06
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Here Side of the square = 6 = diameter of the circle
So radius of circle = 3
Area outside the circle is divided into 4 congruent regions as can clearly be seen. We are supposed to find the area of '2' out of those '4' regions. So our required area will be 'half' of the area lying inside the square but outside the circle.
Now, area of square = 6^2 = 36
Area of circle = π(3)^2 = 9π
Area inside square but outside the circle = 36-9π
Our required area = (36-9π)/2
Hence C
So radius of circle = 3
Area outside the circle is divided into 4 congruent regions as can clearly be seen. We are supposed to find the area of '2' out of those '4' regions. So our required area will be 'half' of the area lying inside the square but outside the circle.
Now, area of square = 6^2 = 36
Area of circle = π(3)^2 = 9π
Area inside square but outside the circle = 36-9π
Our required area = (36-9π)/2
Hence C
